Question

Janta let's find the value of x​

Answer 1

From the fig. on considering $\sf{\triangle ABH,}$ we get,

$\displaystyle\longrightarrow\sf{BH=\sqrt{AB^2-AH^2}}$

$\displaystyle\longrightarrow\sf{BH=\sqrt{a^2-b^2}}$

And,

$\displaystyle\longrightarrow\sf{\cos72^o=\dfrac{b}{a}}$

$\displaystyle\longrightarrow\sf{\dfrac{\sqrt5-1}{4}=\dfrac{b}{a}}$

$\displaystyle\longrightarrow\sf{b=\dfrac{(\sqrt5-1)a}{4}\quad\quad\dots(1)}$

$\displaystyle\longrightarrow\sf{b^2=\dfrac{(6-2\sqrt5)a^2}{16}\quad\quad\dots(2)}$

Then on considering $\sf{\triangle BCH,}$ we get,

$\displaystyle\longrightarrow\sf{\sin x=\dfrac{\sqrt{a^2-b^2}}{a+b}}$

From (1) and (2),

$\displaystyle\longrightarrow\sf{\sin x=\dfrac{\sqrt{a^2-\dfrac{(6-2\sqrt5)a^2}{16}}}{a+\dfrac{(\sqrt5-1)a}{4}}}$

$\displaystyle\longrightarrow\sf{\sin x=\dfrac{\sqrt{\dfrac{16a^2-(6-2\sqrt5)a^2}{16}}}{\dfrac{4a+(\sqrt5-1)a}{4}}}$

$\displaystyle\longrightarrow\sf{\sin x=\dfrac{\dfrac{a}{4}\sqrt{16-6+2\sqrt5}}{\dfrac{a}{4}\left(4+\sqrt5-1\right)}}$

$\displaystyle\longrightarrow\sf{\sin x=\dfrac{\sqrt{10+2\sqrt5}}{3+\sqrt5}}$

$\displaystyle\Longrightarrow\sf{\underline{\underline{x\approx46.6^o}}}$