Question

Draw a circle with centre O having radius 3 cm. Draw tangent segments PA and PB through the point P outside the circle such that ∠APB = 70°.

Answer 1

m∠A = 90°, m∠B = 90°   ...[Tangent theorem]

m∠A + m∠B + m∠APB + m∠AOB = 360°   ...[Angle sum property of quadrilateral]

∴ 90° + 90° + 70° + m∠AOB = 360°

∴ m∠AOB = 360° − 250°

∴ m∠AOB = 110°

Construction:

1. Construct a circle of radius 3 cm with centre O.
2. Join OA, OB, and increase it.
3. Now draw the perpendicular bisectors of these two lines.
4. Join A to P and B to P. They meet at point P, which is 70°.