Given,
•Radius of the circle is 4 cm.
•Angle between the tangents is 60°.
In order to draw the pair of tangents, we follow the following steps.
Steps of construction:
1. Take a point O on the plane of the paper and draw a circle of radius OA=4 cm.
2. Extend the line segment OA to B such that OA=AB=4 cm.
3. Taking A as the centre draw a circle of radius AO=AB=4 cm. This circle intersects the first circle drawn in step 1 at P and Q.
4. Join BP and BQ to get desired tangents.
Justification:
In ∆OAP, we have
OA=OP=4 cm
(Radius)
Also, AP=4 cm
(Radius of circle with centre A)
∴ ∆OAP is equilateral
So, ∠PAO=60°
Now,
∠BAP + ∠PAO = 180° (linear angle)
∠BAP + 60° = 180°
∠ BAP = 60°
In ∆BAP, we have
BA=AP = 4 cm (radii of the circle with centre A)
∠BAP=120°
As two sides BA and AP are equal rABP is isosceles.
So, ∠ABP=∠APB
Let ∠ABP=∠APB = α
As the sum of angles is a triangle is 180°
∠ABP + ∠APB + ∠BAP = 180°
α + α + 120° = 180°
2α = 60°
α = 30°
Therefore ∠ABP=∠APB=30°
Hence ∠PBQ=60°