Mathematics

Class 92025–26 SNC28 Questions SolvedEnglish Medium

Review Exercise 6 — Class 9 Mathematics Solutions

All 28 questions below are worked step by step — every step clearly worked out with mathematical reasons — for the Punjab Board 9th class Mathematics textbook.

Review Exercise 6

6 Questions (22 Sub-parts)

Quick Solution Step by Step

Every step explained, with hints and the method used.

Question 1

EasyUnder Review

Choose the correct option.

Hint

Work each statement out quickly, then match your result with the options.

Solution

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

Quick Solution

Q1Part (i)

EasyUnder Review

The value of $\tan^{- 1} (\frac{1}{2})$ in radians is:

Multiple Choice

$\frac{π}{2}$

$\frac{3 π}{2}$

$0.4636 π$

$0.4636$

Answer

Hint

Put the calculator in radian mode and find the angle whose tangent is 0.5.

Solution

Use the calculator in radian mode.

Set the calculator to RAD and evaluate the inverse tangent of $0.5$ .

$\tan^{- 1} (0.5) \approx 0.4636 rad$
Rule out the other options.

$\frac{π}{2}$ and $\frac{3 π}{2}$ are far too large, and $0.4636 π \approx 1.457$ rad is not the value; the answer is $0.4636$ itself.

Answer

$Option (d) : 0.4636$

Quick Solution

Q1Part (ii)

EasyUnder Review

In a right triangle, the hypotenuse is 13 units and one of the angles is $θ = 30^{\circ}$ . What is the length of the opposite side?

Multiple Choice

6.5 units

Answer

7.5 units

6 units

5 units

Hint

Opposite over hypotenuse is the sine ratio.

Solution

Choose the ratio.

The opposite side and the hypotenuse are linked by sine.

$\sin 30^{\circ} = \frac{opp}{13}$
Solve.

Use $\sin 30^{\circ} = \frac{1}{2}$ .
$\begin{matrix} opp & = 13 \times \frac{1}{2} \\ = 6.5 units \end{matrix}$

Answer

$Option (a) : 6.5 units$

Quick Solution

Q1Part (iii)

EasyUnder Review

A person standing 50 m away from a building sees the top of the building at an angle of elevation of $45^{\circ}$ . Height of the building is:

Multiple Choice

50 m

Answer

25 m

35 m

70 m

Hint

Use tan 45° = height ÷ distance.

Solution

Choose the ratio.

Height is opposite the angle and the 50 m distance is adjacent, so use tangent.

$\tan 45^{\circ} = \frac{h}{50}$
Solve.

Since $\tan 45^{\circ} = 1$ :

$h = 50 m$

Answer

$Option (a) : 50 m$

Quick Solution

Q1Part (iv)

EasyUnder Review

$\sec^{2} θ - \tan^{2} θ =$ ______.

Multiple Choice

$\sin^{2} θ$

$1$

Answer

$\cos^{2} θ$

$\cot^{2} θ$

Hint

Rearrange 1 + tan²θ = sec²θ.

Solution

Recall the identity.

The Pythagorean identity in secant form is $1 + \tan^{2} θ = \sec^{2} θ$ .
Rearrange.

Subtract $\tan^{2} θ$ from both sides.

$\sec^{2} θ - \tan^{2} θ = 1$

Answer

$Option (b) : 1$

Quick Solution

Q1Part (v)

EasyUnder Review

If $\sin θ = \frac{3}{5}$ , and $θ$ is an acute angle, $\cos^{2} θ =$ ______.

Multiple Choice

$\frac{7}{25}$

$\frac{24}{25}$

$\frac{16}{25}$

Answer

$\frac{4}{25}$

Hint

Use cos²θ = 1 − sin²θ.

Solution

Use the identity.

From $\sin^{2} θ + \cos^{2} θ = 1$ :

\begin{matrix} \cos^{2} θ & = 1 - \sin^{2} θ \\ = 1 - \frac{9}{25} \\ = \frac{16}{25} \end{matrix}

Answer

$Option (c) : \frac{16}{25}$

Quick Solution

Q1Part (vi)

EasyUnder Review

$\frac{5 π}{24}$ rad = ______ degrees.

Multiple Choice

$30^{\circ}$

${37.5}^{\circ}$

Answer

$45^{\circ}$

${52.5}^{\circ}$

Hint

Multiply radians by 180°/π.

Solution

Convert.

Multiply by $\frac{180^{\circ}}{π}$ .

\begin{matrix} \frac{5 π}{24} & = \frac{5 \times 180^{\circ}}{24} \\ = {37.5}^{\circ} \end{matrix}

Answer

$Option (b) : {37.5}^{\circ}$

Quick Solution

Q1Part (vii)

EasyUnder Review

${292.5}^{\circ} =$ ______ rad.

Multiple Choice

$\frac{17 π}{6}$

$\frac{17 π}{4}$

$1.6 π$

$1.625 π$

Answer

Hint

Multiply degrees by π/180°.

Solution

Convert.

Multiply by $\frac{π}{180^{\circ}}$ .

\begin{matrix} {292.5}^{\circ} & = \frac{292.5}{180} π \\ = 1.625 π \end{matrix}

Answer

$Option (d) : 1.625 π$

Quick Solution

Q1Part (viii)

EasyUnder Review

Which of the following is a valid identity?

Multiple Choice

$\cos (\frac{π}{2} - θ) = \sin θ$

Answer

$\cos (\frac{π}{2} - θ) = \cos θ$

$\cos (\frac{π}{2} - θ) = \sec θ$

$\cos (\frac{π}{2} - θ) = cosec θ$

Hint

Think of the two acute angles of a right triangle.

Solution

Use complementary angles.

In a right triangle the two acute angles are $θ$ and $90^{\circ} - θ$ ; the side adjacent to one is opposite the other, so the cosine of one equals the sine of the other.

$\cos (\frac{π}{2} - θ) = \sin θ$
Check with a value.

Take $θ = 30^{\circ}$ : $\cos 60^{\circ} = \frac{1}{2} = \sin 30^{\circ}$ , so (a) holds; the others fail.

Answer

$Option (a) : \cos (\frac{π}{2} - θ) = \sin θ$

Quick Solution

Q1Part (ix)

EasyUnder Review

$\sin 60^{\circ} =$ ______.

Multiple Choice

$1$

$\frac{1}{2}$

$\sqrt{(3)^{2}}$

$\frac{\sqrt{3}}{2}$

Answer

Hint

Recall the 30°–60°–90° triangle with sides 1, √3, 2.

Solution

Use the 30°–60°–90° triangle.

With sides $1, \sqrt{3}, 2$ , the side opposite $60^{\circ}$ is $\sqrt{3}$ and the hypotenuse is $2$ .

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

Answer

$Option (d) : \frac{\sqrt{3}}{2}$

Quick Solution

Q1Part (x)

EasyUnder Review

$\cos^{2} 100 π + \sin^{2} 100 π =$ ______.

Multiple Choice

$1$

Answer

$2$

$3$

$4$

Hint

The same angle appears in both terms — which identity is that?

Solution

Apply the identity.

$\sin^{2} θ + \cos^{2} θ = 1$ holds for every angle, including $θ = 100 π$ .

$\cos^{2} 100 π + \sin^{2} 100 π = 1$

Answer

$Option (a) : 1$

Quick Solution

Question 2

EasyUnder Review

Convert the given angles from:

Multiple Choice

degrees to radians giving answer in terms of $π$ .

radians to degrees giving answer in degrees and minutes.

Hint

Use $180^{\circ} = π$ rad: multiply by $\frac{π}{180^{\circ}}$ for radians and by $\frac{180^{\circ}}{π}$ for degrees, with $1^{\circ} = 60^{'}$ .

Solution

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

Quick Solution

Q2Part (a)(i)

EasyUnder Review

Convert $255^{\circ}$ to radians, giving the answer in terms of $π$ .

Hint

Multiply by π/180 and cancel by 15.

Solution

Multiply by the conversion factor.

Since $180^{\circ} = π$ rad, multiply by $\frac{π}{180^{\circ}}$ .

$255^{\circ} = 255 \times \frac{π}{180} rad$
Simplify the fraction.

Divide 255 and 180 by their HCF, 15.
$\begin{matrix} 255 \times \frac{π}{180} & = \frac{255 π}{180} \\ = \frac{17 π}{12} rad \end{matrix}$

Answer

$255^{\circ} = \frac{17 π}{12} rad$

Quick Solution

Q2Part (a)(ii)

MediumUnder Review

Convert $75^{\circ} 45^{'}$ to radians, giving the answer in terms of $π$ .

Hint

First turn 45′ into a fraction of a degree.

Solution

Write the minutes as degrees.

Since $60^{'} = 1^{\circ}$ , $45^{'} = {\frac{45}{60}}^{\circ} = {0.75}^{\circ}$ .
$\begin{matrix} 75^{\circ} 45^{'} & = 75^{\circ} + {(\frac{45}{60})}^{\circ} \\ = {75.75}^{\circ} \\ = {\frac{303}{4}}^{\circ} \end{matrix}$
Convert to radians.

Multiply by $\frac{π}{180^{\circ}}$ and simplify (303 and 180 share the factor 3).
$\begin{matrix} \frac{303}{4} \times \frac{π}{180} & = \frac{303 π}{720} \\ = \frac{101 π}{240} rad \end{matrix}$

Answer

$75^{\circ} 45^{'} = \frac{101 π}{240} rad$

Quick Solution

Q2Part (a)(iii)

EasyUnder Review

Convert ${142.5}^{\circ}$ to radians, giving the answer in terms of $π$ .

Hint

Write 142.5 as 285/2 before multiplying by π/180.

Solution

Write the decimal as a fraction.

$142.5 = \frac{285}{2}$ .

${142.5}^{\circ} = {\frac{285}{2}}^{\circ}$
Convert to radians.

Multiply by $\frac{π}{180^{\circ}}$ and cancel by 15.
$\begin{matrix} \frac{285}{2} \times \frac{π}{180} & = \frac{285 π}{360} \\ = \frac{19 π}{24} rad \end{matrix}$

Answer

${142.5}^{\circ} = \frac{19 π}{24} rad$

Quick Solution

Q2Part (b)(i)

EasyUnder Review

Convert $\frac{17 π}{24}$ rad to degrees and minutes.

Hint

Multiply by 180°/π, then change the decimal part to minutes.

Solution

Convert to degrees.

Multiply by $\frac{180^{\circ}}{π}$ ; the $π$ cancels.
$\begin{matrix} \frac{17 π}{24} & = \frac{17 \times 180^{\circ}}{24} \\ = {127.5}^{\circ} \end{matrix}$
Turn the decimal into minutes.

Multiply the decimal part by 60, since $1^{\circ} = 60^{'}$ .
$\begin{matrix} {127.5}^{\circ} & = 127^{\circ} + (0.5 \times 60)^{'} \\ = 127^{\circ} 30^{'} \end{matrix}$

Answer

$\frac{17 π}{24} rad = 127^{\circ} 30^{'}$

Quick Solution

Q2Part (b)(ii)

EasyUnder Review

Convert $\frac{7 π}{12}$ rad to degrees and minutes.

Hint

Multiply by 180°/π.

Solution

Convert to degrees.

Multiply by $\frac{180^{\circ}}{π}$ .
$\begin{matrix} \frac{7 π}{12} & = \frac{7 \times 180^{\circ}}{12} \\ = 7 \times 15^{\circ} \\ = 105^{\circ} \end{matrix}$
State in degrees and minutes.

There is no fractional part, so the minutes are zero.

$105^{\circ} = 105^{\circ} 0^{'}$

Answer

$\frac{7 π}{12} rad = 105^{\circ}$

Quick Solution

Q2Part (b)(iii)

EasyUnder Review

Convert $\frac{11 π}{16}$ rad to degrees and minutes.

Hint

Multiply by 180°/π, then change 0.75° to minutes.

Solution

Convert to degrees.

Multiply by $\frac{180^{\circ}}{π}$ .
$\begin{matrix} \frac{11 π}{16} & = \frac{11 \times 180^{\circ}}{16} \\ = \frac{495^{\circ}}{4} \\ = {123.75}^{\circ} \end{matrix}$
Turn the decimal into minutes.

Multiply the decimal part by 60.
$\begin{matrix} {123.75}^{\circ} & = 123^{\circ} + (0.75 \times 60)^{'} \\ = 123^{\circ} 45^{'} \end{matrix}$

Answer

$\frac{11 π}{16} rad = 123^{\circ} 45^{'}$

Quick Solution

Question 3

MediumUnder Review

Prove the following trigonometric identities:

Hint

Start from the more complicated side, write everything in terms of $\sin θ$ and $\cos θ$ , and use the Pythagorean identities.

Solution

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

Quick Solution

Q3Part (i)

MediumUnder Review

$\frac{\sin θ}{1 - \cos θ} = \frac{1 + \cos θ}{\sin θ}$

Hint

Multiply top and bottom by 1 + cos θ.

Solution

Multiply by the conjugate.

Multiply the numerator and denominator of the L.H.S. by $1 + \cos θ$ , so the denominator becomes a difference of squares.
$\begin{matrix} L.H.S. & = \frac{\sin θ}{1 - \cos θ} \times \frac{1 + \cos θ}{1 + \cos θ} \\ = \frac{\sin θ (1 + \cos θ)}{1 - \cos^{2} θ} \end{matrix}$
Use the Pythagorean identity.

$1 - \cos^{2} θ = \sin^{2} θ$ ; then cancel one $\sin θ$ .
$\begin{matrix} \frac{\sin θ (1 + \cos θ)}{1 - \cos^{2} θ} & = \frac{\sin θ (1 + \cos θ)}{\sin^{2} θ} \\ = \frac{1 + \cos θ}{\sin θ} \\ = R.H.S. \end{matrix}$

Answer

$\frac{\sin θ}{1 - \cos θ} = \frac{1 + \cos θ}{\sin θ} (proved)$

Quick Solution

Q3Part (ii)

EasyUnder Review

$\sin θ (cosec θ - \sin θ) = \frac{1}{\sec^{2} θ}$

Hint

Expand, then use sin θ · cosec θ = 1.

Solution

Expand the bracket.

Distribute $\sin θ$ and use the reciprocal relation $\sin θ cosec θ = 1$ .
$\begin{matrix} L.H.S. & = \sin θ cosec θ - \sin^{2} θ \\ = 1 - \sin^{2} θ \end{matrix}$
Use the Pythagorean identity.

$1 - \sin^{2} θ = \cos^{2} θ$ , and $\cos θ = \frac{1}{\sec θ}$ .
$\begin{matrix} 1 - \sin^{2} θ & = \cos^{2} θ \\ = \frac{1}{\sec^{2} θ} \\ = R.H.S. \end{matrix}$

Answer

$\sin θ (cosec θ - \sin θ) = \frac{1}{\sec^{2} θ} (proved)$

Quick Solution

Q3Part (iii)

MediumUnder Review

$\frac{cosec θ - \sec θ}{cosec θ + \sec θ} = \frac{1 - \tan θ}{1 + \tan θ}$

Hint

Write cosec and sec as reciprocals, then multiply top and bottom by sin θ.

Solution

Write in sines and cosines.

Use $cosec θ = \frac{1}{\sin θ}$ and $\sec θ = \frac{1}{\cos θ}$ .

$L.H.S. = \frac{\frac{1}{\sin θ} - \frac{1}{\cos θ}}{\frac{1}{\sin θ} + \frac{1}{\cos θ}}$
Multiply numerator and denominator by $\sin θ$ .

This clears the first fraction in each, and $\frac{\sin θ}{\cos θ} = \tan θ$ .
$\begin{matrix} L.H.S. & = \frac{1 - \frac{\sin θ}{\cos θ}}{1 + \frac{\sin θ}{\cos θ}} \\ = \frac{1 - \tan θ}{1 + \tan θ} \\ = R.H.S. \end{matrix}$

Answer

$\frac{cosec θ - \sec θ}{cosec θ + \sec θ} = \frac{1 - \tan θ}{1 + \tan θ} (proved)$

Quick Solution

Q3Part (iv)

EasyUnder Review

$\tan θ + \cot θ = \frac{1}{\sin θ \cos θ}$

Hint

Change tan and cot to sin/cos and add the fractions.

Solution

Write in sines and cosines.

Use $\tan θ = \frac{\sin θ}{\cos θ}$ and $\cot θ = \frac{\cos θ}{\sin θ}$ .

$L.H.S. = \frac{\sin θ}{\cos θ} + \frac{\cos θ}{\sin θ}$
Add over a common denominator.

The common denominator is $\sin θ \cos θ$ ; then use $\sin^{2} θ + \cos^{2} θ = 1$ .
$\begin{matrix} L.H.S. & = \frac{\sin^{2} θ + \cos^{2} θ}{\sin θ \cos θ} \\ = \frac{1}{\sin θ \cos θ} \\ = R.H.S. \end{matrix}$

Answer

$\tan θ + \cot θ = \frac{1}{\sin θ \cos θ} (proved)$

Quick Solution

Q3Part (v)

HardUnder Review

$\frac{\cos θ + \sin θ}{\cos θ - \sin θ} + \frac{\cos θ - \sin θ}{\cos θ + \sin θ} = \frac{2}{1 - 2 \sin^{2} θ}$

Hint

Add the fractions; use (a+b)² + (a−b)² = 2(a² + b²) and (a−b)(a+b) = a² − b².

Solution

Add over a common denominator.

The common denominator is $(\cos θ - \sin θ) (\cos θ + \sin θ) = \cos^{2} θ - \sin^{2} θ$ .

$L.H.S. = \frac{(\cos θ + \sin θ)^{2} + (\cos θ - \sin θ)^{2}}{\cos^{2} θ - \sin^{2} θ}$
Expand the numerator.

The cross terms $+ 2 \sin θ \cos θ$ and $- 2 \sin θ \cos θ$ cancel, leaving $2 (\cos^{2} θ + \sin^{2} θ) = 2$ .
$\begin{matrix} (\cos θ + \sin θ)^{2} + (\cos θ - \sin θ)^{2} & = 2 \cos^{2} θ + 2 \sin^{2} θ \\ = 2 \end{matrix}$
Rewrite the denominator.

Replace $\cos^{2} θ$ by $1 - \sin^{2} θ$ .
$\begin{matrix} L.H.S. & = \frac{2}{(1 - \sin^{2} θ) - \sin^{2} θ} \\ = \frac{2}{1 - 2 \sin^{2} θ} \\ = R.H.S. \end{matrix}$

Answer

$\frac{\cos θ + \sin θ}{\cos θ - \sin θ} + \frac{\cos θ - \sin θ}{\cos θ + \sin θ} = \frac{2}{1 - 2 \sin^{2} θ} (proved)$

Quick Solution

Q3Part (vi)

MediumUnder Review

$\frac{1 + \cos θ}{1 - \cos θ} = (cosec θ + \cot θ)^{2}$

Hint

Start from the right side and write it as one fraction over sin²θ.

Solution

Expand the R.H.S..

Here the R.H.S. is easier to work with. Write $cosec θ = \frac{1}{\sin θ}$ and $\cot θ = \frac{\cos θ}{\sin θ}$ .
$\begin{matrix} R.H.S. & = {(\frac{1}{\sin θ} + \frac{\cos θ}{\sin θ})}^{2} \\ = \frac{(1 + \cos θ)^{2}}{\sin^{2} θ} \end{matrix}$
Use the Pythagorean identity.

Replace $\sin^{2} θ$ by $1 - \cos^{2} θ = (1 - \cos θ) (1 + \cos θ)$ , then cancel $1 + \cos θ$ .
$\begin{matrix} \frac{(1 + \cos θ)^{2}}{\sin^{2} θ} & = \frac{(1 + \cos θ)^{2}}{(1 - \cos θ) (1 + \cos θ)} \\ = \frac{1 + \cos θ}{1 - \cos θ} \\ = L.H.S. \end{matrix}$

Answer

$\frac{1 + \cos θ}{1 - \cos θ} = (cosec θ + \cot θ)^{2} (proved)$

Quick Solution

Question 4

MediumUnder Review

If $\tan θ = \frac{3}{\sqrt{2}}$ , then find the remaining trigonometric ratios when $θ$ lies in the first quadrant.

Hint

Take opposite = 3 and adjacent = √2, find the hypotenuse by Pythagoras, then read off each ratio.

Solution

Draw the reference triangle.

Take a right triangle $A B C$ with $∠ B = 90^{\circ}$ and $∠ A = θ$ . Since $\tan θ = \frac{opposite}{adjacent}$ , let the opposite side be 3 and the adjacent side be $\sqrt{2}$ .
$\begin{matrix} \tan θ & = \frac{opp}{adj} \\ = \frac{3}{\sqrt{2}} \end{matrix}$
Find the hypotenuse.

By Pythagoras' theorem, ${hyp}^{2} = {opp}^{2} + {adj}^{2}$ .
$\begin{matrix} {hyp}^{2} & = 3^{2} + (\sqrt{2})^{2} \\ = 9 + 2 \\ = 11 \\ ⟹ hyp & = \sqrt{11} \end{matrix}$
Write the remaining ratios.

$θ$ is in the first quadrant, so every ratio is positive.
$\begin{matrix} \sin θ & = \frac{3}{\sqrt{11}} \\ \cos θ & = \frac{\sqrt{2}}{\sqrt{11}} \\ \cot θ & = \frac{\sqrt{2}}{3} \\ \sec θ & = \frac{\sqrt{11}}{\sqrt{2}} \\ cosec θ & = \frac{\sqrt{11}}{3} \end{matrix}$

Answer

$\sin θ = \frac{3}{\sqrt{11}}, \cos θ = \frac{\sqrt{2}}{\sqrt{11}}, \cot θ = \frac{\sqrt{2}}{3}, \sec θ = \frac{\sqrt{11}}{\sqrt{2}}, cosec θ = \frac{\sqrt{11}}{3}$

Quick Solution

Question 5

MediumUnder Review

From a point on the ground, the angle of elevation to the top of a 30-meter-high building is $28^{\circ}$ . How far is the point from the base of the building?

Hint

Height is opposite the angle and the required distance is adjacent: use tan 28°.

Solution

Set up the triangle.

Let $B$ be the base and $C$ the top of the building, so $B C = 30$ m stands vertically, and let $A$ be the point on the ground with $A B = x$ . The angle of elevation at $A$ is $28^{\circ}$ and $∠ B = 90^{\circ}$ .
Choose the ratio.

The height is opposite the $28^{\circ}$ angle and $x$ is adjacent, so use tangent.

$\tan 28^{\circ} = \frac{30}{x}$
Solve for $x$ .

Rearrange and use $\tan 28^{\circ} \approx 0.5317$ .
$\begin{matrix} x & = \frac{30}{\tan 28^{\circ}} \\ = \frac{30}{0.5317} \\ \approx 56.42 m \end{matrix}$

Answer

$x \approx 56.42 m$

Quick Solution

Question 6

EasyUnder Review

A ladder leaning against a wall forms an angle of $65^{\circ}$ with the ground. If the ladder is 10 meters long, how high does it reach on the wall?

Hint

The ladder is the hypotenuse and the height is opposite 65°: use sine.

Solution

Set up the triangle.

Let the foot of the ladder be $A$ , the foot of the wall $B$ and the top of the ladder $C$ . Then $A C = 10$ m is the hypotenuse, $∠ A = 65^{\circ}$ , $∠ B = 90^{\circ}$ and the height $B C = h$ is opposite the $65^{\circ}$ angle.
Choose the ratio.

Opposite and hypotenuse are linked by sine.

$\sin 65^{\circ} = \frac{h}{10}$
Solve for $h$ .

Use $\sin 65^{\circ} \approx 0.9063$ .
$\begin{matrix} h & = 10 \sin 65^{\circ} \\ = 10 \times 0.9063 \\ \approx 9.06 m \end{matrix}$

Answer

$h \approx 9.06 m$

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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