Answer:
Refer the digram & explanation.
Step-by-step explanation:
<u>GIVEN :-</u>
- Tangents PQ & PR drawn to a circle with center O from an external point P.
<u>TO PROVE :-</u>
<u>CONSTRUCTION :-</u>
<u>FACTS TO KNOW BEFORE SOLVING :-</u>
- (Angle at which tangents are inclined to each other from the external point) + (Angle the tangents subtend at the center of the circle) = 180°.
<u>PROCEDURE :-</u>
∠QPR + ∠QOR = 180°
⇒ ∠QOR = 180° - ∠QPR
In ΔQOR , OQ = OR (∵ Radii of the circle are equal)
⇒ ΔQOR is an isosceles triangle (∵ OQ = OR i.e. two sides of the triangle are equal.)
⇒ ∠OQR = ∠ORQ (∵ In an isosceles triangle , the base angles are equal)
So,
∠QOR + ∠OQR + ∠ORQ = 180°
⇒ ∠QOR + 2∠OQR = 180° (∵ ∠OQR = ∠ORQ)
⇒ 180° - ∠QPR = 180° - 2∠OQR (∵ ∠QPR + ∠QOR = 180°)
⇒ -∠QPR = -2∠OQR (∵ Cancelling 180° from both the sides.)
⇒ ∠QPR = 2∠OQR (Proved)