Question

Let $AB$ and $BD$ are the tangent and secant to the circle respectively. $C$ is the point of intersection of $BD$ and circle. Also, measure of arc $AD$ and $AC$ are ${100}^o$ and ${30}^o$ respectively. Then, find the measure of $\angle ABD$.

• A. ${30}^o$
• B. ${115}^o$
• C. ${40}^o$
• D. ${45}^o$

Answer 1

The correct option is B ${115}^o$

Given, $\angle AOC={35}^o\angle AOD={100}^o\angle COD={100}^o-{30}^o={70}^{o}OA=OC=OD=radius$

such that

$\angle OAC=\angle OCA\angle OCD=\angle ODC$ (angle opposites to equal sides)

In $△OAC$

$\angle OAC+\angle OCA+\angle AOC={180}^{o}2\angle OCA+{30}^o={180}^o\angle OCA={75}^o$

In $△COD$

$\angle OCD+\angle ODC+\angle COD={180}^{o}2\angle OCD+{70}^o={180}^o\Rightarrow \angle OCD={55}^o$

We know, $\angle OAB={90}^o$

$\angle OAC+\angle CAB={90}^o\angle CAB={90}^o-{75}^o={15}^o$

Also, $\angle OCD+\angle OCA+\angle ACB={180}^o\angle ACB={50}^o$

In $△ABC$

$\angle ACB+\angle CAB+\angle ABC={180}^o$ (sum of angles)

${50}^o+{15}^o+\angle ABC={180}^o\angle ABC={115}^o$