Question

If sin alpha = 1/√10 and sin beeta = 1/√5 , then prove that alpha + beeta = 45°​

Answer 1

Step-by-step explanation:

Given,

sin A = 1/√10 and sin B = 1/√5

We know that sin2x + cos2x = 1

Thus, cos A = √(1 – sin2A) = 3/√10

cos B = √(1 – sin2B) = 2/√5

As we know,

sin(A + B) = sin A cos B + cos A sin B

= (1/√10) (2/√5) + (3/√10) (1/√5)

= (2 + 3)/ √50

= 5/(5√2)

= 1/√2

sin(A + B) = sin π/4 {since A and B are acute angles as per the given}

Therefore, A + B = π/4