This class 10 Maths additional practice question is from the chapter Introduction to Trigonometry. Concept Covered: Computing values of angles from trigonometric identities by solving linear equation in 3 variables.

Question 15 : In an acute angle triangle ABC, sin (A + B - C)= \\frac{\text{1}}{\text{2}}), cot (A - B + C) = 0 and cos (B + C – A) =\\frac{\text{1}}{\text{2}}). What are the values of A, B, C?

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sin (A + B - C) = \\frac{\text{1}}{\text{2}})

cot (A - B + C) = 0

cos (B + C – A) =\\frac{\text{1}}{\text{2}})

We know sin 30° = \\frac{\text{1}}{\text{2}}); given sin (A + B - C) = \\frac{\text{1}}{\text{2}})

**∴ A + B - C = 30°** ......(1)

We know cot 90°= 0; given cot (A – B + C) = 0

**∴ A - B + C = 90°** ......(2)

We know cos 60° = \\frac{\text{1}}{\text{2}}); given cos (B + C – A) = \\frac{\text{1}}{\text{2}})

**∴ B + C – A = 60°** ......(3)

Add ① and ②:

(A + B - C) + (A – B + C) = 30° + 90°

2A = 120°

**Or A = 60°**

Substitute A = 60° in ① and ③

① → 60° + B – C = 30° → B – C = - 30° ......(4)

③ → B + C – 60° = 60° → B + C = 120° ......(5)

Add equations **4** and **5**

(B - C) + (B + C ) = -30° + 120°

2B = 90°

∴ **B = 45°**

Substitute B = 45° in ④

B – C = - 30°

45° – C = - 30° → **C = 75°**

The values are **A = 60°, B = 45°and C = 75° **

Class 10 Maths

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CBSE Class 10 Maths - 2021

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