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 $\Delta MNO$ 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 $d=r\left(\sqrt{3}\right)$ and mark E.
• D. None of the above

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 $\Delta OME$ 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 $\Delta MNO$ is NOT equilateral.
Next, since in $\Delta OME,sin{30}^{\circ }$ =$\frac{1}{2}=\frac{OM}{OE}=\frac{r}{d}$ we have d = 2r and hence option (C) is also ruled out.

Recalling that $\angle MOE={60}^{\circ }$, following are the steps of construction:
1. Draw a ray OX from the centre. This ray contains the point (E) from which tangents are drawn.
2. With O as centre, construct $\angle MOX={60}^{\circ }$ [constructing angle ${60}^{\circ }$ is easy]
3. 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.