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°
• B. 50°
• C. 20°
• D. Can't be determined

Answer 1

The correct option is C

20°

ADB is a straight line

$\angle$ADP + $\angle$PDB = ${180}^{\circ }$

100 + $\angle$PDB = ${180}^{\circ }$

$\angle$PDB = ${80}^{\circ }$

Similarly $\angle$QED = ${80}^{\circ }$

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

∠PED = ${10}^{\circ }$ ($\angle$DPE + $\angle$PED + $\angle$EDP = ${180}^{\circ }$)

Similarly $\angle$QDE = ${10}^{\circ }$

∠DRE = ${160}^{\circ }$ ($\angle$DRE + $\angle$RDE + $\angle$RED = ${180}^{\circ }$)

PRE is a straight line

$\angle$PRD + $\angle$DRE = ${180}^{\circ }$

$\angle$PRD + ${160}^{\circ }$ = ${180}^{\circ }$

$\angle$PRD = ${20}^{\circ }$