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Question
Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other an angle of 60º.
Solution 1
In order to draw the pair of tangents, we follow the following steps.
Step I: Take a point O on the plane of the paper and draw a circle of radius OA = 5 cm.
Step II: Produce OA to B such that OA = AB = 5 cm.
Step III: Taking A as the centre draw a circle of radius AO = AB = 5 cm.
Suppose it cuts the circle drawn in step I at P and Q.
Step IV: Join BP and BQ to get the desired tangents.
Justification: In OAP, we have
OA = OP = 5 cm (= Radius) Also,
AP = 5 cm (= Radius of circle with centre A)
∴ ∆OAP is equilateral.
⇒ ∠PAO = 60º ⇒ ∠BAP = 120º
In ∆BAP, we have
BA = AP and ∠BAP = 120º
∴ ∠ABP = ∠APB = 30º ⇒ ∠PBQ = 60º
Solution 2
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 + ∠APB = 360°
90° + 120° + 90° + ∠APB = 360°
∠APB = 60°
This justifies the construction.
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