Question

sin A + cos B= 1, A = 30° and B is an acute angle, then find the value of B. ​

Answer 1

Answer: The value of B is 60°.

Step-by-step explanation:

• Required trigonmetric identities:

1. $cos(2\theta)=cos^2\theta-sin^2\theta$
2. $sin^2\theta+cos^2\theta=1$

• We have,                 $sinA=1-cosB \\\implies sinA=(sin^2(B/2)+cos^2(B/2))-(cos^2(B/2)-sin^2(B/2))\\\implies sinA=2sin^2(B/2)\\\implies 1/2=2sin^2(B/2) \\\implies sin^2(B/2)=1/4 \\\implies sin(B/2)= \pm 1/2$
• But,B is given to be acute,so is B/2 .Thus sin(B/2) is non-negative,that is,sin(B/2)=1/2.
• Therefore,B/2=30°,that is,B=60°

Answer 2

Answer: B=60°

Step-by-step explanation: Sin A +Cos B =1

Sin 30° +Cos B=1

1/2+CosB=1

Cos B=1/2

B=60°

Therefore, B is an acute angle and equal to 60°(60°<90°)