Mathematics

Class 92025–26 SNC23 Questions SolvedEnglish Medium

Exercise 6.5 — Class 9 Mathematics Solutions

All 23 questions below are worked step by step — every step clearly worked out with mathematical reasons — for the Punjab Board 9th class Mathematics textbook.

Exercise 6.5

9 Questions (14 Sub-parts)

Quick Solution Step by Step

Just the working. Switch to Step by Step for the full explanation.

Question 1

EasyUnder Review

Find the values of $x$ , $y$ and $z$ from the following right-angled triangles.

Hint

Identify the hypotenuse, perpendicular and base relative to the given angle, then use the trigonometric ratio that links the known side to each unknown side.

Solution

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

Step by Step

Q1Part (i)

Easy

In right-angled $△ A B C$ , $m ∠ B = 90^{\circ}$ and $m ∠ A = 30^{\circ}$ . The base $A B = y = 4 cm$ , the side $B C = x$ is opposite $∠ A$ , and the hypotenuse is $A C = z$ . Find $x$ and $z$ .

Quick Solution

Under Review

\begin{matrix} \tan 30^{\circ} & = \frac{x}{4} \\ x & = \frac{4}{\sqrt{3}} \\ = \frac{4 \sqrt{3}}{3} cm \\ \cos 30^{\circ} & = \frac{4}{z} \\ \frac{\sqrt{3}}{2} & = \frac{4}{z} \\ z & = \frac{8}{\sqrt{3}} \\ = \frac{8 \sqrt{3}}{3} cm \end{matrix}

Step by Step

Q1Part (ii)

Easy

In right-angled $△ A B C$ , $m ∠ B = 90^{\circ}$ and $m ∠ A = \frac{π}{4}$ . The side $B C = y = \sqrt{3} cm$ is opposite $∠ A$ , the base is $A B = x$ and the hypotenuse is $A C = z$ . Find $x$ and $z$ .

Quick Solution

Under Review

\begin{matrix} \frac{π}{4} & = 45^{\circ} \\ \tan 45^{\circ} & = \frac{\sqrt{3}}{x} \\ x & = \sqrt{3} cm \\ \sin 45^{\circ} & = \frac{\sqrt{3}}{z} \\ z & = \sqrt{2} \cdot \sqrt{3} \\ = \sqrt{6} cm \end{matrix}

Step by Step

Q1Part (iii)

Easy

In right-angled $△ A B C$ , $m ∠ B = 90^{\circ}$ and $m ∠ C = 60^{\circ}$ . The hypotenuse is $C A = z = 2 cm$ , the base next to $∠ C$ is $C B = x$ , and the side $A B = y$ is opposite $∠ C$ . Find $x$ and $y$ .

Quick Solution

Under Review

\begin{matrix} \cos 60^{\circ} & = \frac{x}{2} \\ x & = 2 \times \frac{1}{2} \\ = 1 cm \\ \sin 60^{\circ} & = \frac{y}{2} \\ y & = 2 \times \frac{\sqrt{3}}{2} \\ = \sqrt{3} cm \end{matrix}

Step by Step

Q1Part (iv)

Easy

In right-angled $△ A B C$ , $m ∠ B = 90^{\circ}$ and $m ∠ A = 45^{\circ}$ . The base is $A B = y = 4 cm$ , the side $B C = x$ is opposite $∠ A$ , and the hypotenuse is $A C = z$ . Find $x$ and $z$ .

Quick Solution

Under Review

\begin{matrix} \tan 45^{\circ} & = \frac{x}{4} \\ x & = 4 cm \\ \cos 45^{\circ} & = \frac{4}{z} \\ z & = 4 \sqrt{2} cm \end{matrix}

Step by Step

Question 2

MediumUnder Review

Find the unknown side and angles of the following triangles.

Hint

Use Pythagoras' theorem for the missing side, an inverse trigonometric ratio for one acute angle, and $m ∠ A + m ∠ C = 90^{\circ}$ for the other.

Solution

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

Step by Step

Q2Part (i)

Medium

In right-angled $△ A B C$ , $m ∠ B = 90^{\circ}$ , $a = B C = \sqrt{3} cm$ and $c = A B = \sqrt{13} cm$ . Find the hypotenuse $b = A C$ , $m ∠ A$ and $m ∠ C$ . (Give angles to one decimal place.)

Quick Solution

Under Review

\begin{matrix} b^{2} & = 3 + 13 \\ = 16 \\ b & = 4 cm \\ \tan A & = \frac{\sqrt{3}}{\sqrt{13}} \\ m ∠ A & \approx {25.7}^{\circ} \\ m ∠ C & = 90^{\circ} - {25.7}^{\circ} \\ \approx {64.3}^{\circ} \end{matrix}

Step by Step

Q2Part (ii)

Easy

In right-angled $△ A B C$ , $m ∠ B = 90^{\circ}$ , $a = B C = 4 cm$ and $c = A B = 4 cm$ . Find the hypotenuse $b = A C$ , $m ∠ A$ and $m ∠ C$ .

Quick Solution

Under Review

\begin{matrix} b^{2} & = 16 + 16 \\ = 32 \\ b & = 4 \sqrt{2} cm \\ \tan A & = \frac{4}{4} \\ = 1 \\ m ∠ A & = 45^{\circ} \\ m ∠ C & = 90^{\circ} - 45^{\circ} \\ = 45^{\circ} \end{matrix}

Step by Step

Question 3

Easy

Each side of a square field is $60 m$ long. Find the lengths of the diagonals of the field.

Quick Solution

Under Review

\begin{matrix} A C^{2} & = 60^{2} + 60^{2} \\ = 7200 \\ A C & = 60 \sqrt{2} \\ \approx 84.85 m \end{matrix}

Step by Step

Question 4

MediumUnder Review

Solve the following triangles when $m ∠ B = 90^{\circ}$ . (In $△ A B C$ , $a = B C$ , $b = C A$ and $c = A B$ ; side $b$ is the hypotenuse.)

Hint

Solving a triangle means finding all its unknown sides and angles: use the angle sum for the third angle, then trigonometric ratios (or Pythagoras' theorem) for the sides.

Solution

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

Step by Step

Q4Part (i)

Easy

$m ∠ C = 60^{\circ}$ , $c = 3 \sqrt{3} cm$

Quick Solution

Under Review

\begin{matrix} m ∠ A & = 180^{\circ} - 150^{\circ} \\ = 30^{\circ} \\ \tan 60^{\circ} & = \frac{3 \sqrt{3}}{a} \\ a & = 3 cm \\ \sin 60^{\circ} & = \frac{3 \sqrt{3}}{b} \\ b & = 6 cm \end{matrix}

Step by Step

Q4Part (ii)

Easy

$m ∠ C = 45^{\circ}$ , $a = 8 cm$

Quick Solution

Under Review

\begin{matrix} m ∠ A & = 180^{\circ} - 135^{\circ} \\ = 45^{\circ} \\ \tan 45^{\circ} & = \frac{c}{8} \\ c & = 8 cm \\ \cos 45^{\circ} & = \frac{8}{b} \\ b & = 8 \sqrt{2} cm \end{matrix}

Step by Step

Q4Part (iii)

Medium

$a = 12 cm$ , $c = 6 cm$

Quick Solution

Under Review

\begin{matrix} b^{2} & = 144 + 36 \\ = 180 \\ b & = 6 \sqrt{5} \\ \approx 13.42 cm \\ \tan A & = \frac{12}{6} \\ = 2 \\ m ∠ A & \approx {63.4}^{\circ} \\ m ∠ C & = 90^{\circ} - {63.4}^{\circ} \\ \approx {26.6}^{\circ} \end{matrix}

Step by Step

Q4Part (iv)

Easy

$m ∠ A = 60^{\circ}$ , $c = 4 cm$

Quick Solution

Under Review

\begin{matrix} m ∠ C & = 180^{\circ} - 150^{\circ} \\ = 30^{\circ} \\ \tan 60^{\circ} & = \frac{a}{4} \\ a & = 4 \sqrt{3} cm \\ \cos 60^{\circ} & = \frac{4}{b} \\ b & = 8 cm \end{matrix}

Step by Step

Q4Part (v)

Easy

$m ∠ A = 30^{\circ}$ , $c = 4 cm$

Quick Solution

Under Review

\begin{matrix} m ∠ C & = 180^{\circ} - 120^{\circ} \\ = 60^{\circ} \\ \tan 30^{\circ} & = \frac{a}{4} \\ a & = \frac{4}{\sqrt{3}} \\ = \frac{4 \sqrt{3}}{3} cm \\ \cos 30^{\circ} & = \frac{4}{b} \\ b & = \frac{8}{\sqrt{3}} \\ = \frac{8 \sqrt{3}}{3} cm \end{matrix}

Step by Step

Q4Part (vi)

Medium

$b = 10 cm$ , $a = 6 cm$

Quick Solution

Under Review

\begin{matrix} c^{2} & = 100 - 36 \\ = 64 \\ c & = 8 cm \\ \sin A & = \frac{6}{10} \\ = 0.6 \\ m ∠ A & \approx {36.9}^{\circ} \\ m ∠ C & = 90^{\circ} - {36.9}^{\circ} \\ \approx {53.1}^{\circ} \end{matrix}

Step by Step

Question 5

Medium

Let $Q$ and $R$ be two points on the same bank of a canal. The point $P$ is placed on the other bank, straight across from $R$ , so that $P R$ is perpendicular to the bank and $m ∠ P R Q = 90^{\circ}$ . If $Q R = 5 m$ and $P Q = 13 m$ , find the width of the canal and $m ∠ P Q R$ in radians.

Quick Solution

Under Review

\begin{matrix} P R^{2} & = 169 - 25 \\ = 144 \\ P R & = 12 m \\ \cos θ & = \frac{5}{13} \\ θ & \approx {67.38}^{\circ} \\ \approx 1.18 rad \end{matrix}

Step by Step

Question 6

Medium

Calculate the length $x$ in the figure. The figure is a quadrilateral $A B C D$ with diagonal $B D$ . In $△ A B D$ , $m ∠ A B D = 90^{\circ}$ , $A B = 10 cm$ and $A D = 17 cm$ . In $△ B C D$ , $m ∠ B C D = 90^{\circ}$ , $C D = 8 cm$ and $B C = x$ .

Quick Solution

Under Review

\begin{matrix} B D^{2} & = 17^{2} - 10^{2} \\ = 189 \\ x^{2} & = 189 - 8^{2} \\ = 125 \\ x & = 5 \sqrt{5} \\ \approx 11.18 cm \end{matrix}

Step by Step

Question 7

Easy

A ladder is placed against a wall such that the foot of the ladder is $2 m$ away from the wall. If the length of the ladder is $8 m$ , find the height of the wall (the height at which the top of the ladder touches the wall).

Quick Solution

Under Review

\begin{matrix} B C^{2} & = 8^{2} - 2^{2} \\ = 60 \\ B C & = \sqrt{60} \\ \approx 7.75 m \end{matrix}

Step by Step

Question 8

Medium

The diagonal of a rectangular field $A B C D$ is $(x + 9) m$ and the sides are $(x + 7) m$ and $x m$ . Find the value of $x$ .

Quick Solution

Under Review

\begin{matrix} (x + 7)^{2} + x^{2} & = (x + 9)^{2} \\ x^{2} - 4 x - 32 & = 0 \\ (x - 8) (x + 4) & = 0 \\ ⟹ x & = 8 (x \neq - 4) \end{matrix}

Step by Step

Question 9

EasyUnder Review

Calculate the value of $x$ in each case.

Hint

Identify the hypotenuse in each right triangle and apply Pythagoras' theorem.

Solution

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

Step by Step

Q9Part (i)

Easy

In right-angled $△ A B C$ , $m ∠ B = 90^{\circ}$ , $A B = x$ , $B C = 12 cm$ and the hypotenuse $A C = 20 cm$ .

Quick Solution

Under Review

\begin{matrix} x^{2} & = 20^{2} - 12^{2} \\ = 256 \\ x & = 16 cm \end{matrix}

Step by Step

Q9Part (ii)

Easy

In $△ A D C$ , $D B$ is perpendicular to $A C$ at the point $B$ , and $B$ is the midpoint of $A C$ , so $A B = B C$ . Given $B C = 4 cm$ , $D C = 5 cm$ and $A D = x$ .

Quick Solution

Under Review

\begin{matrix} D B^{2} & = 5^{2} - 4^{2} \\ = 9 \\ x^{2} & = 4^{2} + D B^{2} \\ = 16 + 9 \\ = 25 \\ x & = 5 cm \end{matrix}

Step by Step

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.

Rights holder with a concern about this page? copyright@prepsure.pk