Important Question For Class 10 Math | Chapter 8 | Q15

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.

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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Explanatory Answer | Important Trigonometry Question 15

sin (A + B - C) = \\frac{\text{1}}{\text{2}})
cot (A - B + C) = 0
cos (B + C – A) =\\frac{\text{1}}{\text{2}})

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)

Solving Equations Using Elimination Method

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°

 

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