Question

In the given circle with centre $O$, $\angle ABC={100}^{\circ }$, $\angle ACD={40}^{\circ }$ and $CT$ is a tangent to the circle at $C$. Find $\angle ADC$ and $\angle DCT$.

Answer 1

Given:$\angle$ABC$={100}^o$
We know that, $\angle$ABC$+\angle$ADC$={180}^o$ (The sum of opposite angles in a cyclic quadrilateral $={180}^o$)
$\therefore {100}^o+\angle$ADC$={180}^o$
$\angle$ADC$={180}^o-{100}^o$
$\angle$ADC$={80}^o$
Join OA and OC, we have a isosceles $\Delta$ OAC,
$\because$OA$=$OC (Radii of a circle)
$\therefore$ $\angle$AOC$=2\times \angle$ADC (by theorem)
or $\angle$AOC$=2\times {80}^o={160}^o$
In $\Delta$AOC,
$\angle$AOC$+\angle$OAC$+\angle$OCA$={180}^o$
${160}^o+\angle$OCA$+\angle$OCA$={180}^o[\because \angle$OAC$=\angle$OCA]
$2\angle$OCA$={20}^o$
$\angle$OCA$={10}^o$
$\angle$OCA$+\angle$OCD$={40}^o$
${10}^o+\angle$OCD$={40}^o$
$\therefore$ $\angle$OCD$={30}^o$
Hence, $\angle$OCD$+\angle$DCT$=\angle$OCT
$\because \angle$OCT$={90}^o$
(The tangent at a point to circle is $\perp$ to the radius through the point to constant)
${30}^o+\angle$DCT$={90}^o$
$\therefore$ $\angle$DCT$={60}^o$.