Question

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

• A. ${100}^o$
• B. ${180}^o$
• C. ${80}^o$
• D. None of these

Answer 1

Answer: (C)

${80}^o$

Tangent is perpendicular to radius at the point of contact.

Therefore $\angle OQP=\angle ORP={90}^{\circ }$

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

So in quadrilateral QORP

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

${90}^{\circ }+\angle O+{90}^{\circ }+\frac{4}{5}\angle O={360}^{\circ }$ [Given $\angle P=\frac{4}{5}\angle O$]

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

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

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

$\angle QPR=\angle P=\frac{4}{5}{100}^{\circ }$

Therefore $\angle QPR={80}^{\circ }$

So, option c is the answer.