Question

ABC is an isosceles triangle and AC, BC are the tangents at M and N respectively. DE is the diameter of the circle. $\angle$ADP = $\angle$BEQ = ${100}^{\circ }$. What is value of $\angle$PRD?

• A. ${60}^{\circ }$
• B. ${50}^{\circ }$
• C. ${20}^{\circ }$
• D. Can't be determined

Answer 1

The correct option is C

${20}^{\circ }$

ADB is a straight line. By linear pair axiom,
$\angle$ADP + $\angle$PDB = ${180}^{\circ }$
100 + $\angle$PDB = ${180}^{\circ }$
$\angle$PDB = ${80}^{\circ }$
Similarly $\angle$QED = ${80}^{\circ }$

We have, ∠DPE = ${90}^{\circ }$ (angle subtended by a diameter)

In $△DPE$,
$\angle$DPE + $\angle$PED + $\angle$EDP = 180
[Angle sum property of a triangle]
∠PED = ${10}^{\circ }$
Similarly $\angle$QDE = ${10}^{\circ }$

In $△DRE$,
$\angle$DRE + $\angle$RDE + $\angle$RED = 180
[Angle sum property of a triangle]
∠DRE = ${160}^{\circ }$

PRE is a straight line. By linear pair axiom,
$\angle$PRD + $\angle$DRE = ${180}^{\circ }$
$\angle$PRD + 160 = ${180}^{\circ }$
$\angle$PRD = ${20}^{\circ }$