Question

Draw a pair of tangents to a circle of radius 6 cm which are inclined to each other at an angle of 60°. Also justify the construction

Answer 1

Construction:
1) Draw a circle of radius 6 cm with centre O.
2) Take a point A on the circle. Join OA.
3) Draw a perpendicular to OA at A.
4) Draw a radius OB, making an angle of 120° (180° – 60°) with OA.
5) Draw a perpendicular to OB at point B. Let these perpendiculars intersect at P.
6) PA and PB are the required tangents inclined at angle of 60°

Answer 2

We know that radius of the circle is perpendicular to the tangents (∠OPN = ∠ORN = 90°)
Sum of all the 4 angles of quadrilateral = 360°
Angle between the radius (∠POR)  = 360° – (90° + 90° + 60°) = 120°

GIVEN: A circle of radius 5 centimetre and a pair of tangents inclined to each other at an angle of 60°.

STEPS OF CONSTRUCTION:
1.Draw a circle with OS centre and radius 6 centimetre.

2. Draw any diameter POQ of the circle.

3.Draw the radius over means the circle at are such that ∠QOR = 60°.

4.Draw PD ⟂ PQ & RE ⟂ OR.
Let PD &RE  intersect each other at point N. Then NP & NR are the required tangents to the given circle inclined to each other at an angle of 60°.

JUSTIFICATION:
By construction, ∠OPN = 90° and OP is radius  PN is tangent to the circle.
Similarly, NR is a tangent to the circle.

Now, ∠POR = 180° - 60° = 120°

[POQ is a straight line and ∠QOR= 60°]

In quadrilateral OPNR,
∠OPN = 90°, ∠POR = 120° & ∠ORN = 90°.
∠PNR= 360° - ( 90° +120° +90°)

∠PNR = 360° 300° = 60°

HOPE THIS WILL HELP YOU ..