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

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

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.)

Step by Step

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

$Option (d) : 0.4636$

Step by Step

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

$Option (a) : 6.5 units$

Step by Step

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

$Option (a) : 50 m$

Step by Step

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

$Option (b) : 1$

Step by Step

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

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

Step by Step

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

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

Step by Step

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

$Option (d) : 1.625 π$

Step by Step

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

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

Step by Step

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

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

Step by Step

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

$Option (a) : 1$

Step by Step

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.)

Step by Step

Q2Part (a)(i)

Easy

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

Quick Solution

Under Review

\begin{matrix} 255^{\circ} & = 255 \times \frac{π}{180} \\ = \frac{17 π}{12} rad \end{matrix}

Step by Step

Q2Part (a)(ii)

Medium

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

Quick Solution

Under Review

\begin{matrix} 75^{\circ} 45^{'} & = {\frac{303}{4}}^{\circ} \\ = \frac{303}{4} \times \frac{π}{180} \\ = \frac{101 π}{240} rad \end{matrix}

Step by Step

Q2Part (a)(iii)

Easy

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

Quick Solution

Under Review

\begin{matrix} {142.5}^{\circ} & = \frac{285}{2} \times \frac{π}{180} \\ = \frac{19 π}{24} rad \end{matrix}

Step by Step

Q2Part (b)(i)

Easy

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

Quick Solution

Under Review

\begin{matrix} \frac{17 π}{24} & = \frac{17 \times 180^{\circ}}{24} \\ = {127.5}^{\circ} \\ = 127^{\circ} 30^{'} \end{matrix}

Step by Step

Q2Part (b)(ii)

Easy

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

Quick Solution

Under Review

\begin{matrix} \frac{7 π}{12} & = \frac{7 \times 180^{\circ}}{12} \\ = 105^{\circ} \end{matrix}

Step by Step

Q2Part (b)(iii)

Easy

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

Quick Solution

Under Review

\begin{matrix} \frac{11 π}{16} & = \frac{11 \times 180^{\circ}}{16} \\ = {123.75}^{\circ} \\ = 123^{\circ} 45^{'} \end{matrix}

Step by Step

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.)

Step by Step

Q3Part (i)

Medium

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

Quick Solution

Under Review

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

Step by Step

Q3Part (ii)

Easy

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

Quick Solution

Under Review

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

Step by Step

Q3Part (iii)

Medium

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

Quick Solution

Under Review

\begin{matrix} L.H.S. & = \frac{\frac{1}{\sin θ} - \frac{1}{\cos θ}}{\frac{1}{\sin θ} + \frac{1}{\cos θ}} \times \frac{\sin θ}{\sin θ} \\ = \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}

Step by Step

Q3Part (iv)

Easy

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

Quick Solution

Under Review

\begin{matrix} L.H.S. & = \frac{\sin θ}{\cos θ} + \frac{\cos θ}{\sin θ} \\ = \frac{\sin^{2} θ + \cos^{2} θ}{\sin θ \cos θ} \\ = \frac{1}{\sin θ \cos θ} \\ = R.H.S. \end{matrix}

Step by Step

Q3Part (v)

Hard

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

Quick Solution

Under Review

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

Step by Step

Q3Part (vi)

Medium

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

Quick Solution

Under Review

\begin{matrix} R.H.S. & = {(\frac{1 + \cos θ}{\sin θ})}^{2} \\ = \frac{(1 + \cos θ)^{2}}{1 - \cos^{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}

Step by Step

Question 4

Medium

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

Quick Solution

Under Review

\begin{matrix} hyp & = \sqrt{3^{2} + (\sqrt{2})^{2}} \\ = \sqrt{11} \\ \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}

Step by Step

Question 5

Medium

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?

Quick Solution

Under Review

\begin{matrix} \tan 28^{\circ} & = \frac{30}{x} \\ x & = \frac{30}{\tan 28^{\circ}} \\ = \frac{30}{0.5317} \\ \approx 56.42 m \end{matrix}

Step by Step

Question 6

Easy

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?

Quick Solution

Under Review

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

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