Draw a pair of tangents to a circle or radius 3 cm which are inclined to each other at an angle of 60 degree.
The tangents can be constructed in the following manner:
Draw a circle of radius 5 cm and with centre as O.
Take a point A on the circumference of the circle and join OA. Draw a perpendicular to OA at point A.
Draw a radius OB, making an angle of 120° (180° − 60°) with OA.
Draw a perpendicular to OB at point B. Let both the perpendiculars intersect at point P. PA and PB are the required tangents at an angle of 60°.
The construction can be justified by proving that ∠APB = 60°
By our construction
∠OAP = 90°
∠OBP = 90°
And ∠AOB = 120°
We know that the sum of all interior angles of a quadrilateral = 360°
∠OAP + ∠AOB + ∠OBP + ∠APB = 360°
90° + 120° + 90° + ∠APB = 360°
∠APB = 60°
This justifies the construction.