In the given figure, O is the centre of the circle with PA and PB as tangents. If measure of ZADB = 60°, then show that APAB is an equilateral triangle.
Answers
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Given :-
- O is the centre of the circle with PA and PB as tangents .
- ∠ADB = 60° .
To show :-
- ∆PAB is an equilateral triangle.
Solution :-
we know that,
- The angle at the centre is double the angle at the circumference.
- Tangents is perpendicular to radius .
so,
→ ∠AOB = 2 * ∠ADB
→ ∠AOB = 2 * 60°
→ ∠AOB = 120° .
now,
→ ∠OAP = ∠OBP = 90° (Tangents.)
then, in quadrilateral APBO, by angle sum Property of angles ,
→ ∠AOB + ∠OAP + ∠OBP + ∠APB = 360°
→ 120° + 90° + 90° + ∠APB = 360°
→ 300° + ∠APB = 360°
→ ∠APB = 360° - 300°
→ ∠APB = 60° .
now, we also know that,
- The lengths of tangents drawn from an external point to a circle are equal.
- Angle Opposite to equal sides are equal .
therefore, in ∆PAB
→ PA = PB . (Tangents)
→ ∠PAB = ∠PBA .
hence,
→ ∠APB + ∠PAB + ∠PBA = 180° (Angle sum property)
→ 60° + 2∠PAB = 180°
→ 2∠PAB = 180° - 60°
→ 2∠PAB = 120°
→ ∠PAB = 60° = ∠PBA .
Since all three angles of ∆PAB are 60° . Hence, we can conclude that ∆PAB is an equilateral triangle.
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