NCERT Mathematics · Class 10 · English · 2026-27

Introduction to Trigonometry — Exercise 8.1

Independently written, step-by-step solutions. Prompts below are concise paraphrases; use the official textbook for the complete question wording.

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Question 1

ABC is right-angled at B, with AB=24 cm and BC=7 cm. Find the sine and cosine of A and C.

Step-by-step solution

AC=√(24²+7²)=25. For angle A, opposite=BC and adjacent=AB; for C the roles interchange.

Answer: sinA=7/25, cosA=24/25; sinC=24/25, cosC=7/25.

Resource ID: CB-10-M-CH8-EX8.1-Q1

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Question 2

In the right triangle of Fig. 8.13, evaluate tanP−cotR.

Step-by-step solution

PQ=12, PR=13, so QR=√(13²−12²)=5. tanP=QR/PQ=5/12. cotR=QR/PQ=5/12, so subtracting gives zero.

Answer: 0.

Resource ID: CB-10-M-CH8-EX8.1-Q2

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Question 3

For acute A with sinA=3/4, calculate cosA and tanA.

Step-by-step solution

Choose opposite=3 and hypotenuse=4. The adjacent side is √(16−9)=√7. Divide adjacent by hypotenuse for cosine and opposite by adjacent for tangent.

Answer: cosA=√7/4; tanA=3/√7=3√7/7.

Resource ID: CB-10-M-CH8-EX8.1-Q3

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Question 4

Given 15cotA=8 for acute A, find sinA and secA.

Step-by-step solution

cotA=8/15, so take adjacent=8, opposite=15. Hypotenuse=√(64+225)=17.

Answer: sinA=15/17; secA=17/8.

Resource ID: CB-10-M-CH8-EX8.1-Q4

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Question 5

Given secθ=13/12 for acute θ, find the five remaining ratios.

Step-by-step solution

Take hypotenuse=13 and adjacent=12. Opposite=√(169−144)=5. Use the side definitions and reciprocals.

Answer: sinθ=5/13; cosθ=12/13; tanθ=5/12; cotθ=12/5; cosecθ=13/5.

Resource ID: CB-10-M-CH8-EX8.1-Q5

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Question 6

Show that acute angles A and B are equal if cosA=cosB.

Step-by-step solution

Construct right triangles representing A and B and scale them to equal hypotenuse. Equality of cosines gives equal adjacent legs. Their remaining legs are equal by Pythagoras. The right triangles are congruent by RHS, so the corresponding acute angles are equal.

Answer: A=B.

Resource ID: CB-10-M-CH8-EX8.1-Q6

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Question 7

Given cotθ=7/8, evaluate the product quotient in (i) and cot²θ in (ii).

Step-by-step solution

In (i), (1+sinθ)(1−sinθ)=cos²θ and (1+cosθ)(1−cosθ)=sin²θ. Their ratio is cot²θ. Squaring 7/8 gives the same result for both parts.

Answer: (i) 49/64; (ii) 49/64.

Resource ID: CB-10-M-CH8-EX8.1-Q7

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Question 8

If 3cotA=4, compare (1−tan²A)/(1+tan²A) with cos²A−sin²A.

Step-by-step solution

cotA=4/3; use adjacent=4, opposite=3, hypotenuse=5. LHS=(1−9/16)/(1+9/16)=7/25. RHS=(4/5)²−(3/5)²=7/25.

Answer: They are equal; both are 7/25.

Resource ID: CB-10-M-CH8-EX8.1-Q8

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Question 9

In right triangle ABC with tanA=1/√3, evaluate the two expressions involving A and C.

Step-by-step solution

Since A is acute, A=30°; hence C=60°. (i) sinA cosC+cosA sinC=(1/2)(1/2)+(√3/2)(√3/2)=1. (ii) cosA cosC−sinA sinC=(√3/2)(1/2)−(1/2)(√3/2)=0.

Answer: (i) 1; (ii) 0.

Resource ID: CB-10-M-CH8-EX8.1-Q9

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Question 10

PQR is right-angled at Q; PR+QR=25 cm and PQ=5 cm. Find sinP,cosP,tanP.

Step-by-step solution

Let PR=H and QR=x. Pythagoras gives H²−x²=25. Since H+x=25, (H−x)25=25, so H−x=1. Solving gives H=13,x=12. Relative to P, opposite=12 and adjacent=5.

Answer: sinP=12/13; cosP=5/13; tanP=12/5.

Resource ID: CB-10-M-CH8-EX8.1-Q10

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Question 11

Assess the five statements about trigonometric notation and ranges.

Step-by-step solution

(i) False: tan60°=√3>1. (ii) True: secA=12/5 is possible for a triangle with adjacent=5 and hypotenuse=12. (iii) False: cos abbreviates cosine; cosec abbreviates cosecant. (iv) False: cotA means a function of A, not multiplication. (v) False: sine cannot exceed 1, while 4/3>1.

Answer: False, True, False, False, False.

Resource ID: CB-10-M-CH8-EX8.1-Q11

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