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Question
Draw a circle of radius 3 cm. Draw a pair of tangents to this circle, which are inclined to each other at an angle of 60º.
Sum
Solution
Steps of construction
Step I: Draw a circle with O as centre and radius = 3 cm.
Step II: Draw any diameter AOB of this circle.
Step III: Construct ∠BOC = 60º such that radius OC meets the circle at C.
Step IV: Draw AM ⊥ AB and CN ⊥OC.
Let AM and CN intersect each other at P.
Then, PA and PC are the desired tangents to the given circle, inclined at an angle of 60º
Proof: ∠AOC = (180º – 60º) = 120º
In quad. OAPC, we have
∠OAP = 90º, ∠AOC = 120º, ∠OCP = 90º.
∴∠APC = [360º – (90º + 120º + 90º)] = 60º
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