# Important Question For Class 10 Math | Q15

#### Chapter 8 | Trigonometry | Trigonometric Ratios & Linear Equations | Hard Question

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.

cos (B + C – A) =$$frac{\text{1}}{\text{2}}$ #### Step 1: Frame 3 Linear Equations in 3 Variables We know sin 30° = $\frac{\text{1}}{\text{2}}$; given sin$A + B - C) = $$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)

#### Step 2: Solve Equations Using Elimination Method

(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)

(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°

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