Question

If $sin(A-B)=\frac{1}{2}$ and $cos(A+B)=\frac{1}{2}$, 0° < A+B ≤ 90°, A>B, Find A and B.​

Answer 1

EXPLANATION.

⇒ sin(A - B) = 1/2.

⇒ cos(A + B) = 1/2.

⇒ 0° < A + B ≤ 90°.   &   A > B.

As we know that,

We can write equation as,

⇒ sin(A - B) = sin30°.  - - - - - (1).

⇒ cos(A + B) = cos60°. - - - - - (2).

We can write equation as,

⇒ A - B = 30°.

⇒ A + B = 60°.

We get,

⇒ 2A = 90°.

⇒ A = 45°.

Put the value of A = 45° in equation (2), we get.

⇒ A + B = 60°.

⇒ 45 + B = 60°.

⇒ B = 60° - 45°.

⇒ B = 15°.

Value of A = 45° & B = 15°.

                                                                                                                             

MORE INFORMATION.

Fundamental trigonometric identities.

(1) = sin²θ + cos²θ = 1.

(2) = 1 + tan²θ = sec²θ.

(3) = 1 + cot²θ = cosec²θ.