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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)

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

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∘, m∠A=θ and m∠C=ϕ:

Multiple Choice

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.)

Step by Step

Q1Part (a)(i)

Easy

In right-angled triangle ABC, m∠B=90∘, mBC=4, mAB=3, mAC=5, m∠A=θ and m∠C=ϕ. Find sin⁡θ.

Quick Solution

Under Review

sin⁡θ=BCAC=45
Step by Step

Q1Part (a)(ii)

Easy

In right-angled triangle ABC, m∠B=90∘, mBC=4, mAB=3, mAC=5, m∠A=θ and m∠C=ϕ. Find cos⁡θ.

Quick Solution

Under Review

cos⁡θ=ABAC=35
Step by Step

Q1Part (a)(iii)

Easy

In right-angled triangle ABC, m∠B=90∘, mBC=4, mAB=3, mAC=5, m∠A=θ and m∠C=ϕ. Find tan⁡θ.

Quick Solution

Under Review

tan⁡θ=BCAB=43
Step by Step

Q1Part (a)(iv)

Easy

In right-angled triangle ABC, m∠B=90∘, mBC=4, mAB=3, mAC=5, m∠A=θ and m∠C=ϕ. Find sec⁡θ.

Quick Solution

Under Review

sec⁡θ=ACAB=53
Step by Step

Q1Part (a)(v)

Easy

In right-angled triangle ABC, m∠B=90∘, mBC=4, mAB=3, mAC=5, m∠A=θ and m∠C=ϕ. Find cosec⁡θ.

Quick Solution

Under Review

cosec⁡θ=ACBC=54
Step by Step

Q1Part (a)(vi)

Easy

In right-angled triangle ABC, m∠B=90∘, mBC=4, mAB=3, mAC=5, m∠A=θ and m∠C=ϕ. Find cot⁡ϕ.

Quick Solution

Under Review

cot⁡ϕ=BCAB=43
Step by Step

Q1Part (a)(vii)

Easy

In right-angled triangle ABC, m∠B=90∘, mBC=4, mAB=3, mAC=5, m∠A=θ and m∠C=ϕ. Find tan⁡ϕ.

Quick Solution

Under Review

tan⁡ϕ=ABBC=34
Step by Step

Q1Part (a)(viii)

Easy

In right-angled triangle ABC, m∠B=90∘, mBC=4, mAB=3, mAC=5, m∠A=θ and m∠C=ϕ. Find cosec⁡ϕ.

Quick Solution

Under Review

cosec⁡ϕ=ACAB=53
Step by Step

Q1Part (a)(ix)

Easy

In right-angled triangle ABC, m∠B=90∘, mBC=4, mAB=3, mAC=5, m∠A=θ and m∠C=ϕ. Find sec⁡ϕ.

Quick Solution

Under Review

sec⁡ϕ=ACBC=54
Step by Step

Q1Part (a)(x)

Easy

In right-angled triangle ABC, m∠B=90∘, mBC=4, mAB=3, mAC=5, m∠A=θ and m∠C=ϕ. Find cos⁡ϕ.

Quick Solution

Under Review

cos⁡ϕ=BCAC=45
Step by Step

Q1Part (b)(i)

Easy

In right-angled triangle ABC, m∠B=90∘, mBC=8, mAB=15, mAC=17, m∠A=θ and m∠C=ϕ. Find sin⁡θ.

Quick Solution

Under Review

sin⁡θ=BCAC=817
Step by Step

Q1Part (b)(ii)

Easy

In right-angled triangle ABC, m∠B=90∘, mBC=8, mAB=15, mAC=17, m∠A=θ and m∠C=ϕ. Find cos⁡θ.

Quick Solution

Under Review

cos⁡θ=ABAC=1517
Step by Step

Q1Part (b)(iii)

Easy

In right-angled triangle ABC, m∠B=90∘, mBC=8, mAB=15, mAC=17, m∠A=θ and m∠C=ϕ. Find tan⁡θ.

Quick Solution

Under Review

tan⁡θ=BCAB=815
Step by Step

Q1Part (b)(iv)

Easy

In right-angled triangle ABC, m∠B=90∘, mBC=8, mAB=15, mAC=17, m∠A=θ and m∠C=ϕ. Find sec⁡θ.

Quick Solution

Under Review

sec⁡θ=ACAB=1715
Step by Step

Q1Part (b)(v)

Easy

In right-angled triangle ABC, m∠B=90∘, mBC=8, mAB=15, mAC=17, m∠A=θ and m∠C=ϕ. Find cosec⁡θ.

Quick Solution

Under Review

cosec⁡θ=ACBC=178
Step by Step

Q1Part (b)(vi)

Easy

In right-angled triangle ABC, m∠B=90∘, mBC=8, mAB=15, mAC=17, m∠A=θ and m∠C=ϕ. Find cot⁡ϕ.

Quick Solution

Under Review

cot⁡ϕ=BCAB=815
Step by Step

Q1Part (b)(vii)

Easy

In right-angled triangle ABC, m∠B=90∘, mBC=8, mAB=15, mAC=17, m∠A=θ and m∠C=ϕ. Find tan⁡ϕ.

Quick Solution

Under Review

tan⁡ϕ=ABBC=158
Step by Step

Q1Part (b)(viii)

Easy

In right-angled triangle ABC, m∠B=90∘, mBC=8, mAB=15, mAC=17, m∠A=θ and m∠C=ϕ. Find cosec⁡ϕ.

Quick Solution

Under Review

cosec⁡ϕ=ACAB=1715
Step by Step

Q1Part (b)(ix)

Easy

In right-angled triangle ABC, m∠B=90∘, mBC=8, mAB=15, mAC=17, m∠A=θ and m∠C=ϕ. Find sec⁡ϕ.

Quick Solution

Under Review

sec⁡ϕ=ACBC=178
Step by Step

Q1Part (b)(x)

Easy

In right-angled triangle ABC, m∠B=90∘, mBC=8, mAB=15, mAC=17, m∠A=θ and m∠C=ϕ. Find cos⁡ϕ.

Quick Solution

Under Review

cos⁡ϕ=BCAC=817
Step by Step

Q1Part (c)(i)

Easy

In right-angled triangle ABC, m∠B=90∘, mBC=5, mAB=12, mAC=13, m∠A=θ and m∠C=ϕ. Find sin⁡θ.

Quick Solution

Under Review

sin⁡θ=BCAC=513
Step by Step

Q1Part (c)(ii)

Easy

In right-angled triangle ABC, m∠B=90∘, mBC=5, mAB=12, mAC=13, m∠A=θ and m∠C=ϕ. Find cos⁡θ.

Quick Solution

Under Review

cos⁡θ=ABAC=1213
Step by Step

Q1Part (c)(iii)

Easy

In right-angled triangle ABC, m∠B=90∘, mBC=5, mAB=12, mAC=13, m∠A=θ and m∠C=ϕ. Find tan⁡θ.

Quick Solution

Under Review

tan⁡θ=BCAB=512
Step by Step

Q1Part (c)(iv)

Easy

In right-angled triangle ABC, m∠B=90∘, mBC=5, mAB=12, mAC=13, m∠A=θ and m∠C=ϕ. Find sec⁡θ.

Quick Solution

Under Review

sec⁡θ=ACAB=1312
Step by Step

Q1Part (c)(v)

Easy

In right-angled triangle ABC, m∠B=90∘, mBC=5, mAB=12, mAC=13, m∠A=θ and m∠C=ϕ. Find cosec⁡θ.

Quick Solution

Under Review

cosec⁡θ=ACBC=135
Step by Step

Q1Part (c)(vi)

Easy

In right-angled triangle ABC, m∠B=90∘, mBC=5, mAB=12, mAC=13, m∠A=θ and m∠C=ϕ. Find cot⁡ϕ.

Quick Solution

Under Review

cot⁡ϕ=BCAB=512
Step by Step

Q1Part (c)(vii)

Easy

In right-angled triangle ABC, m∠B=90∘, mBC=5, mAB=12, mAC=13, m∠A=θ and m∠C=ϕ. Find tan⁡ϕ.

Quick Solution

Under Review

tan⁡ϕ=ABBC=125
Step by Step

Q1Part (c)(viii)

Easy

In right-angled triangle ABC, m∠B=90∘, mBC=5, mAB=12, mAC=13, m∠A=θ and m∠C=ϕ. Find cosec⁡ϕ.

Quick Solution

Under Review

cosec⁡ϕ=ACAB=1312
Step by Step

Q1Part (c)(ix)

Easy

In right-angled triangle ABC, m∠B=90∘, mBC=5, mAB=12, mAC=13, m∠A=θ and m∠C=ϕ. Find sec⁡ϕ.

Quick Solution

Under Review

sec⁡ϕ=ACBC=135
Step by Step

Q1Part (c)(x)

Easy

In right-angled triangle ABC, m∠B=90∘, mBC=5, mAB=12, mAC=13, m∠A=θ and m∠C=ϕ. Find cos⁡ϕ.

Quick Solution

Under Review

cos⁡ϕ=BCAC=513
Step by Step

Question 2

EasyUnder Review

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

In right-angled triangle ABC, m∠B=90∘, mBC=a, mAB=c, mAC=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.)

Step by Step

Q2Part (i)

Easy

In right-angled triangle ABC, m∠B=90∘, mBC=a, mAB=c, mAC=b, m∠A=ϕ and m∠C=θ. Find sin⁡θ.

Quick Solution

Under Review

sin⁡θ=ABAC=cb
Step by Step

Q2Part (ii)

Easy

In right-angled triangle ABC, m∠B=90∘, mBC=a, mAB=c, mAC=b, m∠A=ϕ and m∠C=θ. Find cos⁡θ.

Quick Solution

Under Review

cos⁡θ=BCAC=ab
Step by Step

Q2Part (iii)

Easy

In right-angled triangle ABC, m∠B=90∘, mBC=a, mAB=c, mAC=b, m∠A=ϕ and m∠C=θ. Find tan⁡θ.

Quick Solution

Under Review

tan⁡θ=ABBC=ca
Step by Step

Q2Part (iv)

Easy

In right-angled triangle ABC, m∠B=90∘, mBC=a, mAB=c, mAC=b, m∠A=ϕ and m∠C=θ. Find sin⁡ϕ.

Quick Solution

Under Review

sin⁡ϕ=BCAC=ab
Step by Step

Q2Part (v)

Easy

In right-angled triangle ABC, m∠B=90∘, mBC=a, mAB=c, mAC=b, m∠A=ϕ and m∠C=θ. Find cos⁡ϕ.

Quick Solution

Under Review

cos⁡ϕ=ABAC=cb
Step by Step

Q2Part (vi)

Easy

In right-angled triangle ABC, m∠B=90∘, mBC=a, mAB=c, mAC=b, m∠A=ϕ and m∠C=θ. Find tan⁡ϕ.

Quick Solution

Under Review

tan⁡ϕ=BCAB=ac
Step by Step

Question 3

EasyUnder Review

Considering the adjoining triangle, verify that:

In right-angled triangle ABC, m∠B=90∘, mAB=4, mBC=3, mAC=5 and m∠A=θ.

(i) sin⁡θcosec⁡θ=1 (ii) cos⁡θsec⁡θ=1 (iii) tan⁡θcot⁡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.)

Step by Step

Q3Part (i)

Easy

In right-angled triangle ABC, m∠B=90∘, mAB=4, mBC=3, mAC=5 and m∠A=θ. Verify that sin⁡θcosec⁡θ=1.

Quick Solution

Under Review

sin⁡θcosec⁡θ=35×53=1
Step by Step

Q3Part (ii)

Easy

In right-angled triangle ABC, m∠B=90∘, mAB=4, mBC=3, mAC=5 and m∠A=θ. Verify that cos⁡θsec⁡θ=1.

Quick Solution

Under Review

cos⁡θsec⁡θ=45×54=1
Step by Step

Q3Part (iii)

Easy

In right-angled triangle ABC, m∠B=90∘, mAB=4, mBC=3, mAC=5 and m∠A=θ. Verify that tan⁡θcot⁡θ=1.

Quick Solution

Under Review

tan⁡θcot⁡θ=34×43=1
Step by Step

Question 4

EasyUnder Review

Fill in the blanks.

(i) sin⁡30∘=sin⁡(90∘−60∘)= ______

(ii) cos⁡30∘=cos⁡(90∘−60∘)= ______

(iii) tan⁡30∘=tan⁡(90∘−60∘)= ______

(iv) tan⁡60∘=tan⁡(90∘−30∘)= ______

(v) sin⁡60∘=sin⁡(90∘−30∘)= ______

(vi) cos⁡60∘=cos⁡(90∘−30∘)= ______

(vii) sin⁡45∘=sin⁡(90∘−45∘)= ______

(viii) tan⁡45∘=tan⁡(90∘−45∘)= ______

(ix) cos⁡45∘=cos⁡(90∘−45∘)= ______

Hint

Use the complementary-angle rule: a ratio of 90∘−α equals the co-ratio of α.

Solution

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

Step by Step

Q4Part (i)

Easy

Fill in the blank:

sin⁡30∘=sin⁡(90∘−60∘)= ______

Quick Solution

Under Review

sin⁡30∘=sin⁡(90∘−60∘)=cos⁡60∘
Step by Step

Q4Part (ii)

Easy

Fill in the blank:

cos⁡30∘=cos⁡(90∘−60∘)= ______

Quick Solution

Under Review

cos⁡30∘=cos⁡(90∘−60∘)=sin⁡60∘
Step by Step

Q4Part (iii)

Easy

Fill in the blank:

tan⁡30∘=tan⁡(90∘−60∘)= ______

Quick Solution

Under Review

tan⁡30∘=tan⁡(90∘−60∘)=cot⁡60∘
Step by Step

Q4Part (iv)

Easy

Fill in the blank:

tan⁡60∘=tan⁡(90∘−30∘)= ______

Quick Solution

Under Review

tan⁡60∘=tan⁡(90∘−30∘)=cot⁡30∘
Step by Step

Q4Part (v)

Easy

Fill in the blank:

sin⁡60∘=sin⁡(90∘−30∘)= ______

Quick Solution

Under Review

sin⁡60∘=sin⁡(90∘−30∘)=cos⁡30∘
Step by Step

Q4Part (vi)

Easy

Fill in the blank:

cos⁡60∘=cos⁡(90∘−30∘)= ______

Quick Solution

Under Review

cos⁡60∘=cos⁡(90∘−30∘)=sin⁡30∘
Step by Step

Q4Part (vii)

Easy

Fill in the blank:

sin⁡45∘=sin⁡(90∘−45∘)= ______

Quick Solution

Under Review

sin⁡45∘=sin⁡(90∘−45∘)=cos⁡45∘
Step by Step

Q4Part (viii)

Easy

Fill in the blank:

tan⁡45∘=tan⁡(90∘−45∘)= ______

Quick Solution

Under Review

tan⁡45∘=tan⁡(90∘−45∘)=cot⁡45∘
Step by Step

Q4Part (ix)

Easy

Fill in the blank:

cos⁡45∘=cos⁡(90∘−45∘)= ______

Quick Solution

Under Review

cos⁡45∘=cos⁡(90∘−45∘)=sin⁡45∘
Step by Step

Question 5

MediumUnder Review

If in a right-angled triangle ABC, m∠B=90∘ and C is an acute angle of 60∘. Also sin⁡m∠A=ab, then find the following trigonometric ratios.

(Figure: mBC=a, mAB=c, mAC=b.)

(i) mBCmAB (ii) cos⁡60∘ (iii) tan⁡60∘ (iv) cosec⁡π3 (v) cot⁡60∘ (vi) sin⁡30∘ (vii) cos⁡30∘ (viii) tan⁡π6 (ix) sec⁡30∘ (x) cot⁡30∘

Hint

Find m∠A=30∘ 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.)

Step by Step

Q5Part (i)

Easy

In right-angled triangle ABC, m∠B=90∘ and C is an acute angle of 60∘. The sides are mBC=a, mAB=c, mAC=b, and sin⁡m∠A=ab. Find mBCmAB.

Quick Solution

Under Review

mBCmAB=ac
Step by Step

Q5Part (ii)

Easy

In right-angled triangle ABC, m∠B=90∘ and C is an acute angle of 60∘. The sides are mBC=a, mAB=c, mAC=b, and sin⁡m∠A=ab. Find cos⁡60∘.

Quick Solution

Under Review

cos⁡60∘=BCAC=ab
Step by Step

Q5Part (iii)

Easy

In right-angled triangle ABC, m∠B=90∘ and C is an acute angle of 60∘. The sides are mBC=a, mAB=c, mAC=b, and sin⁡m∠A=ab. Find tan⁡60∘.

Quick Solution

Under Review

tan⁡60∘=ABBC=ca
Step by Step

Q5Part (iv)

Easy

In right-angled triangle ABC, m∠B=90∘ and C is an acute angle of 60∘. The sides are mBC=a, mAB=c, mAC=b, and sin⁡m∠A=ab. Find cosec⁡π3.

Quick Solution

Under Review

cosec⁡π3=cosec⁡60∘=ACAB=bc
Step by Step

Q5Part (v)

Easy

In right-angled triangle ABC, m∠B=90∘ and C is an acute angle of 60∘. The sides are mBC=a, mAB=c, mAC=b, and sin⁡m∠A=ab. Find cot⁡60∘.

Quick Solution

Under Review

cot⁡60∘=BCAB=ac
Step by Step

Q5Part (vi)

Easy

In right-angled triangle ABC, m∠B=90∘ and C is an acute angle of 60∘. The sides are mBC=a, mAB=c, mAC=b, and sin⁡m∠A=ab. Find sin⁡30∘.

Quick Solution

Under Review

sin⁡30∘=BCAC=ab
Step by Step

Q5Part (vii)

Easy

In right-angled triangle ABC, m∠B=90∘ and C is an acute angle of 60∘. The sides are mBC=a, mAB=c, mAC=b, and sin⁡m∠A=ab. Find cos⁡30∘.

Quick Solution

Under Review

cos⁡30∘=ABAC=cb
Step by Step

Q5Part (viii)

Easy

In right-angled triangle ABC, m∠B=90∘ and C is an acute angle of 60∘. The sides are mBC=a, mAB=c, mAC=b, and sin⁡m∠A=ab. Find tan⁡π6.

Quick Solution

Under Review

tan⁡π6=tan⁡30∘=BCAB=ac
Step by Step

Q5Part (ix)

Easy

In right-angled triangle ABC, m∠B=90∘ and C is an acute angle of 60∘. The sides are mBC=a, mAB=c, mAC=b, and sin⁡m∠A=ab. Find sec⁡30∘.

Quick Solution

Under Review

sec⁡30∘=ACAB=bc
Step by Step

Q5Part (x)

Easy

In right-angled triangle ABC, m∠B=90∘ and C is an acute angle of 60∘. The sides are mBC=a, mAB=c, mAC=b, and sin⁡m∠A=ab. Find cot⁡30∘.

Quick Solution

Under Review

cot⁡30∘=ABBC=ca
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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