Question

sin (A+B) = 1; cos(A-B) = 3​

Answer 1

Correct Question :-

If sin(A + B) = 1 and cos(A - B) = √3/2 , find the value of A and B.

Answer :-

• A = 60

• B = 30

Solution :-

$\longrightarrow \sin(a + b) = 1 \\ \\ \longrightarrow \sin(a + b) = sin \: 90 \: (as \: sin \: 90 = 1) \\ \\ \longrightarrow (a + b) = 90 \\ \\ \longrightarrow b = 90 - a - - - - (i)$

$\longrightarrow \cos(a - b) = \frac{ \sqrt{3} }{2} \\ \\ \longrightarrow \cos(a - b) = cos \: 30 \: \: (as \: \: cos \: 30 = \frac{ \sqrt{3} }{2} ) \\ \\\longrightarrow a - b= 30$

Now putting the value of b from equation (i) →

$\longrightarrow a - b = 30 \\ \\ \longrightarrow a - (90 - a) = 30 \\ \\ \longrightarrow a - 90 + a = 30 \\ \\ \longrightarrow 2a - 90 = 30 \\ \\\longrightarrow 2a = 90 + 30 \\ \\\longrightarrow 2a = 120 \\ \\ \longrightarrow a = \frac{120}{2} \\ \\ \longrightarrow a = 60$

Now, put the value of a into equation (i) →

$\longrightarrow b = 90 - a \\ \\\longrightarrow b = 90 - 60 \\ \\ \longrightarrow b = 30$

Hence, The value of A and B is 60 and 30.