Question

if sin(A+B)=√3/2 and cos(A-B)=√3/2 where 0°<A+B_<90° and a>b find a and b​

Answer 1

Step-by-step explanation:

We know sin60°=√3/2

hence A+B=60                   ---i

and,

cos30°=√3/2

hence A-B=30                   ----ii

On adding i and ii;

we get,

(A+A)+(B-B)= 60+90

therefore,

2A=90

A=45°

From i

A+B=60

45+B=60

B=60-45

B=15°

HOPE IT HELPS!

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