Question

In the given figure, O is the centre of the circle with PA and PB as tangents. If measure of angle ADB=60°, then show that ∆PAB is an equilateral triangle.​

Answer 1

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.

Learn more :-

In ABC, AD is angle bisector,

angle BAC = 111 and AB+BD=AC find the value of angle ACB=?

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