Question

Sin (A+B)=^3/2,cos(A,B)=^3/2 0°<90° dind the angles A and B
​

Answer 1

Answer:

shsbvdgnrbrgrhrhebgr

ebbebehejnehejenevebe

ejhevegrgrbtbrhhththr

rhbrbrbrb

Answer 2

Answer:

$\huge{\boxed{\rm{\red{A=45°,B=15°}}}}$

Step-by-step explanation:

Given:-

Sin(A+B)=√3/2

Sin(A-B)=√3/2,0<A,B<90°

To find:-

Find the angles A and B

Solution:-

Sin(A+B)=√3/2

=>Sin(A+B)=Sin60°

A+B=60°--------(1)

Cos(A-B)=√3/2

Cos(A-B)=Cos 30°

A-B=30°--------(2)

From (1)&(2)

A+B=60°

A-B=30°

(+)________

2A+0=90°

__________

=>2A=90°

=>A=90°/2=45°

Therefore, A=45°

From(1)

45°+B=60°

=>B=60°-45°

=>B=15°

Answer:-

The values of A=45° and B=15°

Check:-

Sin(A+B)

=>Sin(45°+15°)

=>Sin60°

=>√3/2

Cos(A-B)

=>Cos(45°-15°)

=>Cos30°

=>√3/2

Verified the given relations