Question

if sin (A - B) =√3/2,cos (A + B) = √3/2 , then find A and B values​

Answer 1

CORRECT QUESTION:

if sin (A + B) =√3/2, cos (A - B) = √3/2 , then find A and B values​

Answer:

A = 45° and B = 15°

Step-by-step explanation:

Given that:

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

$\implies \sf{sin(A+B) = sin \: 60^{\circ}}$

Taking sin inverse on both sides,

$\sf{(A+B)=60^{\circ}\qquad---[Eq^{n}\:1]}$

It is also given that:

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

$\implies \sf{cos(A-B) = cos\: 30^{\circ}}$

Taking cos inverse on both sides,

$\sf{(A-B)=30^{\circ}\qquad---[Eq^{n}\:2]}$

Adding Equation 1 and Equation 2;

        A + B = 60°

 {+}   A - B = 30°

          2A = 90°

⇒ A = 45°

Substitute the value of A in Equation 1 to find the value of B.

A + B = 60°

⇒ 45° + B = 60°

⇒ B = 15°

∴ A = 45° and B = 15°