Question

Tan a minus b equals to 1 by root 3 and sin a equal to root 3 by 2 find cos B​

Answer 1

Given :

$\sf \tan(a - b) = \frac{1}{ \sqrt{3} } \\ \\ \sf \sin a = \frac{ \sqrt{3} }{2}$

Solution :

$\sf \sin a = \frac{ \sqrt{3} }{2} \\ \\ \sf \sin a = \sin 60° \\ \\ \boxed{\sf a = 60°} \\ \\ \sf \tan(a - b) = \frac{1}{ \sqrt{3} } \\ \\ \sf \tan(60 - b ) = \tan30 \\ \\ \sf 60 - b = 30 \\ \\ \sf - b = 30 - 60 \\ \\ \boxed{ \sf b = 30°} \\ \\ \boxed{ \sf \green{ \cos b = \cos30 = \frac{\sqrt{3}}{2} }}$

Answer 2

$\huge\bold{\sf{Solution}}$

GIVEN-

$\tan(a-b)$ = $\dfrac{1}{\sqrt3}$

$\sin(a)$ = $\dfrac{\sqrt3}{2}$

TO FIND-

$\cos(b)$

Here we go,

→$\sin(a)$ = $\dfrac{\sqrt3}{2}$

$\sin(a)$ = $\sin60°$

$\sin(a)$ = $60°$

→$\tan(a-b)$ = $\dfrac{1}{\sqrt(3)}$

$\tan(a-b)$ = $\tan30°$

$\tan(60°-b)$ = $\tan30°$

$60°-b$ = $30°$

$b$ = $60°-30°$

$b$ = $30°$

→Now, $\cosb$ = $\cos30°$ = $\dfrac{\sqrt3}{2}$

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