Question

if sin(A-B) =1/2,cos(A+B)=1/2,
${0}^{0} < a + b \leqslant {90}^{0}$
, A>B, then find 'A'
​

Answer 1

Answer

45°

$\rule{200}2$

Explanation

Given, that

sin(A - B) = 1/2 and cos(A + B) = 1/2

Given that, 0° < A + B ≤ 90°

This means that both A and B are acute angles.

We know the trigonometric ratios for acute angles.

We know that, sin30° = 1/2 and cos60° = 1/2

Hence, sin(A - B) = 1/2

→ Sin(A - B) = sin(30°)

Comparing the two angles, as both are acute,

→ A - B = 30° .........(1)

And, cos(A + B) = 1/2

→ cos(A + B) = cos(60°)

→ A + B = 60° ........(2)

Adding (1) and (2), we get

2A = 90°

→ A = 45°

Putting the value of A in any of the equation (1) or (2) gives B = 15°

Hence, angle A = 45° and angle B = 15°

Answer 2

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Given,

sin(A - B) = ½ & cos(A + B) = ½,

0°< (A + B) ≤ 90°

This means, both A and B are acute angles.

We know the Trigonometric ratios for acute angles.

We know that,

sin30° = ½ & cos 60° = ½

∵ sin(A - B) = ½

➡️ sin(A - B) = sin30°

Comparing the two angles, as both are acute,

➡️ A - B = 30° —————(i)

And, cos(A + B) = 60°

➡️ A + B = 60° ———–—(ii)

Adding equation--(i) and (ii), we get,

2A = 90°

➡️ A = 45°

Putting the value of A in any of the equation-(i) or (ii), we get, B = 15°

∴ A = 45° & B = 15°