Question

If sin(A+B)= 1|√2 and cos(A-B)= 1|√2,then angle B=?

Answer 1

Answer:

B = 45°

Step-by-step explanation:

sin (A + B) = 1/2 and cos (A - B) = 1/2

sin (A + B) = sin 30° and cos (A - B) = cos 60°

A + B = 30° and A - B = 60°

A + B = 30°

A - B = 60°

____________

2A = 90°

A = 90°/2 = 45°

45° + B = 90°

B =90° - 45°

B = 45°

Answer 2

Answer: A=45° and B=0°

Step-by-step explanation:

• Given, sin(A+B)=1/√2

                ⇒cos(90°-A-B)=1/√2

• cos(A-B)=1/√2

• According to question,

       cos(90°-A-B)=cos(A-B)

     ⇒ 90°-A-B=A-B

     ⇒2A=90°

     ∴ A=45°

• Again,sin(A+B)=1/√2

               ⇒ sin(A+B)=sin 45°

               ⇒ A+B=45°

               ⇒ 45°+B=45°

                ∴ B=0

• Hence,A=45° and B=0°.