Question

$PQ$ and $PR$ are the tangents to a circle with centre $O$. If $\angle P=\frac{4}{5}\angle O.$ Find $\angle QOR$.

• A. ${90}^o$
• B. ${80}^o$
• C. ${110}^o$
• D. ${100}^o$

Answer 1

Answer: (D)

${100}^o$

Given,

$PQ$ and $PR$ are tangents to the circle.

And $\angle P=\frac{4}{5}\angle O$

Tangents are perpendicular to radius at the point of contact.

Therefore, $\angle PQO=\angle ORP={90}^{\circ }$

Sum of all the four angles of a quadrilateral is ${360}^{\circ }$

So in quadrilateral $QORP$,

$\angle PQO+\angle QOR+\angle ORP+\angle RPQ={360}^{\circ }$

$\Rightarrow {90}^{\circ }+\angle O+{90}^{\circ }+\angle P={360}^{\circ }$

$\Rightarrow {180}^{\circ }+\angle O+\frac{4}{5}\angle O={360}^{\circ }$

$\Rightarrow \frac{9}{5}\angle O={360}^{\circ }-{180}^{\circ }$

$\Rightarrow \frac{9}{5}\angle O={180}^{\circ }$

$\Rightarrow \angle O=\frac{5}{9}\times {180}^{\circ }$

$\Rightarrow \angle O={100}^{\circ }$

$\Rightarrow \angle QOR=\angle O={100}^{\circ }$

Hence, $Op-D$ is correct.