Question

If sin (A+B)=√3/2 & sin(A-B)=1/√2 then find the values of A nd B​

Answer 1

Answer:

Step-by-step explanation:

Sin ( a+ b) = 60'

And sin(a-b)=90'

Because sin root3/2 = 60' and sin1= 90'

On solving further

A+b=60

A-b= 90

We obtain, b= 15 and a= 45

Hope this helps

Answer 2

Answer: The values of A and B are 45° and 15° respectively.

Step-by-step explanation:

Since we have given that

sin (A+B) = $\frac{\sqrt{3}}{2}$  and

sin (A-B) = $\frac{1}{\sqrt{2}}$

Since we know that

$\sin 60^\circ=\frac{\sqrt{3}}{2}$

Similarly,

$\sin 30^\circ=\frac{1}{\sqrt{2}}$

Now, we get that

$\sin (A+B)=\sin 60^\circ\\\\\implies A+B=60^\circ-----------(1)\\\\\sin(A-B)=\sin 30^\circ\\\\\implies A-B=30----------(2)$

So, by elimination method, we get that,

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

So, B=60°-45°= 15°

Hence, the values of A and B are 45° and 15° respectively.