Question

To draw a pair of tangents to a circle which are inclined to each other at an angle of ${60}^0$, it is required to draw tangents at endpoints of those two radii of the circle, the angle between them should be

• A. ${135}^0$
• B. ${90}^0$
• C. ${60}^0$
• D. ${120}^0$

Answer 1

The correct option is D ${120}^0$

Given:

$PA$ & $PB$ are two tangents drawn from P to a circle with centre O at A & B respectively.

$\angle APB={60}^o$

To find out-

$\angle AOB$

Solution-

$PA$ & $PB$ are two tangents drawn from P to the circle at A & B, respectively.

$\therefore PA=PB⟹\Delta PAB$ is isosceles.

i.e $\angle PAB=\angle PBA.$

or $\angle PAB+\angle PBA=2\angle PAB$ .....(i)

Now, $\angle PAB+\angle PBA+\angle APB={180}^o$ ....(angle sum property of triangles)

$⟹\angle PAB+\angle PBA{+60}^o={180}^O$

$⟹2\angle PAB={120}^o$ ...(from i)

$\therefore \angle PAB={60}^o=\angle PBA$ .........(ii)

Again, $\angle OAP={90}^o$ ....(angle between a radius and the tangent at the point of contact.)

$\therefore \angle OAB=\angle OAP-\angle PAB={90}^o-{60}^o$ .... (from ii) .........(iii)

Since, $OA=OB$ ...(radii of the same circle)

$\therefore \Delta OAB$ is isosceles

$⟹\angle OAB=\angle OBA={30}^o$ ....(from iii)

So, $\angle AOB={180}^o-(\angle OAB+\angle OBA)={180}^o-{30}^o-{30}^O={120}^O$ ...(angle sum property of triangles)