Let $p$ and $q$ be the length of two chords of a circle which subtend angles ${36}^{\mathrm{o}}$ and ${60}^{\mathrm{o}}$ respectively at the centre of the circle. Then, the angle (in radian) subtended by the chord of length $p+q$ at the centre of the circle is

(use $\mathrm{\pi }=3.1$)

Solution

If $l$ is the length of the chord, then $l=2r\mathrm{s}\mathrm{i}\mathrm{n}\frac{\theta }{2}$ .
$p=2r\mathrm{s}\mathrm{i}\mathrm{n }{18}^o, q=2r\sin {30}^{\mathrm{o}}$
$\Rightarrow p+q=2r\left(\sin {18}^{\mathrm{o}}+\sin {30}^{\mathrm{o}}\right)$
$=2r\left(\frac{\sqrt{5}-1}{4}+\frac{1}{2}\right)=2r\left(\frac{\sqrt{5}+1}{4}\right)$
$=2r\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{}{36}^{\mathrm{o}}=2r\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{}{54}^{\mathrm{o}}=2r\mathrm{s}\mathrm{i}\mathrm{n}\left(\frac{{108}^{\mathrm{o}}}{2}\right)$
$\Rightarrow p+q$ subtends an angle ${108}^{\mathrm{o}}$ at the centre
${108}^{\mathrm{o}}=1.86$ radian

Hence, the correct answer is 1.86.