Question

If sin (A+B ) = 1 , cos ( A+ B ) = √3/2 0° <A+ B<90° then find A and B

Answer 1

As sin(90) =1

We have A+B =90°

As cos (30) =√3/2

So we also have A+B=30,

So as both cannot happen at the same time, I think your question is


If sin (A+B ) = 1 , cos ( A-B ) = √3/2 0° <A+ B<90° then find A and B

Sin(A+B)=1

cos(A-B)=√3/2

We know siin90=1

A+B=90 _ _ _ _(1)

cos (A-B)=√3/2

A-B=cos-¹ (√3/2)


We know cos-¹ (√3/2)=30

A-B=30 _ _ _ _ _ (2)

Solve (1) & (2)

A+B=90----(1)

A-B=30 ----(2)

add (1) and (2)

2A=120

A=60°

Sub A in (1)

=>B =90 -A =90-60
=30°