Question

In the given figure, O is the centre of circle, ∠ABO = 20° and ∠ACO = 30°. If ∠BOC = x°, find the value of x.

Answer 1

In $△$AOB,
AO = BO (radius of the circle)
So, $\angle$OBA = $\angle$BAO = 20° (Angles opposite equal sides are equal)
In $△$AOC,
AO = CO (Radii of the circle)
So, $\angle$OCA = $\angle$CAO = 30° (Angles opposite equal sides are equal)
$\angle$BAC = $\angle$CAO + $\angle$BAO
= 30°+ 20°
= 50°

$$\begin{gathered} \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{angle}\mathrm{subtended}\mathrm{by}\mathrm{an}\mathrm{arc}\mathrm{of}\mathrm{a}\mathrm{circle}\mathrm{at}\mathrm{the}\mathrm{centre}\mathrm{is} \\ \mathrm{double}\mathrm{the}\mathrm{angle}\mathrm{subtended}\mathrm{by}\mathrm{it}\mathrm{at}\mathrm{any}\mathrm{point}\text{in}\mathrm{the}\mathrm{remaining}\mathrm{part}\mathrm{of}\mathrm{the}\mathrm{circle}. \\ \mathrm{So},\angle BOC=2\angle BAC \\ \Rightarrow \angle BOC=2\times 50°=100° \end{gathered}$$

$\angle$BOC = 100°
$\Rightarrow$x° = 100°
$\Rightarrow$x = 100