Question

If sin (A-B) = 1/2 and cos (A+B) = 1/2 , 0°< (A+B) <90° and A > B then find A and B.




ANSWER FAST GUYS

Answer 1

We are familiar with the trigonometric identity given below.

sin θ = cos(90 - θ)

And we know the trigonometric value given below.

sin 30° = cos 60° = 1/2

Given,

sin (A - B) = 1/2     ⇒     A - B = 30°   →   (1)

cos (A + B) = 1/2   ⇒     A + B = 60°  →   (2)

Since  0° < (A + B) < 90°,  we can take  A + B = 60°.

From the above trigonometric identity, we get,

A + B = 90° - (A - B)     ⇒     (A + B) + (A - B) = 90°

Or by adding (1) and (2), we also get,

    (A - B) + (A + B) = 30° + 60° = 90°

⇒  A - B + A + B = 90°

⇒  2A = 90°

⇒  A = 45°

Okay, from (1),

A - B = 30°     ⇒     B = A - 30° = 45° - 30° = 15°

Or from (2),

A + B = 60°     ⇒     B = 60° - A = 60° - 45° = 15°

Hence,

A = 45°   ;   B = 15°

Answer 2

Trigonometry :

sin ( A - B ) = $\mathsf{\dfrac{1}{2}}$

cos ( A + B ) = $\mathsf{\dfrac{1}{2}}$

sin ( A - B ) = sin 30° [ sin 30° = 1/2 ]

( A - B ) = 30° --> ( i )

Also,

cos ( A + B ) = cos 60° [ cos 60° = 1/2 ]

( A + B ) = 60° --> ( ii )

Adding equation (i) and (ii),

2A = 60° + 30°

2A = 90°

A = $\mathsf{\dfrac{90}{2}}$

A = 45°

Putting value of A in equation ( i ),

( A - B ) = 30°

45° - B = 30°

B = 45° - 30°

B = 15°

Verification :

Adding (A + B),

45° + 15° = 60°

1 ). That is, 0° < {( A + B ) = 60°} < 90°. This is true.

2 ). Also, A > B

45° > 15° which is true.