Question

find measure of the angles A and B, if cos(A-B)= root3/2 and sin(A+B)=root3/2

Answer 1

hope it will help you?

Answer 2

Given:

$cos(A-B)=\frac{\sqrt{3}}{2}$

$sin(A+B)=\frac{\sqrt{3}}{2}$

To find:

Angle A and B

Solution:

We know that

$sin60^{\circ}=\frac{\sqrt{3}}{2}$

$cos30^{\circ}=\frac{\sqrt{3}}{2}$

Using the values

$cos(A-B)=cos30^{\circ}$

$A-B=30$...(1)

$sin(A+B)=sin60^{\circ}$

$A+B=60$...(2)

Adding equation (1) and (2) we get

$2A=90$

$A=45^{\circ}$

Substitute the value of A in equation (1)

$45-B=30$

$B=45-30=15^{\circ}$

Hence, the measure of angles

$A=45^{\circ}$ and $B=15^{\circ}$