Question

If the tangents RA and RB from a point R to a circle with centre O are inclined to each other at an angle $\theta $ and $\angle AOB = {40^{\text{o}}}$, then find the value of $\theta $.

Answer 1

Hint: We need to use the properties of tangents in this question. The following properties will be used-
1. The radius of the circle from centre to the point of tangency is perpendicular to the tangent.
2. The quadrilateral formed by the tangents and the two radii has the sum of opposite angles 180.
3. The two tangents are equally inclined in opposite sides with respect to the line joining the external point and the centre of the circle.

Complete step by step answer:

In the given circle and tangents, using property 1, we can write that-
$\angle ARB = {{\theta }},\;\angle RBO = {90^{\text{o}}},\;\angle AOB = {40^{\text{o}}}\left( {given} \right),\;\angle OAB = {90^{\text{o}}}$

In the quadrilateral AOBR, we know that the sum of the angles is $360^{o}$, because the sum of the angles in a quadrilateral is always $360^{o}$. So we can write that-
$\angle ARB + \angle RBO + \angle AOB + \angle OAB = {360^{\text{o}}}$
${{\theta }} + 90^{o} + 40^{o} + 90^{o} = 360^{o}$
${{\theta }} + 220^{o} = 360^{o}$
${{\theta }} = {140^{\text{o}}}$
This is the required answer.

Note: Instead of using the angle sum property of a quadrilateral, we can also use property 2, that is, the sum of the opposite angles is $180^{o}$. So, the value of $\theta $ can be directly written as-
In quadrilateral OABR-
$\begin{align}
  &Sum\;of\;opposite\;angles = {180^{\text{o}}} \\
  &\angle ARB + \angle AOB = {180^{\text{o}}} \\
  &\angle ARB + {40^{\text{o}}} = {180^{\text{o}}} \\
  &\angle ARB = {140^{\text{o}}} \\
\end{align} $