Question

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

Answer 1

Solution:-

Given

$\rm\cos(A + B) = 0$

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

To find

$\rm \: value \: of \: B$

Using trigonometry identity

$\rm \: \cos(A + B) = \cos A. \cos B - \sin A . \sin B$

if

$\rm \cos A = \dfrac{ \sqrt{3} }{2} = \dfrac{b}{h}$

using pythagoras theorem

$\rm {h}^{2} = {p}^{2} + {b}^{2}$

$\rm \: 4 = {p}^{2} + 3$

$\rm \: p = 1$

So

$\rm\sin A = \dfrac{1}{2} = \dfrac{p}{h}$

put the value

$\rm \: \cos(A + B) = \cos A. \cos B - \sin A . \sin B$

$\rm \dfrac{ \sqrt{3} }{2} \times \cos B - \dfrac{1}{2} \times \sin B = 0$

$\dfrac{ \sqrt{3} \cos B }{2} = \dfrac{ \sin B}{2}$

$\dfrac{ \sqrt{3} \cos B }{ \not2} = \dfrac{ \sin B}{ \not2}$

$\rm{ \sqrt{3} \cos B }= \sin B$

$\rm \dfrac{ \sin B}{ \cos B } = \sqrt{3}$

$\rm\tan B = \sqrt{3}$

$\rm \: B = \tan {}^{ - 1} \sqrt{3}$

$\rm \: B = 60 \degree$