The tangents can be constructed in the following manner:
Step 1
Draw a circle of radius 5 cm and with centre as O.
Step 2
Take a point A on the circumference of the circle and join OA. Draw a perpendicular to OA at point A.
Step 3
Draw a radius OB, making an angle of 120° (180° – 60°) with OA.
Step 4
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°.
Justification
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 + ∠ABP = 360° ⇒ 90° + 120° + 90° + ∠APB = 360°
∠APB = 60°
This justifies the construction.