Mathematics

Class 92025–26 SNC63 Questions SolvedEnglish Medium

Exercise 6.2 — Class 9 Mathematics Solutions

All 63 questions below are worked step by step — every step clearly worked out with mathematical reasons — for the Punjab Board 9th class Mathematics textbook.

Exercise 6.2

5 Questions (58 Sub-parts)

Quick Solution Step by Step

Every step explained, with hints and the method used.

Question 1

EasyUnder Review

For each of the following right-angled triangles, find the trigonometric ratios:

(i) $\sin θ$ (ii) $\cos θ$ (iii) $\tan θ$ (iv) $\sec θ$ (v) $cosec θ$ (vi) $\cot ϕ$ (vii) $\tan ϕ$ (viii) $cosec ϕ$ (ix) $\sec ϕ$ (x) $\cos ϕ$

In each triangle $m ∠ B = 90^{\circ}$ , $m ∠ A = θ$ and $m ∠ C = ϕ$ :

Multiple Choice

$m B C = 4$ , $m A B = 3$ , $m A C = 5$

$m B C = 8$ , $m A B = 15$ , $m A C = 17$

$m B C = 5$ , $m A B = 12$ , $m A C = 13$

Hint

For each angle, mark which side is opposite it (perpendicular), which is next to it (base), and which is the hypotenuse, then use the ratio definitions.

Solution

(This question consists of multiple subparts. See solutions below.)

Quick Solution

Q1Part (a)(i)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ , $m B C = 4$ , $m A B = 3$ , $m A C = 5$ , $m ∠ A = θ$ and $m ∠ C = ϕ$ . Find $\sin θ$ .

Hint

Use $\sin = \frac{Perp.}{Hyp.}$ , with the sides read from angle $θ$ at $A$ .

Solution

Name the sides relative to $θ$ .

The angle $θ$ is at vertex $A$ . The side opposite it, $B C$ , is the perpendicular; the side next to it (other than the hypotenuse), $A B$ , is the base; $A C$ , opposite the right angle, is the hypotenuse.

$Perp. = B C = 4, Base = A B = 3, Hyp. = A C = 5$
Apply the definition of $\sin$ .

By definition $\sin = \frac{Perp.}{Hyp.}$ .
$\begin{matrix} \sin θ & = \frac{Perp.}{Hyp.} \\ = \frac{B C}{A C} \\ = \frac{4}{5} \end{matrix}$

Answer

$\sin θ = \frac{4}{5}$

Quick Solution

Q1Part (a)(ii)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ , $m B C = 4$ , $m A B = 3$ , $m A C = 5$ , $m ∠ A = θ$ and $m ∠ C = ϕ$ . Find $\cos θ$ .

Hint

Use $\cos = \frac{Base}{Hyp.}$ , with the sides read from angle $θ$ at $A$ .

Solution

Name the sides relative to $θ$ .

The angle $θ$ is at vertex $A$ . The side opposite it, $B C$ , is the perpendicular; the side next to it (other than the hypotenuse), $A B$ , is the base; $A C$ , opposite the right angle, is the hypotenuse.

$Perp. = B C = 4, Base = A B = 3, Hyp. = A C = 5$
Apply the definition of $\cos$ .

By definition $\cos = \frac{Base}{Hyp.}$ .
$\begin{matrix} \cos θ & = \frac{Base}{Hyp.} \\ = \frac{A B}{A C} \\ = \frac{3}{5} \end{matrix}$

Answer

$\cos θ = \frac{3}{5}$

Quick Solution

Q1Part (a)(iii)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ , $m B C = 4$ , $m A B = 3$ , $m A C = 5$ , $m ∠ A = θ$ and $m ∠ C = ϕ$ . Find $\tan θ$ .

Hint

Use $\tan = \frac{Perp.}{Base}$ , with the sides read from angle $θ$ at $A$ .

Solution

Name the sides relative to $θ$ .

The angle $θ$ is at vertex $A$ . The side opposite it, $B C$ , is the perpendicular; the side next to it (other than the hypotenuse), $A B$ , is the base; $A C$ , opposite the right angle, is the hypotenuse.

$Perp. = B C = 4, Base = A B = 3, Hyp. = A C = 5$
Apply the definition of $\tan$ .

By definition $\tan = \frac{Perp.}{Base}$ .
$\begin{matrix} \tan θ & = \frac{Perp.}{Base} \\ = \frac{B C}{A B} \\ = \frac{4}{3} \end{matrix}$

Answer

$\tan θ = \frac{4}{3}$

Quick Solution

Q1Part (a)(iv)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ , $m B C = 4$ , $m A B = 3$ , $m A C = 5$ , $m ∠ A = θ$ and $m ∠ C = ϕ$ . Find $\sec θ$ .

Hint

Use $\sec = \frac{Hyp.}{Base}$ , with the sides read from angle $θ$ at $A$ .

Solution

Name the sides relative to $θ$ .

The angle $θ$ is at vertex $A$ . The side opposite it, $B C$ , is the perpendicular; the side next to it (other than the hypotenuse), $A B$ , is the base; $A C$ , opposite the right angle, is the hypotenuse.

$Perp. = B C = 4, Base = A B = 3, Hyp. = A C = 5$
Apply the definition of $\sec$ .

By definition $\sec = \frac{Hyp.}{Base}$ .
$\begin{matrix} \sec θ & = \frac{Hyp.}{Base} \\ = \frac{A C}{A B} \\ = \frac{5}{3} \end{matrix}$

Answer

$\sec θ = \frac{5}{3}$

Quick Solution

Q1Part (a)(v)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ , $m B C = 4$ , $m A B = 3$ , $m A C = 5$ , $m ∠ A = θ$ and $m ∠ C = ϕ$ . Find $cosec θ$ .

Hint

Use $cosec = \frac{Hyp.}{Perp.}$ , with the sides read from angle $θ$ at $A$ .

Solution

Name the sides relative to $θ$ .

The angle $θ$ is at vertex $A$ . The side opposite it, $B C$ , is the perpendicular; the side next to it (other than the hypotenuse), $A B$ , is the base; $A C$ , opposite the right angle, is the hypotenuse.

$Perp. = B C = 4, Base = A B = 3, Hyp. = A C = 5$
Apply the definition of $cosec$ .

By definition $cosec = \frac{Hyp.}{Perp.}$ .
$\begin{matrix} cosec θ & = \frac{Hyp.}{Perp.} \\ = \frac{A C}{B C} \\ = \frac{5}{4} \end{matrix}$

Answer

$cosec θ = \frac{5}{4}$

Quick Solution

Q1Part (a)(vi)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ , $m B C = 4$ , $m A B = 3$ , $m A C = 5$ , $m ∠ A = θ$ and $m ∠ C = ϕ$ . Find $\cot ϕ$ .

Hint

Use $\cot = \frac{Base}{Perp.}$ , with the sides read from angle $ϕ$ at $C$ .

Solution

Name the sides relative to $ϕ$ .

The angle $ϕ$ is at vertex $C$ . The side opposite it, $A B$ , is the perpendicular; the side next to it (other than the hypotenuse), $B C$ , is the base; $A C$ , opposite the right angle, is the hypotenuse.

$Perp. = A B = 3, Base = B C = 4, Hyp. = A C = 5$
Apply the definition of $\cot$ .

By definition $\cot = \frac{Base}{Perp.}$ .
$\begin{matrix} \cot ϕ & = \frac{Base}{Perp.} \\ = \frac{B C}{A B} \\ = \frac{4}{3} \end{matrix}$

Answer

$\cot ϕ = \frac{4}{3}$

Quick Solution

Q1Part (a)(vii)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ , $m B C = 4$ , $m A B = 3$ , $m A C = 5$ , $m ∠ A = θ$ and $m ∠ C = ϕ$ . Find $\tan ϕ$ .

Hint

Use $\tan = \frac{Perp.}{Base}$ , with the sides read from angle $ϕ$ at $C$ .

Solution

Name the sides relative to $ϕ$ .

The angle $ϕ$ is at vertex $C$ . The side opposite it, $A B$ , is the perpendicular; the side next to it (other than the hypotenuse), $B C$ , is the base; $A C$ , opposite the right angle, is the hypotenuse.

$Perp. = A B = 3, Base = B C = 4, Hyp. = A C = 5$
Apply the definition of $\tan$ .

By definition $\tan = \frac{Perp.}{Base}$ .
$\begin{matrix} \tan ϕ & = \frac{Perp.}{Base} \\ = \frac{A B}{B C} \\ = \frac{3}{4} \end{matrix}$

Answer

$\tan ϕ = \frac{3}{4}$

Quick Solution

Q1Part (a)(viii)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ , $m B C = 4$ , $m A B = 3$ , $m A C = 5$ , $m ∠ A = θ$ and $m ∠ C = ϕ$ . Find $cosec ϕ$ .

Hint

Use $cosec = \frac{Hyp.}{Perp.}$ , with the sides read from angle $ϕ$ at $C$ .

Solution

Name the sides relative to $ϕ$ .

The angle $ϕ$ is at vertex $C$ . The side opposite it, $A B$ , is the perpendicular; the side next to it (other than the hypotenuse), $B C$ , is the base; $A C$ , opposite the right angle, is the hypotenuse.

$Perp. = A B = 3, Base = B C = 4, Hyp. = A C = 5$
Apply the definition of $cosec$ .

By definition $cosec = \frac{Hyp.}{Perp.}$ .
$\begin{matrix} cosec ϕ & = \frac{Hyp.}{Perp.} \\ = \frac{A C}{A B} \\ = \frac{5}{3} \end{matrix}$

Answer

$cosec ϕ = \frac{5}{3}$

Quick Solution

Q1Part (a)(ix)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ , $m B C = 4$ , $m A B = 3$ , $m A C = 5$ , $m ∠ A = θ$ and $m ∠ C = ϕ$ . Find $\sec ϕ$ .

Hint

Use $\sec = \frac{Hyp.}{Base}$ , with the sides read from angle $ϕ$ at $C$ .

Solution

Name the sides relative to $ϕ$ .

The angle $ϕ$ is at vertex $C$ . The side opposite it, $A B$ , is the perpendicular; the side next to it (other than the hypotenuse), $B C$ , is the base; $A C$ , opposite the right angle, is the hypotenuse.

$Perp. = A B = 3, Base = B C = 4, Hyp. = A C = 5$
Apply the definition of $\sec$ .

By definition $\sec = \frac{Hyp.}{Base}$ .
$\begin{matrix} \sec ϕ & = \frac{Hyp.}{Base} \\ = \frac{A C}{B C} \\ = \frac{5}{4} \end{matrix}$

Answer

$\sec ϕ = \frac{5}{4}$

Quick Solution

Q1Part (a)(x)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ , $m B C = 4$ , $m A B = 3$ , $m A C = 5$ , $m ∠ A = θ$ and $m ∠ C = ϕ$ . Find $\cos ϕ$ .

Hint

Use $\cos = \frac{Base}{Hyp.}$ , with the sides read from angle $ϕ$ at $C$ .

Solution

Name the sides relative to $ϕ$ .

The angle $ϕ$ is at vertex $C$ . The side opposite it, $A B$ , is the perpendicular; the side next to it (other than the hypotenuse), $B C$ , is the base; $A C$ , opposite the right angle, is the hypotenuse.

$Perp. = A B = 3, Base = B C = 4, Hyp. = A C = 5$
Apply the definition of $\cos$ .

By definition $\cos = \frac{Base}{Hyp.}$ .
$\begin{matrix} \cos ϕ & = \frac{Base}{Hyp.} \\ = \frac{B C}{A C} \\ = \frac{4}{5} \end{matrix}$

Answer

$\cos ϕ = \frac{4}{5}$

Quick Solution

Q1Part (b)(i)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ , $m B C = 8$ , $m A B = 15$ , $m A C = 17$ , $m ∠ A = θ$ and $m ∠ C = ϕ$ . Find $\sin θ$ .

Hint

Use $\sin = \frac{Perp.}{Hyp.}$ , with the sides read from angle $θ$ at $A$ .

Solution

Name the sides relative to $θ$ .

The angle $θ$ is at vertex $A$ . The side opposite it, $B C$ , is the perpendicular; the side next to it (other than the hypotenuse), $A B$ , is the base; $A C$ , opposite the right angle, is the hypotenuse.

$Perp. = B C = 8, Base = A B = 15, Hyp. = A C = 17$
Apply the definition of $\sin$ .

By definition $\sin = \frac{Perp.}{Hyp.}$ .
$\begin{matrix} \sin θ & = \frac{Perp.}{Hyp.} \\ = \frac{B C}{A C} \\ = \frac{8}{17} \end{matrix}$

Answer

$\sin θ = \frac{8}{17}$

Quick Solution

Q1Part (b)(ii)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ , $m B C = 8$ , $m A B = 15$ , $m A C = 17$ , $m ∠ A = θ$ and $m ∠ C = ϕ$ . Find $\cos θ$ .

Hint

Use $\cos = \frac{Base}{Hyp.}$ , with the sides read from angle $θ$ at $A$ .

Solution

Name the sides relative to $θ$ .

The angle $θ$ is at vertex $A$ . The side opposite it, $B C$ , is the perpendicular; the side next to it (other than the hypotenuse), $A B$ , is the base; $A C$ , opposite the right angle, is the hypotenuse.

$Perp. = B C = 8, Base = A B = 15, Hyp. = A C = 17$
Apply the definition of $\cos$ .

By definition $\cos = \frac{Base}{Hyp.}$ .
$\begin{matrix} \cos θ & = \frac{Base}{Hyp.} \\ = \frac{A B}{A C} \\ = \frac{15}{17} \end{matrix}$

Answer

$\cos θ = \frac{15}{17}$

Quick Solution

Q1Part (b)(iii)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ , $m B C = 8$ , $m A B = 15$ , $m A C = 17$ , $m ∠ A = θ$ and $m ∠ C = ϕ$ . Find $\tan θ$ .

Hint

Use $\tan = \frac{Perp.}{Base}$ , with the sides read from angle $θ$ at $A$ .

Solution

Name the sides relative to $θ$ .

The angle $θ$ is at vertex $A$ . The side opposite it, $B C$ , is the perpendicular; the side next to it (other than the hypotenuse), $A B$ , is the base; $A C$ , opposite the right angle, is the hypotenuse.

$Perp. = B C = 8, Base = A B = 15, Hyp. = A C = 17$
Apply the definition of $\tan$ .

By definition $\tan = \frac{Perp.}{Base}$ .
$\begin{matrix} \tan θ & = \frac{Perp.}{Base} \\ = \frac{B C}{A B} \\ = \frac{8}{15} \end{matrix}$

Answer

$\tan θ = \frac{8}{15}$

Quick Solution

Q1Part (b)(iv)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ , $m B C = 8$ , $m A B = 15$ , $m A C = 17$ , $m ∠ A = θ$ and $m ∠ C = ϕ$ . Find $\sec θ$ .

Hint

Use $\sec = \frac{Hyp.}{Base}$ , with the sides read from angle $θ$ at $A$ .

Solution

Name the sides relative to $θ$ .

The angle $θ$ is at vertex $A$ . The side opposite it, $B C$ , is the perpendicular; the side next to it (other than the hypotenuse), $A B$ , is the base; $A C$ , opposite the right angle, is the hypotenuse.

$Perp. = B C = 8, Base = A B = 15, Hyp. = A C = 17$
Apply the definition of $\sec$ .

By definition $\sec = \frac{Hyp.}{Base}$ .
$\begin{matrix} \sec θ & = \frac{Hyp.}{Base} \\ = \frac{A C}{A B} \\ = \frac{17}{15} \end{matrix}$

Answer

$\sec θ = \frac{17}{15}$

Quick Solution

Q1Part (b)(v)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ , $m B C = 8$ , $m A B = 15$ , $m A C = 17$ , $m ∠ A = θ$ and $m ∠ C = ϕ$ . Find $cosec θ$ .

Hint

Use $cosec = \frac{Hyp.}{Perp.}$ , with the sides read from angle $θ$ at $A$ .

Solution

Name the sides relative to $θ$ .

The angle $θ$ is at vertex $A$ . The side opposite it, $B C$ , is the perpendicular; the side next to it (other than the hypotenuse), $A B$ , is the base; $A C$ , opposite the right angle, is the hypotenuse.

$Perp. = B C = 8, Base = A B = 15, Hyp. = A C = 17$
Apply the definition of $cosec$ .

By definition $cosec = \frac{Hyp.}{Perp.}$ .
$\begin{matrix} cosec θ & = \frac{Hyp.}{Perp.} \\ = \frac{A C}{B C} \\ = \frac{17}{8} \end{matrix}$

Answer

$cosec θ = \frac{17}{8}$

Quick Solution

Q1Part (b)(vi)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ , $m B C = 8$ , $m A B = 15$ , $m A C = 17$ , $m ∠ A = θ$ and $m ∠ C = ϕ$ . Find $\cot ϕ$ .

Hint

Use $\cot = \frac{Base}{Perp.}$ , with the sides read from angle $ϕ$ at $C$ .

Solution

Name the sides relative to $ϕ$ .

The angle $ϕ$ is at vertex $C$ . The side opposite it, $A B$ , is the perpendicular; the side next to it (other than the hypotenuse), $B C$ , is the base; $A C$ , opposite the right angle, is the hypotenuse.

$Perp. = A B = 15, Base = B C = 8, Hyp. = A C = 17$
Apply the definition of $\cot$ .

By definition $\cot = \frac{Base}{Perp.}$ .
$\begin{matrix} \cot ϕ & = \frac{Base}{Perp.} \\ = \frac{B C}{A B} \\ = \frac{8}{15} \end{matrix}$

Answer

$\cot ϕ = \frac{8}{15}$

Quick Solution

Q1Part (b)(vii)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ , $m B C = 8$ , $m A B = 15$ , $m A C = 17$ , $m ∠ A = θ$ and $m ∠ C = ϕ$ . Find $\tan ϕ$ .

Hint

Use $\tan = \frac{Perp.}{Base}$ , with the sides read from angle $ϕ$ at $C$ .

Solution

Name the sides relative to $ϕ$ .

The angle $ϕ$ is at vertex $C$ . The side opposite it, $A B$ , is the perpendicular; the side next to it (other than the hypotenuse), $B C$ , is the base; $A C$ , opposite the right angle, is the hypotenuse.

$Perp. = A B = 15, Base = B C = 8, Hyp. = A C = 17$
Apply the definition of $\tan$ .

By definition $\tan = \frac{Perp.}{Base}$ .
$\begin{matrix} \tan ϕ & = \frac{Perp.}{Base} \\ = \frac{A B}{B C} \\ = \frac{15}{8} \end{matrix}$

Answer

$\tan ϕ = \frac{15}{8}$

Quick Solution

Q1Part (b)(viii)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ , $m B C = 8$ , $m A B = 15$ , $m A C = 17$ , $m ∠ A = θ$ and $m ∠ C = ϕ$ . Find $cosec ϕ$ .

Hint

Use $cosec = \frac{Hyp.}{Perp.}$ , with the sides read from angle $ϕ$ at $C$ .

Solution

Name the sides relative to $ϕ$ .

The angle $ϕ$ is at vertex $C$ . The side opposite it, $A B$ , is the perpendicular; the side next to it (other than the hypotenuse), $B C$ , is the base; $A C$ , opposite the right angle, is the hypotenuse.

$Perp. = A B = 15, Base = B C = 8, Hyp. = A C = 17$
Apply the definition of $cosec$ .

By definition $cosec = \frac{Hyp.}{Perp.}$ .
$\begin{matrix} cosec ϕ & = \frac{Hyp.}{Perp.} \\ = \frac{A C}{A B} \\ = \frac{17}{15} \end{matrix}$

Answer

$cosec ϕ = \frac{17}{15}$

Quick Solution

Q1Part (b)(ix)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ , $m B C = 8$ , $m A B = 15$ , $m A C = 17$ , $m ∠ A = θ$ and $m ∠ C = ϕ$ . Find $\sec ϕ$ .

Hint

Use $\sec = \frac{Hyp.}{Base}$ , with the sides read from angle $ϕ$ at $C$ .

Solution

Name the sides relative to $ϕ$ .

The angle $ϕ$ is at vertex $C$ . The side opposite it, $A B$ , is the perpendicular; the side next to it (other than the hypotenuse), $B C$ , is the base; $A C$ , opposite the right angle, is the hypotenuse.

$Perp. = A B = 15, Base = B C = 8, Hyp. = A C = 17$
Apply the definition of $\sec$ .

By definition $\sec = \frac{Hyp.}{Base}$ .
$\begin{matrix} \sec ϕ & = \frac{Hyp.}{Base} \\ = \frac{A C}{B C} \\ = \frac{17}{8} \end{matrix}$

Answer

$\sec ϕ = \frac{17}{8}$

Quick Solution

Q1Part (b)(x)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ , $m B C = 8$ , $m A B = 15$ , $m A C = 17$ , $m ∠ A = θ$ and $m ∠ C = ϕ$ . Find $\cos ϕ$ .

Hint

Use $\cos = \frac{Base}{Hyp.}$ , with the sides read from angle $ϕ$ at $C$ .

Solution

Name the sides relative to $ϕ$ .

The angle $ϕ$ is at vertex $C$ . The side opposite it, $A B$ , is the perpendicular; the side next to it (other than the hypotenuse), $B C$ , is the base; $A C$ , opposite the right angle, is the hypotenuse.

$Perp. = A B = 15, Base = B C = 8, Hyp. = A C = 17$
Apply the definition of $\cos$ .

By definition $\cos = \frac{Base}{Hyp.}$ .
$\begin{matrix} \cos ϕ & = \frac{Base}{Hyp.} \\ = \frac{B C}{A C} \\ = \frac{8}{17} \end{matrix}$

Answer

$\cos ϕ = \frac{8}{17}$

Quick Solution

Q1Part (c)(i)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ , $m B C = 5$ , $m A B = 12$ , $m A C = 13$ , $m ∠ A = θ$ and $m ∠ C = ϕ$ . Find $\sin θ$ .

Hint

Use $\sin = \frac{Perp.}{Hyp.}$ , with the sides read from angle $θ$ at $A$ .

Solution

Name the sides relative to $θ$ .

The angle $θ$ is at vertex $A$ . The side opposite it, $B C$ , is the perpendicular; the side next to it (other than the hypotenuse), $A B$ , is the base; $A C$ , opposite the right angle, is the hypotenuse.

$Perp. = B C = 5, Base = A B = 12, Hyp. = A C = 13$
Apply the definition of $\sin$ .

By definition $\sin = \frac{Perp.}{Hyp.}$ .
$\begin{matrix} \sin θ & = \frac{Perp.}{Hyp.} \\ = \frac{B C}{A C} \\ = \frac{5}{13} \end{matrix}$

Answer

$\sin θ = \frac{5}{13}$

Quick Solution

Q1Part (c)(ii)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ , $m B C = 5$ , $m A B = 12$ , $m A C = 13$ , $m ∠ A = θ$ and $m ∠ C = ϕ$ . Find $\cos θ$ .

Hint

Use $\cos = \frac{Base}{Hyp.}$ , with the sides read from angle $θ$ at $A$ .

Solution

Name the sides relative to $θ$ .

The angle $θ$ is at vertex $A$ . The side opposite it, $B C$ , is the perpendicular; the side next to it (other than the hypotenuse), $A B$ , is the base; $A C$ , opposite the right angle, is the hypotenuse.

$Perp. = B C = 5, Base = A B = 12, Hyp. = A C = 13$
Apply the definition of $\cos$ .

By definition $\cos = \frac{Base}{Hyp.}$ .
$\begin{matrix} \cos θ & = \frac{Base}{Hyp.} \\ = \frac{A B}{A C} \\ = \frac{12}{13} \end{matrix}$

Answer

$\cos θ = \frac{12}{13}$

Quick Solution

Q1Part (c)(iii)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ , $m B C = 5$ , $m A B = 12$ , $m A C = 13$ , $m ∠ A = θ$ and $m ∠ C = ϕ$ . Find $\tan θ$ .

Hint

Use $\tan = \frac{Perp.}{Base}$ , with the sides read from angle $θ$ at $A$ .

Solution

Name the sides relative to $θ$ .

The angle $θ$ is at vertex $A$ . The side opposite it, $B C$ , is the perpendicular; the side next to it (other than the hypotenuse), $A B$ , is the base; $A C$ , opposite the right angle, is the hypotenuse.

$Perp. = B C = 5, Base = A B = 12, Hyp. = A C = 13$
Apply the definition of $\tan$ .

By definition $\tan = \frac{Perp.}{Base}$ .
$\begin{matrix} \tan θ & = \frac{Perp.}{Base} \\ = \frac{B C}{A B} \\ = \frac{5}{12} \end{matrix}$

Answer

$\tan θ = \frac{5}{12}$

Quick Solution

Q1Part (c)(iv)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ , $m B C = 5$ , $m A B = 12$ , $m A C = 13$ , $m ∠ A = θ$ and $m ∠ C = ϕ$ . Find $\sec θ$ .

Hint

Use $\sec = \frac{Hyp.}{Base}$ , with the sides read from angle $θ$ at $A$ .

Solution

Name the sides relative to $θ$ .

The angle $θ$ is at vertex $A$ . The side opposite it, $B C$ , is the perpendicular; the side next to it (other than the hypotenuse), $A B$ , is the base; $A C$ , opposite the right angle, is the hypotenuse.

$Perp. = B C = 5, Base = A B = 12, Hyp. = A C = 13$
Apply the definition of $\sec$ .

By definition $\sec = \frac{Hyp.}{Base}$ .
$\begin{matrix} \sec θ & = \frac{Hyp.}{Base} \\ = \frac{A C}{A B} \\ = \frac{13}{12} \end{matrix}$

Answer

$\sec θ = \frac{13}{12}$

Quick Solution

Q1Part (c)(v)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ , $m B C = 5$ , $m A B = 12$ , $m A C = 13$ , $m ∠ A = θ$ and $m ∠ C = ϕ$ . Find $cosec θ$ .

Hint

Use $cosec = \frac{Hyp.}{Perp.}$ , with the sides read from angle $θ$ at $A$ .

Solution

Name the sides relative to $θ$ .

The angle $θ$ is at vertex $A$ . The side opposite it, $B C$ , is the perpendicular; the side next to it (other than the hypotenuse), $A B$ , is the base; $A C$ , opposite the right angle, is the hypotenuse.

$Perp. = B C = 5, Base = A B = 12, Hyp. = A C = 13$
Apply the definition of $cosec$ .

By definition $cosec = \frac{Hyp.}{Perp.}$ .
$\begin{matrix} cosec θ & = \frac{Hyp.}{Perp.} \\ = \frac{A C}{B C} \\ = \frac{13}{5} \end{matrix}$

Answer

$cosec θ = \frac{13}{5}$

Quick Solution

Q1Part (c)(vi)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ , $m B C = 5$ , $m A B = 12$ , $m A C = 13$ , $m ∠ A = θ$ and $m ∠ C = ϕ$ . Find $\cot ϕ$ .

Hint

Use $\cot = \frac{Base}{Perp.}$ , with the sides read from angle $ϕ$ at $C$ .

Solution

Name the sides relative to $ϕ$ .

The angle $ϕ$ is at vertex $C$ . The side opposite it, $A B$ , is the perpendicular; the side next to it (other than the hypotenuse), $B C$ , is the base; $A C$ , opposite the right angle, is the hypotenuse.

$Perp. = A B = 12, Base = B C = 5, Hyp. = A C = 13$
Apply the definition of $\cot$ .

By definition $\cot = \frac{Base}{Perp.}$ .
$\begin{matrix} \cot ϕ & = \frac{Base}{Perp.} \\ = \frac{B C}{A B} \\ = \frac{5}{12} \end{matrix}$

Answer

$\cot ϕ = \frac{5}{12}$

Quick Solution

Q1Part (c)(vii)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ , $m B C = 5$ , $m A B = 12$ , $m A C = 13$ , $m ∠ A = θ$ and $m ∠ C = ϕ$ . Find $\tan ϕ$ .

Hint

Use $\tan = \frac{Perp.}{Base}$ , with the sides read from angle $ϕ$ at $C$ .

Solution

Name the sides relative to $ϕ$ .

The angle $ϕ$ is at vertex $C$ . The side opposite it, $A B$ , is the perpendicular; the side next to it (other than the hypotenuse), $B C$ , is the base; $A C$ , opposite the right angle, is the hypotenuse.

$Perp. = A B = 12, Base = B C = 5, Hyp. = A C = 13$
Apply the definition of $\tan$ .

By definition $\tan = \frac{Perp.}{Base}$ .
$\begin{matrix} \tan ϕ & = \frac{Perp.}{Base} \\ = \frac{A B}{B C} \\ = \frac{12}{5} \end{matrix}$

Answer

$\tan ϕ = \frac{12}{5}$

Quick Solution

Q1Part (c)(viii)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ , $m B C = 5$ , $m A B = 12$ , $m A C = 13$ , $m ∠ A = θ$ and $m ∠ C = ϕ$ . Find $cosec ϕ$ .

Hint

Use $cosec = \frac{Hyp.}{Perp.}$ , with the sides read from angle $ϕ$ at $C$ .

Solution

Name the sides relative to $ϕ$ .

The angle $ϕ$ is at vertex $C$ . The side opposite it, $A B$ , is the perpendicular; the side next to it (other than the hypotenuse), $B C$ , is the base; $A C$ , opposite the right angle, is the hypotenuse.

$Perp. = A B = 12, Base = B C = 5, Hyp. = A C = 13$
Apply the definition of $cosec$ .

By definition $cosec = \frac{Hyp.}{Perp.}$ .
$\begin{matrix} cosec ϕ & = \frac{Hyp.}{Perp.} \\ = \frac{A C}{A B} \\ = \frac{13}{12} \end{matrix}$

Answer

$cosec ϕ = \frac{13}{12}$

Quick Solution

Q1Part (c)(ix)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ , $m B C = 5$ , $m A B = 12$ , $m A C = 13$ , $m ∠ A = θ$ and $m ∠ C = ϕ$ . Find $\sec ϕ$ .

Hint

Use $\sec = \frac{Hyp.}{Base}$ , with the sides read from angle $ϕ$ at $C$ .

Solution

Name the sides relative to $ϕ$ .

The angle $ϕ$ is at vertex $C$ . The side opposite it, $A B$ , is the perpendicular; the side next to it (other than the hypotenuse), $B C$ , is the base; $A C$ , opposite the right angle, is the hypotenuse.

$Perp. = A B = 12, Base = B C = 5, Hyp. = A C = 13$
Apply the definition of $\sec$ .

By definition $\sec = \frac{Hyp.}{Base}$ .
$\begin{matrix} \sec ϕ & = \frac{Hyp.}{Base} \\ = \frac{A C}{B C} \\ = \frac{13}{5} \end{matrix}$

Answer

$\sec ϕ = \frac{13}{5}$

Quick Solution

Q1Part (c)(x)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ , $m B C = 5$ , $m A B = 12$ , $m A C = 13$ , $m ∠ A = θ$ and $m ∠ C = ϕ$ . Find $\cos ϕ$ .

Hint

Use $\cos = \frac{Base}{Hyp.}$ , with the sides read from angle $ϕ$ at $C$ .

Solution

Name the sides relative to $ϕ$ .

The angle $ϕ$ is at vertex $C$ . The side opposite it, $A B$ , is the perpendicular; the side next to it (other than the hypotenuse), $B C$ , is the base; $A C$ , opposite the right angle, is the hypotenuse.

$Perp. = A B = 12, Base = B C = 5, Hyp. = A C = 13$
Apply the definition of $\cos$ .

By definition $\cos = \frac{Base}{Hyp.}$ .
$\begin{matrix} \cos ϕ & = \frac{Base}{Hyp.} \\ = \frac{B C}{A C} \\ = \frac{5}{13} \end{matrix}$

Answer

$\cos ϕ = \frac{5}{13}$

Quick Solution

Question 2

EasyUnder Review

For the following right-angled triangle $A B C$ , find the trigonometric ratios for which $m ∠ A = ϕ$ and $m ∠ C = θ$ .

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ , $m B C = a$ , $m A B = c$ , $m A C = b$ , $m ∠ A = ϕ$ and $m ∠ C = θ$ .

(i) $\sin θ$ (ii) $\cos θ$ (iii) $\tan θ$ (iv) $\sin ϕ$ (v) $\cos ϕ$ (vi) $\tan ϕ$

Hint

Read the perpendicular, base and hypotenuse from the angle in question, then write the ratio in terms of $a$ , $b$ , $c$ .

Solution

(This question consists of multiple subparts. See solutions below.)

Quick Solution

Q2Part (i)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ , $m B C = a$ , $m A B = c$ , $m A C = b$ , $m ∠ A = ϕ$ and $m ∠ C = θ$ . Find $\sin θ$ .

Hint

Use $\sin = \frac{Perp.}{Hyp.}$ for angle $θ$ at $C$ .

Solution

Name the sides relative to $θ$ .

The angle $θ$ is at vertex $C$ . The side opposite it, $A B$ , is the perpendicular; the side next to it (other than the hypotenuse), $B C$ , is the base; $A C$ , opposite the right angle, is the hypotenuse.

$Perp. = A B = c, Base = B C = a, Hyp. = A C = b$
Apply the definition of $\sin$ .

By definition $\sin = \frac{Perp.}{Hyp.}$ .
$\begin{matrix} \sin θ & = \frac{Perp.}{Hyp.} \\ = \frac{A B}{A C} \\ = \frac{c}{b} \end{matrix}$

Answer

$\sin θ = \frac{c}{b}$

Quick Solution

Q2Part (ii)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ , $m B C = a$ , $m A B = c$ , $m A C = b$ , $m ∠ A = ϕ$ and $m ∠ C = θ$ . Find $\cos θ$ .

Hint

Use $\cos = \frac{Base}{Hyp.}$ for angle $θ$ at $C$ .

Solution

Name the sides relative to $θ$ .

The angle $θ$ is at vertex $C$ . The side opposite it, $A B$ , is the perpendicular; the side next to it (other than the hypotenuse), $B C$ , is the base; $A C$ , opposite the right angle, is the hypotenuse.

$Perp. = A B = c, Base = B C = a, Hyp. = A C = b$
Apply the definition of $\cos$ .

By definition $\cos = \frac{Base}{Hyp.}$ .
$\begin{matrix} \cos θ & = \frac{Base}{Hyp.} \\ = \frac{B C}{A C} \\ = \frac{a}{b} \end{matrix}$

Answer

$\cos θ = \frac{a}{b}$

Quick Solution

Q2Part (iii)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ , $m B C = a$ , $m A B = c$ , $m A C = b$ , $m ∠ A = ϕ$ and $m ∠ C = θ$ . Find $\tan θ$ .

Hint

Use $\tan = \frac{Perp.}{Base}$ for angle $θ$ at $C$ .

Solution

Name the sides relative to $θ$ .

The angle $θ$ is at vertex $C$ . The side opposite it, $A B$ , is the perpendicular; the side next to it (other than the hypotenuse), $B C$ , is the base; $A C$ , opposite the right angle, is the hypotenuse.

$Perp. = A B = c, Base = B C = a, Hyp. = A C = b$
Apply the definition of $\tan$ .

By definition $\tan = \frac{Perp.}{Base}$ .
$\begin{matrix} \tan θ & = \frac{Perp.}{Base} \\ = \frac{A B}{B C} \\ = \frac{c}{a} \end{matrix}$

Answer

$\tan θ = \frac{c}{a}$

Quick Solution

Q2Part (iv)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ , $m B C = a$ , $m A B = c$ , $m A C = b$ , $m ∠ A = ϕ$ and $m ∠ C = θ$ . Find $\sin ϕ$ .

Hint

Use $\sin = \frac{Perp.}{Hyp.}$ for angle $ϕ$ at $A$ .

Solution

Name the sides relative to $ϕ$ .

The angle $ϕ$ is at vertex $A$ . The side opposite it, $B C$ , is the perpendicular; the side next to it (other than the hypotenuse), $A B$ , is the base; $A C$ , opposite the right angle, is the hypotenuse.

$Perp. = B C = a, Base = A B = c, Hyp. = A C = b$
Apply the definition of $\sin$ .

By definition $\sin = \frac{Perp.}{Hyp.}$ .
$\begin{matrix} \sin ϕ & = \frac{Perp.}{Hyp.} \\ = \frac{B C}{A C} \\ = \frac{a}{b} \end{matrix}$

Answer

$\sin ϕ = \frac{a}{b}$

Quick Solution

Q2Part (v)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ , $m B C = a$ , $m A B = c$ , $m A C = b$ , $m ∠ A = ϕ$ and $m ∠ C = θ$ . Find $\cos ϕ$ .

Hint

Use $\cos = \frac{Base}{Hyp.}$ for angle $ϕ$ at $A$ .

Solution

Name the sides relative to $ϕ$ .

The angle $ϕ$ is at vertex $A$ . The side opposite it, $B C$ , is the perpendicular; the side next to it (other than the hypotenuse), $A B$ , is the base; $A C$ , opposite the right angle, is the hypotenuse.

$Perp. = B C = a, Base = A B = c, Hyp. = A C = b$
Apply the definition of $\cos$ .

By definition $\cos = \frac{Base}{Hyp.}$ .
$\begin{matrix} \cos ϕ & = \frac{Base}{Hyp.} \\ = \frac{A B}{A C} \\ = \frac{c}{b} \end{matrix}$

Answer

$\cos ϕ = \frac{c}{b}$

Quick Solution

Q2Part (vi)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ , $m B C = a$ , $m A B = c$ , $m A C = b$ , $m ∠ A = ϕ$ and $m ∠ C = θ$ . Find $\tan ϕ$ .

Hint

Use $\tan = \frac{Perp.}{Base}$ for angle $ϕ$ at $A$ .

Solution

Name the sides relative to $ϕ$ .

The angle $ϕ$ is at vertex $A$ . The side opposite it, $B C$ , is the perpendicular; the side next to it (other than the hypotenuse), $A B$ , is the base; $A C$ , opposite the right angle, is the hypotenuse.

$Perp. = B C = a, Base = A B = c, Hyp. = A C = b$
Apply the definition of $\tan$ .

By definition $\tan = \frac{Perp.}{Base}$ .
$\begin{matrix} \tan ϕ & = \frac{Perp.}{Base} \\ = \frac{B C}{A B} \\ = \frac{a}{c} \end{matrix}$

Answer

$\tan ϕ = \frac{a}{c}$

Quick Solution

Question 3

EasyUnder Review

Considering the adjoining triangle, verify that:

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ , $m A B = 4$ , $m B C = 3$ , $m A C = 5$ and $m ∠ A = θ$ .

(i) $\sin θ cosec θ = 1$ (ii) $\cos θ \sec θ = 1$ (iii) $\tan θ \cot θ = 1$

Hint

Find each ratio and its reciprocal from the triangle, then multiply them.

Solution

(This question consists of multiple subparts. See solutions below.)

Quick Solution

Q3Part (i)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ , $m A B = 4$ , $m B C = 3$ , $m A C = 5$ and $m ∠ A = θ$ . Verify that $\sin θ cosec θ = 1$ .

Hint

Find $\sin θ$ and $cosec θ$ from the sides, then multiply.

Solution

Name the sides relative to $θ$ .

The angle $θ$ is at $A$ : $B C$ is opposite it (perpendicular), $A B$ is next to it (base), and $A C$ is the hypotenuse.

$Perp. = 3, Base = 4, Hyp. = 5$
Write $\sin θ$ and $cosec θ$ .

$\sin θ = \frac{Perp.}{Hyp.}$ and its reciprocal $cosec θ = \frac{Hyp.}{Perp.}$ .
$\begin{matrix} \sin θ & = \frac{B C}{A C} \\ = \frac{3}{5} \end{matrix}$ $\begin{matrix} cosec θ & = \frac{A C}{B C} \\ = \frac{5}{3} \end{matrix}$
Multiply and compare with the right-hand side.

The two fractions are reciprocals, so the product cancels to 1.
$\begin{matrix} LHS & = \sin θ cosec θ \\ = \frac{3}{5} \times \frac{5}{3} \\ = 1 \\ = RHS \end{matrix}$

Answer

$\sin θ cosec θ = 1 (verified)$

Quick Solution

Q3Part (ii)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ , $m A B = 4$ , $m B C = 3$ , $m A C = 5$ and $m ∠ A = θ$ . Verify that $\cos θ \sec θ = 1$ .

Hint

Find $\cos θ$ and $\sec θ$ from the sides, then multiply.

Solution

Name the sides relative to $θ$ .

The angle $θ$ is at $A$ : $B C$ is opposite it (perpendicular), $A B$ is next to it (base), and $A C$ is the hypotenuse.

$Perp. = 3, Base = 4, Hyp. = 5$
Write $\cos θ$ and $\sec θ$ .

$\cos θ = \frac{Base}{Hyp.}$ and its reciprocal $\sec θ = \frac{Hyp.}{Base}$ .
$\begin{matrix} \cos θ & = \frac{A B}{A C} \\ = \frac{4}{5} \end{matrix}$ $\begin{matrix} \sec θ & = \frac{A C}{A B} \\ = \frac{5}{4} \end{matrix}$
Multiply and compare with the right-hand side.

The two fractions are reciprocals, so the product cancels to 1.
$\begin{matrix} LHS & = \cos θ \sec θ \\ = \frac{4}{5} \times \frac{5}{4} \\ = 1 \\ = RHS \end{matrix}$

Answer

$\cos θ \sec θ = 1 (verified)$

Quick Solution

Q3Part (iii)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ , $m A B = 4$ , $m B C = 3$ , $m A C = 5$ and $m ∠ A = θ$ . Verify that $\tan θ \cot θ = 1$ .

Hint

Find $\tan θ$ and $\cot θ$ from the sides, then multiply.

Solution

Name the sides relative to $θ$ .

The angle $θ$ is at $A$ : $B C$ is opposite it (perpendicular), $A B$ is next to it (base), and $A C$ is the hypotenuse.

$Perp. = 3, Base = 4, Hyp. = 5$
Write $\tan θ$ and $\cot θ$ .

$\tan θ = \frac{Perp.}{Base}$ and its reciprocal $\cot θ = \frac{Base}{Perp.}$ .
$\begin{matrix} \tan θ & = \frac{B C}{A B} \\ = \frac{3}{4} \end{matrix}$ $\begin{matrix} \cot θ & = \frac{A B}{B C} \\ = \frac{4}{3} \end{matrix}$
Multiply and compare with the right-hand side.

The two fractions are reciprocals, so the product cancels to 1.
$\begin{matrix} LHS & = \tan θ \cot θ \\ = \frac{3}{4} \times \frac{4}{3} \\ = 1 \\ = RHS \end{matrix}$

Answer

$\tan θ \cot θ = 1 (verified)$

Quick Solution

Question 4

EasyUnder Review

Fill in the blanks.

(i) $\sin 30^{\circ} = \sin (90^{\circ} - 60^{\circ}) =$ ______

(ii) $\cos 30^{\circ} = \cos (90^{\circ} - 60^{\circ}) =$ ______

(iii) $\tan 30^{\circ} = \tan (90^{\circ} - 60^{\circ}) =$ ______

(iv) $\tan 60^{\circ} = \tan (90^{\circ} - 30^{\circ}) =$ ______

(v) $\sin 60^{\circ} = \sin (90^{\circ} - 30^{\circ}) =$ ______

(vi) $\cos 60^{\circ} = \cos (90^{\circ} - 30^{\circ}) =$ ______

(vii) $\sin 45^{\circ} = \sin (90^{\circ} - 45^{\circ}) =$ ______

(viii) $\tan 45^{\circ} = \tan (90^{\circ} - 45^{\circ}) =$ ______

(ix) $\cos 45^{\circ} = \cos (90^{\circ} - 45^{\circ}) =$ ______

Hint

Use the complementary-angle rule: a ratio of $90^{\circ} - α$ equals the co-ratio of $α$ .

Solution

(This question consists of multiple subparts. See solutions below.)

Quick Solution

Q4Part (i)

EasyUnder Review

Fill in the blank:

$\sin 30^{\circ} = \sin (90^{\circ} - 60^{\circ}) =$ ______

Hint

$\sin (90^{\circ} - α) = \cos α$ .

Solution

Recall the complementary-angle rule.

In a right-angled triangle the two acute angles add up to $90^{\circ}$ , so the side opposite one acute angle is the side next to the other. Hence the perpendicular and base swap roles, giving

$\sin (90^{\circ} - α) = \cos α, \cos (90^{\circ} - α) = \sin α, \tan (90^{\circ} - α) = \cot α$
Apply it with $α = 60^{\circ}$ .

Here $30^{\circ} = 90^{\circ} - 60^{\circ}$ , so the sin of $30^{\circ}$ becomes the cos of $60^{\circ}$ .
$\begin{matrix} \sin 30^{\circ} & = \sin (90^{\circ} - 60^{\circ}) \\ = \cos 60^{\circ} \end{matrix}$
Check with the known values.

Both sides have the same value, confirming the blank.

$\sin 30^{\circ} = \frac{1}{2}, \cos 60^{\circ} = \frac{1}{2}$

Answer

$\sin 30^{\circ} = \cos 60^{\circ}$

Quick Solution

Q4Part (ii)

EasyUnder Review

Fill in the blank:

$\cos 30^{\circ} = \cos (90^{\circ} - 60^{\circ}) =$ ______

Hint

$\cos (90^{\circ} - α) = \sin α$ .

Solution

Recall the complementary-angle rule.

In a right-angled triangle the two acute angles add up to $90^{\circ}$ , so the side opposite one acute angle is the side next to the other. Hence the perpendicular and base swap roles, giving

$\sin (90^{\circ} - α) = \cos α, \cos (90^{\circ} - α) = \sin α, \tan (90^{\circ} - α) = \cot α$
Apply it with $α = 60^{\circ}$ .

Here $30^{\circ} = 90^{\circ} - 60^{\circ}$ , so the cos of $30^{\circ}$ becomes the sin of $60^{\circ}$ .
$\begin{matrix} \cos 30^{\circ} & = \cos (90^{\circ} - 60^{\circ}) \\ = \sin 60^{\circ} \end{matrix}$
Check with the known values.

Both sides have the same value, confirming the blank.

$\cos 30^{\circ} = \frac{\sqrt{3}}{2}, \sin 60^{\circ} = \frac{\sqrt{3}}{2}$

Answer

$\cos 30^{\circ} = \sin 60^{\circ}$

Quick Solution

Q4Part (iii)

EasyUnder Review

Fill in the blank:

$\tan 30^{\circ} = \tan (90^{\circ} - 60^{\circ}) =$ ______

Hint

$\tan (90^{\circ} - α) = \cot α$ .

Solution

Recall the complementary-angle rule.

In a right-angled triangle the two acute angles add up to $90^{\circ}$ , so the side opposite one acute angle is the side next to the other. Hence the perpendicular and base swap roles, giving

$\sin (90^{\circ} - α) = \cos α, \cos (90^{\circ} - α) = \sin α, \tan (90^{\circ} - α) = \cot α$
Apply it with $α = 60^{\circ}$ .

Here $30^{\circ} = 90^{\circ} - 60^{\circ}$ , so the tan of $30^{\circ}$ becomes the cot of $60^{\circ}$ .
$\begin{matrix} \tan 30^{\circ} & = \tan (90^{\circ} - 60^{\circ}) \\ = \cot 60^{\circ} \end{matrix}$
Check with the known values.

Both sides have the same value, confirming the blank.

$\tan 30^{\circ} = \frac{1}{\sqrt{3}}, \cot 60^{\circ} = \frac{1}{\sqrt{3}}$

Answer

$\tan 30^{\circ} = \cot 60^{\circ}$

Quick Solution

Q4Part (iv)

EasyUnder Review

Fill in the blank:

$\tan 60^{\circ} = \tan (90^{\circ} - 30^{\circ}) =$ ______

Hint

$\tan (90^{\circ} - α) = \cot α$ .

Solution

Recall the complementary-angle rule.

In a right-angled triangle the two acute angles add up to $90^{\circ}$ , so the side opposite one acute angle is the side next to the other. Hence the perpendicular and base swap roles, giving

$\sin (90^{\circ} - α) = \cos α, \cos (90^{\circ} - α) = \sin α, \tan (90^{\circ} - α) = \cot α$
Apply it with $α = 30^{\circ}$ .

Here $60^{\circ} = 90^{\circ} - 30^{\circ}$ , so the tan of $60^{\circ}$ becomes the cot of $30^{\circ}$ .
$\begin{matrix} \tan 60^{\circ} & = \tan (90^{\circ} - 30^{\circ}) \\ = \cot 30^{\circ} \end{matrix}$
Check with the known values.

Both sides have the same value, confirming the blank.

$\tan 60^{\circ} = \sqrt{3}, \cot 30^{\circ} = \sqrt{3}$

Answer

$\tan 60^{\circ} = \cot 30^{\circ}$

Quick Solution

Q4Part (v)

EasyUnder Review

Fill in the blank:

$\sin 60^{\circ} = \sin (90^{\circ} - 30^{\circ}) =$ ______

Hint

$\sin (90^{\circ} - α) = \cos α$ .

Solution

Recall the complementary-angle rule.

In a right-angled triangle the two acute angles add up to $90^{\circ}$ , so the side opposite one acute angle is the side next to the other. Hence the perpendicular and base swap roles, giving

$\sin (90^{\circ} - α) = \cos α, \cos (90^{\circ} - α) = \sin α, \tan (90^{\circ} - α) = \cot α$
Apply it with $α = 30^{\circ}$ .

Here $60^{\circ} = 90^{\circ} - 30^{\circ}$ , so the sin of $60^{\circ}$ becomes the cos of $30^{\circ}$ .
$\begin{matrix} \sin 60^{\circ} & = \sin (90^{\circ} - 30^{\circ}) \\ = \cos 30^{\circ} \end{matrix}$
Check with the known values.

Both sides have the same value, confirming the blank.

$\sin 60^{\circ} = \frac{\sqrt{3}}{2}, \cos 30^{\circ} = \frac{\sqrt{3}}{2}$

Answer

$\sin 60^{\circ} = \cos 30^{\circ}$

Quick Solution

Q4Part (vi)

EasyUnder Review

Fill in the blank:

$\cos 60^{\circ} = \cos (90^{\circ} - 30^{\circ}) =$ ______

Hint

$\cos (90^{\circ} - α) = \sin α$ .

Solution

Recall the complementary-angle rule.

In a right-angled triangle the two acute angles add up to $90^{\circ}$ , so the side opposite one acute angle is the side next to the other. Hence the perpendicular and base swap roles, giving

$\sin (90^{\circ} - α) = \cos α, \cos (90^{\circ} - α) = \sin α, \tan (90^{\circ} - α) = \cot α$
Apply it with $α = 30^{\circ}$ .

Here $60^{\circ} = 90^{\circ} - 30^{\circ}$ , so the cos of $60^{\circ}$ becomes the sin of $30^{\circ}$ .
$\begin{matrix} \cos 60^{\circ} & = \cos (90^{\circ} - 30^{\circ}) \\ = \sin 30^{\circ} \end{matrix}$
Check with the known values.

Both sides have the same value, confirming the blank.

$\cos 60^{\circ} = \frac{1}{2}, \sin 30^{\circ} = \frac{1}{2}$

Answer

$\cos 60^{\circ} = \sin 30^{\circ}$

Quick Solution

Q4Part (vii)

EasyUnder Review

Fill in the blank:

$\sin 45^{\circ} = \sin (90^{\circ} - 45^{\circ}) =$ ______

Hint

$\sin (90^{\circ} - α) = \cos α$ .

Solution

Recall the complementary-angle rule.

In a right-angled triangle the two acute angles add up to $90^{\circ}$ , so the side opposite one acute angle is the side next to the other. Hence the perpendicular and base swap roles, giving

$\sin (90^{\circ} - α) = \cos α, \cos (90^{\circ} - α) = \sin α, \tan (90^{\circ} - α) = \cot α$
Apply it with $α = 45^{\circ}$ .

Here $45^{\circ} = 90^{\circ} - 45^{\circ}$ , so the sin of $45^{\circ}$ becomes the cos of $45^{\circ}$ .
$\begin{matrix} \sin 45^{\circ} & = \sin (90^{\circ} - 45^{\circ}) \\ = \cos 45^{\circ} \end{matrix}$
Check with the known values.

Both sides have the same value, confirming the blank.

$\sin 45^{\circ} = \frac{1}{\sqrt{2}}, \cos 45^{\circ} = \frac{1}{\sqrt{2}}$

Answer

$\sin 45^{\circ} = \cos 45^{\circ}$

Quick Solution

Q4Part (viii)

EasyUnder Review

Fill in the blank:

$\tan 45^{\circ} = \tan (90^{\circ} - 45^{\circ}) =$ ______

Hint

$\tan (90^{\circ} - α) = \cot α$ .

Solution

Recall the complementary-angle rule.

In a right-angled triangle the two acute angles add up to $90^{\circ}$ , so the side opposite one acute angle is the side next to the other. Hence the perpendicular and base swap roles, giving

$\sin (90^{\circ} - α) = \cos α, \cos (90^{\circ} - α) = \sin α, \tan (90^{\circ} - α) = \cot α$
Apply it with $α = 45^{\circ}$ .

Here $45^{\circ} = 90^{\circ} - 45^{\circ}$ , so the tan of $45^{\circ}$ becomes the cot of $45^{\circ}$ .
$\begin{matrix} \tan 45^{\circ} & = \tan (90^{\circ} - 45^{\circ}) \\ = \cot 45^{\circ} \end{matrix}$
Check with the known values.

Both sides have the same value, confirming the blank.

$\tan 45^{\circ} = 1, \cot 45^{\circ} = 1$

Answer

$\tan 45^{\circ} = \cot 45^{\circ}$

Quick Solution

Q4Part (ix)

EasyUnder Review

Fill in the blank:

$\cos 45^{\circ} = \cos (90^{\circ} - 45^{\circ}) =$ ______

Hint

$\cos (90^{\circ} - α) = \sin α$ .

Solution

Recall the complementary-angle rule.

In a right-angled triangle the two acute angles add up to $90^{\circ}$ , so the side opposite one acute angle is the side next to the other. Hence the perpendicular and base swap roles, giving

$\sin (90^{\circ} - α) = \cos α, \cos (90^{\circ} - α) = \sin α, \tan (90^{\circ} - α) = \cot α$
Apply it with $α = 45^{\circ}$ .

Here $45^{\circ} = 90^{\circ} - 45^{\circ}$ , so the cos of $45^{\circ}$ becomes the sin of $45^{\circ}$ .
$\begin{matrix} \cos 45^{\circ} & = \cos (90^{\circ} - 45^{\circ}) \\ = \sin 45^{\circ} \end{matrix}$
Check with the known values.

Both sides have the same value, confirming the blank.

$\cos 45^{\circ} = \frac{1}{\sqrt{2}}, \sin 45^{\circ} = \frac{1}{\sqrt{2}}$

Answer

$\cos 45^{\circ} = \sin 45^{\circ}$

Quick Solution

Question 5

MediumUnder Review

If in a right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ and $C$ is an acute angle of $60^{\circ}$ . Also $\sin m ∠ A = \frac{a}{b}$ , then find the following trigonometric ratios.

(Figure: $m B C = a$ , $m A B = c$ , $m A C = b$ .)

(i) $\frac{m B C}{m A B}$ (ii) $\cos 60^{\circ}$ (iii) $\tan 60^{\circ}$ (iv) $cosec \frac{π}{3}$ (v) $\cot 60^{\circ}$ (vi) $\sin 30^{\circ}$ (vii) $\cos 30^{\circ}$ (viii) $\tan \frac{π}{6}$ (ix) $\sec 30^{\circ}$ (x) $\cot 30^{\circ}$

Hint

Find $m ∠ A = 30^{\circ}$ first, then read each ratio from the 60° angle at $C$ or the 30° angle at $A$ .

Solution

(This question consists of multiple subparts. See solutions below.)

Quick Solution

Q5Part (i)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ and $C$ is an acute angle of $60^{\circ}$ . The sides are $m B C = a$ , $m A B = c$ , $m A C = b$ , and $\sin m ∠ A = \frac{a}{b}$ . Find $\frac{m B C}{m A B}$ .

Hint

Substitute the side lengths shown in the figure.

Solution

Read the sides from the figure.

$B C$ has length $a$ and $A B$ has length $c$ .
$\begin{matrix} \frac{m B C}{m A B} & = \frac{a}{c} \end{matrix}$
Interpret it as a ratio.

For the $30^{\circ}$ angle at $A$ , $B C$ is the perpendicular and $A B$ the base, so this is $\tan 30^{\circ}$ (equivalently $\cot 60^{\circ}$ ).

$\tan 30^{\circ} = \frac{a}{c}$

Answer

$\frac{m B C}{m A B} = \frac{a}{c}$

Quick Solution

Q5Part (ii)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ and $C$ is an acute angle of $60^{\circ}$ . The sides are $m B C = a$ , $m A B = c$ , $m A C = b$ , and $\sin m ∠ A = \frac{a}{b}$ . Find $\cos 60^{\circ}$ .

Hint

Use angle $60^{\circ}$ at $C$ and $\cos = \frac{Base}{Hyp.}$ .

Solution

Name the sides relative to $60^{\circ}$ .

The angle $60^{\circ}$ is at vertex $C$ . The side opposite it, $A B$ , is the perpendicular; the side next to it (other than the hypotenuse), $B C$ , is the base; $A C$ , opposite the right angle, is the hypotenuse.

$Perp. = A B = c, Base = B C = a, Hyp. = A C = b$
Apply the definition of $\cos$ .

By definition $\cos = \frac{Base}{Hyp.}$ .
$\begin{matrix} \cos 60^{\circ} & = \frac{Base}{Hyp.} \\ = \frac{B C}{A C} \\ = \frac{a}{b} \end{matrix}$

Answer

$\cos 60^{\circ} = \frac{a}{b}$

Quick Solution

Q5Part (iii)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ and $C$ is an acute angle of $60^{\circ}$ . The sides are $m B C = a$ , $m A B = c$ , $m A C = b$ , and $\sin m ∠ A = \frac{a}{b}$ . Find $\tan 60^{\circ}$ .

Hint

Use angle $60^{\circ}$ at $C$ and $\tan = \frac{Perp.}{Base}$ .

Solution

Name the sides relative to $60^{\circ}$ .

The angle $60^{\circ}$ is at vertex $C$ . The side opposite it, $A B$ , is the perpendicular; the side next to it (other than the hypotenuse), $B C$ , is the base; $A C$ , opposite the right angle, is the hypotenuse.

$Perp. = A B = c, Base = B C = a, Hyp. = A C = b$
Apply the definition of $\tan$ .

By definition $\tan = \frac{Perp.}{Base}$ .
$\begin{matrix} \tan 60^{\circ} & = \frac{Perp.}{Base} \\ = \frac{A B}{B C} \\ = \frac{c}{a} \end{matrix}$

Answer

$\tan 60^{\circ} = \frac{c}{a}$

Quick Solution

Q5Part (iv)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ and $C$ is an acute angle of $60^{\circ}$ . The sides are $m B C = a$ , $m A B = c$ , $m A C = b$ , and $\sin m ∠ A = \frac{a}{b}$ . Find $cosec \frac{π}{3}$ .

Hint

Change the radian angle to $60^{\circ}$ first, then use $cosec = \frac{Hyp.}{Perp.}$ at $C$ .

Solution

Convert the angle to degrees.

Since $π$ radians $= 180^{\circ}$ ,
$\begin{matrix} \frac{π}{3} & = \frac{180^{\circ}}{3} \\ = 60^{\circ} \end{matrix}$
Name the sides relative to $60^{\circ}$ .

The angle $60^{\circ}$ is at vertex $C$ . The side opposite it, $A B$ , is the perpendicular; the side next to it (other than the hypotenuse), $B C$ , is the base; $A C$ , opposite the right angle, is the hypotenuse.

$Perp. = A B = c, Base = B C = a, Hyp. = A C = b$
Apply the definition of $cosec$ .

By definition $cosec = \frac{Hyp.}{Perp.}$ .
$\begin{matrix} cosec 60^{\circ} & = \frac{Hyp.}{Perp.} \\ = \frac{A C}{A B} \\ = \frac{b}{c} \end{matrix}$

Answer

$cosec \frac{π}{3} = \frac{b}{c}$

Quick Solution

Q5Part (v)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ and $C$ is an acute angle of $60^{\circ}$ . The sides are $m B C = a$ , $m A B = c$ , $m A C = b$ , and $\sin m ∠ A = \frac{a}{b}$ . Find $\cot 60^{\circ}$ .

Hint

Use angle $60^{\circ}$ at $C$ and $\cot = \frac{Base}{Perp.}$ .

Solution

Name the sides relative to $60^{\circ}$ .

The angle $60^{\circ}$ is at vertex $C$ . The side opposite it, $A B$ , is the perpendicular; the side next to it (other than the hypotenuse), $B C$ , is the base; $A C$ , opposite the right angle, is the hypotenuse.

$Perp. = A B = c, Base = B C = a, Hyp. = A C = b$
Apply the definition of $\cot$ .

By definition $\cot = \frac{Base}{Perp.}$ .
$\begin{matrix} \cot 60^{\circ} & = \frac{Base}{Perp.} \\ = \frac{B C}{A B} \\ = \frac{a}{c} \end{matrix}$

Answer

$\cot 60^{\circ} = \frac{a}{c}$

Quick Solution

Q5Part (vi)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ and $C$ is an acute angle of $60^{\circ}$ . The sides are $m B C = a$ , $m A B = c$ , $m A C = b$ , and $\sin m ∠ A = \frac{a}{b}$ . Find $\sin 30^{\circ}$ .

Hint

Use angle $30^{\circ}$ at $A$ and $\sin = \frac{Perp.}{Hyp.}$ .

Solution

Find the third angle.

The angles of a triangle add up to $180^{\circ}$ .
$\begin{matrix} m ∠ A & = 180^{\circ} - 90^{\circ} - 60^{\circ} \\ = 30^{\circ} \end{matrix}$
Name the sides relative to $30^{\circ}$ .

The angle $30^{\circ}$ is at vertex $A$ . The side opposite it, $B C$ , is the perpendicular; the side next to it (other than the hypotenuse), $A B$ , is the base; $A C$ , opposite the right angle, is the hypotenuse.

$Perp. = B C = a, Base = A B = c, Hyp. = A C = b$
Apply the definition of $\sin$ .

By definition $\sin = \frac{Perp.}{Hyp.}$ .
$\begin{matrix} \sin 30^{\circ} & = \frac{Perp.}{Hyp.} \\ = \frac{B C}{A C} \\ = \frac{a}{b} \end{matrix}$

Answer

$\sin 30^{\circ} = \frac{a}{b}$

Quick Solution

Q5Part (vii)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ and $C$ is an acute angle of $60^{\circ}$ . The sides are $m B C = a$ , $m A B = c$ , $m A C = b$ , and $\sin m ∠ A = \frac{a}{b}$ . Find $\cos 30^{\circ}$ .

Hint

Use angle $30^{\circ}$ at $A$ and $\cos = \frac{Base}{Hyp.}$ .

Solution

Find the third angle.

The angles of a triangle add up to $180^{\circ}$ .
$\begin{matrix} m ∠ A & = 180^{\circ} - 90^{\circ} - 60^{\circ} \\ = 30^{\circ} \end{matrix}$
Name the sides relative to $30^{\circ}$ .

The angle $30^{\circ}$ is at vertex $A$ . The side opposite it, $B C$ , is the perpendicular; the side next to it (other than the hypotenuse), $A B$ , is the base; $A C$ , opposite the right angle, is the hypotenuse.

$Perp. = B C = a, Base = A B = c, Hyp. = A C = b$
Apply the definition of $\cos$ .

By definition $\cos = \frac{Base}{Hyp.}$ .
$\begin{matrix} \cos 30^{\circ} & = \frac{Base}{Hyp.} \\ = \frac{A B}{A C} \\ = \frac{c}{b} \end{matrix}$

Answer

$\cos 30^{\circ} = \frac{c}{b}$

Quick Solution

Q5Part (viii)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ and $C$ is an acute angle of $60^{\circ}$ . The sides are $m B C = a$ , $m A B = c$ , $m A C = b$ , and $\sin m ∠ A = \frac{a}{b}$ . Find $\tan \frac{π}{6}$ .

Hint

Change the radian angle to $30^{\circ}$ first, then use $\tan = \frac{Perp.}{Base}$ at $A$ .

Solution

Convert the angle to degrees.

Since $π$ radians $= 180^{\circ}$ ,
$\begin{matrix} \frac{π}{6} & = \frac{180^{\circ}}{6} \\ = 30^{\circ} \end{matrix}$
Find the third angle.

The angles of a triangle add up to $180^{\circ}$ .
$\begin{matrix} m ∠ A & = 180^{\circ} - 90^{\circ} - 60^{\circ} \\ = 30^{\circ} \end{matrix}$
Name the sides relative to $30^{\circ}$ .

The angle $30^{\circ}$ is at vertex $A$ . The side opposite it, $B C$ , is the perpendicular; the side next to it (other than the hypotenuse), $A B$ , is the base; $A C$ , opposite the right angle, is the hypotenuse.

$Perp. = B C = a, Base = A B = c, Hyp. = A C = b$
Apply the definition of $\tan$ .

By definition $\tan = \frac{Perp.}{Base}$ .
$\begin{matrix} \tan 30^{\circ} & = \frac{Perp.}{Base} \\ = \frac{B C}{A B} \\ = \frac{a}{c} \end{matrix}$

Answer

$\tan \frac{π}{6} = \frac{a}{c}$

Quick Solution

Q5Part (ix)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ and $C$ is an acute angle of $60^{\circ}$ . The sides are $m B C = a$ , $m A B = c$ , $m A C = b$ , and $\sin m ∠ A = \frac{a}{b}$ . Find $\sec 30^{\circ}$ .

Hint

Use angle $30^{\circ}$ at $A$ and $\sec = \frac{Hyp.}{Base}$ .

Solution

Find the third angle.

The angles of a triangle add up to $180^{\circ}$ .
$\begin{matrix} m ∠ A & = 180^{\circ} - 90^{\circ} - 60^{\circ} \\ = 30^{\circ} \end{matrix}$
Name the sides relative to $30^{\circ}$ .

The angle $30^{\circ}$ is at vertex $A$ . The side opposite it, $B C$ , is the perpendicular; the side next to it (other than the hypotenuse), $A B$ , is the base; $A C$ , opposite the right angle, is the hypotenuse.

$Perp. = B C = a, Base = A B = c, Hyp. = A C = b$
Apply the definition of $\sec$ .

By definition $\sec = \frac{Hyp.}{Base}$ .
$\begin{matrix} \sec 30^{\circ} & = \frac{Hyp.}{Base} \\ = \frac{A C}{A B} \\ = \frac{b}{c} \end{matrix}$

Answer

$\sec 30^{\circ} = \frac{b}{c}$

Quick Solution

Q5Part (x)

EasyUnder Review

In right-angled triangle $A B C$ , $m ∠ B = 90^{\circ}$ and $C$ is an acute angle of $60^{\circ}$ . The sides are $m B C = a$ , $m A B = c$ , $m A C = b$ , and $\sin m ∠ A = \frac{a}{b}$ . Find $\cot 30^{\circ}$ .

Hint

Use angle $30^{\circ}$ at $A$ and $\cot = \frac{Base}{Perp.}$ .

Solution

Find the third angle.

The angles of a triangle add up to $180^{\circ}$ .
$\begin{matrix} m ∠ A & = 180^{\circ} - 90^{\circ} - 60^{\circ} \\ = 30^{\circ} \end{matrix}$
Name the sides relative to $30^{\circ}$ .

The angle $30^{\circ}$ is at vertex $A$ . The side opposite it, $B C$ , is the perpendicular; the side next to it (other than the hypotenuse), $A B$ , is the base; $A C$ , opposite the right angle, is the hypotenuse.

$Perp. = B C = a, Base = A B = c, Hyp. = A C = b$
Apply the definition of $\cot$ .

By definition $\cot = \frac{Base}{Perp.}$ .
$\begin{matrix} \cot 30^{\circ} & = \frac{Base}{Perp.} \\ = \frac{A B}{B C} \\ = \frac{c}{a} \end{matrix}$

Answer

$\cot 30^{\circ} = \frac{c}{a}$

Quick Solution

The exercises and question numbering reproduced on these pages are from Mathematics for Class 9 (National Curriculum of Pakistan 2023), published by the Punjab Curriculum and Textbook Board (PCTB), Lahore, authored by Muhammad Akhtar Shirani, Madiha Mahmood, and Ghulam Murtaza. PCTB holds the copyright in the original textbook. PrepSure is not affiliated with, endorsed by, or sponsored by PCTB. The worked solutions, explanations, hints and method notes are PrepSure's own original work, written and reviewed by our team. If you hold rights in this material and believe anything here exceeds fair use, write to us and we will take it down.

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