Question

In the given figure PQ is a diameter of a circle with centre O and PT is a tangent at P. QT meets the circle at R. If ∠POR = ${72}^{\circ }$, find $\angle$PTR.

• A. 44°
• B. 54°
• C. 64°
• D. 34°

Answer 1

The correct option is B

54°

$\angle$OQR = $\angle$ORQ (OR=OQ (radius))

But $\angle$ROP = $\angle$OQR + $\angle$ORQ (Exterior angle theorem)

${72}^{\circ }$ = 2 $\angle$OQR

$\angle$OQR = ${36}^{\circ }$

In $△$QPT

$\angle$PTQ = ${180}^{\circ }$ - $\angle$PQT - $\angle$QPT

$\angle$PTQ = ${180}^{\circ }$ – ${36}^{\circ }$ - ${90}^{\circ }$

$\angle$PTQ = ${54}^{\circ }$

Therefore, $\angle$PTR = ${54}^{\circ }$