Question

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

Answer 1

Answer:

Given information:

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

Step-by-step explanation:

Since,

$sin(A+B)=\frac{\sqrt{3}}{2}\\(A+B)=sin^{-1}(\frac{\sqrt{3}}{2}\\A+B=60^\circ...(1)$

And,

$cos(A-B)=\frac{\sqrt{3}}{2}\\(A-B)=cos^{-1}(\frac{\sqrt{3}}{2})\\A-B=30^\circ...(2)$

Adding equation (1) and (2)

$(A+B)+(A-B)=60^\circ-30^\circ\\2A=90^\circ\\A=\frac{90^\circ}{2}\\=45^\circ$

Therefore, the value of A is 45°. Hence, option B is the correct choice.