Question

Find the value of
$h=a+x$ where

$x=\dfrac{2a\tan\alpha}{\tan\beta-\tan\alpha}$​

Answer 1

Here's we are given that,

$\sf :\implies \red { h=a+x}$

$\sf :\implies \red {x=\dfrac{2a\tan\alpha}{\tan\beta-\tan\alpha}}$

⠀━━━━━━━━━━━━━━━━━━━━━⠀⠀

$\sf :\implies \purple { h=a+x}$

$\sf :\implies h=a+\sf{ \dfrac{2a\tan\alpha}{\tan\beta-\tan\alpha}}$

$\sf :\implies h=\dfrac{a\tan\alpha+a\tan\beta}{\tan\beta-\tan\alpha}$

$\sf :\implies h=\dfrac{\left(\dfrac{a\sin\alpha}{\cos\alpha}+\dfrac{a\sin\beta}{\cos\beta}\right)}{\left(\dfrac{\sin\beta}{\cos\beta}-\dfrac{\sin\alpha}{\cos\alpha}\right)}$

$\sf :\implies h=\dfrac{\left(\dfrac{a\sin\alpha\cos\beta+a\cos\alpha\sin\beta}{\cos\alpha\cos\beta}\right)}{\left(\dfrac{\sin\beta\cos\alpha-\cos\beta\sin\alpha}{\cos\alpha\cos\beta}\right)}$

$\sf :\implies h=\dfrac{a(\sin\alpha\cos\beta+\cos\alpha\sin\beta)}{\sin\beta\cos\alpha-\cos\beta\sin\alpha}$

$\sf{ :\implies \boxed{\underline\purple{{h=\dfrac{a\sin(\alpha+\beta)}{\sin(\beta-\alpha)}}}}}$