Question

You are given a circle with radius ‘r’ and centre O. You are asked to draw a pair of tangents which are inclined at an angle of ${60}^{\circ }$ with each other. Refer to the figure and select the option which would lead us to the required construction. d is the distance OE.

• A. Construct the $△$MNO as it is equilateral
• B. Mark M and N on the circle such that $\angle$MOE = ${60}^{\circ }$ and $\angle$NOE = ${60}^{\circ }$
• C. Using trigonometry, arrive at d = r$\sqrt{3}$ and mark E.
• D. Using trigonometry, arrive at d = r$\sqrt{5}$ and mark E.

Answer 1

The correct option is B

Mark M and N on the circle such that $\angle$MOE = ${60}^{\circ }$ and $\angle$NOE = ${60}^{\circ }$

Since the angle between the tangents is ${60}^{\circ }$ and OE bisects $\angle$MEN, $\angle$MEO = ${30}^{\circ }$ .

Now, since $△$OME is a right angled triangle, right angled at M, we realise that the $\angle$MOE = ${60}^{\circ }$ . Since $\angle$MOE = ${60}^{\circ }$ , we must have $\angle$NOE =${60}^{\circ }$ and hence $\angle$MON = ${120}^{\circ }$ . Hence $△$MNO is NOT equilateral.

Next, since in $△$OME, $\sin {30}^{\circ }$= $\frac{1}{2}$ = $\frac{OM}{OE}$ = $\frac{r}{d}$ , we have d = 2r.

Recalling that $\angle$MOE = ${60}^{\circ }$ , following are the steps of construction:

1. Draw a ray from the centre O.

2. With O as centre, construct $\angle$MOE = ${60}^{\circ }$ [constructing angle ${60}^{\circ }$ is easy]

3. Now extend OM and from M, draw a line perpendicular to OM. This intersects the ray at E. This is the point from where the tangents should be drawn, EM is one tangent.

4. Similarly, EN is another tangent.