Question

cos(A+B) =0;cosA=√3/2 then find value of A and B​

Answer 1

$\cos(90) = 0 \\ \cos(60) = \sqrt{3 \div 2 } \\ now \: accordingly \: you \: can \: solve \: this$

Answer 2

Given,

• $\sf cos(A+B)=0$

• $\sf cos\:A= \dfrac{\sqrt 3}{2}$

If,

$\sf cos\:A= \dfrac{\sqrt 3}{2}$

$\implies\sf A=cos^{-1}(\dfrac{\sqrt 3}{2})$

$\implies\sf A=30^o$

We know,

$\sf cos(A+B)=0 \:\:when\:\: A+B=90\:\:because\:\:cos\:90^o=0$

So,

⇒ A + B = 90

⇒ 30 + B = 90

⇒ B = 90 - 30

⇒ B = 60°

Therefore,

$\huge\boxed{\bf A=30^o\:\:\:and\:\:\:\:B=60^o}$