Question

In the figure, from an external point P, two tangents PT and PS are drawn to a circle with centre O and radius $r$. If OP $=2r$, show that $\angle OTS=\angle OST={{30}^{\circ }}$.

Answer 1

Hint: Use R.H.S congruence criteria to prove that triangle OPT and triangle OPS are congruent. Then find the angle TOP by using the cosine of the angle. Then, in the isosceles triangle SOT, find angle OTS and angle OST.

Complete step by step answer:

R.H.S is a congruence criteria in which R, H and S stands for right angle, hypotenuse and side respectively. It is used to prove two triangles congruent when they have a right angle, their hypotenuses are of equal length and one of the remaining two sides are also equal.
Now, let us come to the question.

In triangle OPT and OPS:
OT = OS (radius of the circle)
OP = OP (common side in both triangles)
$\angle OTP=\angle OSP={{90}^{\circ }}$ (Angle subtended between the radius and tangent line at the point of contact)
Therefore, triangle OPT and OPS are congruent by R.H.S congruence criteria.
Hence, $\angle POT=\angle POS...................(i)$
Now, in triangle POT, assume that $\angle POT=\theta$. We know that, $\cos \theta =\dfrac{\text{base}}{\text{Hypotenuse}}$.
$\begin{align}
  & \therefore \cos \theta =\dfrac{OT}{OP}=\dfrac{r}{2r}=\dfrac{1}{2} \\
 & \therefore \theta ={{60}^{\circ }} \\
\end{align}$
By using relation (i), we get,
$\angle POT=\angle POS={{60}^{\circ }}$
Now, in triangle TOS:
OT = OS = radius. Therefore, $\angle OTS=\angle OST$. Now, let us assume, $\angle OTS=\angle OST$$=x$.
$\begin{align}
  & \angle TOS+\angle OST+\angle OTS={{180}^{\circ }} \\
 & {{120}^{\circ }}+x+x={{180}^{\circ }} \\
 & 2x={{60}^{\circ }} \\
 & x={{30}^{\circ }} \\
\end{align}$
Hence, $\angle OTS=\angle OST$$={{30}^{\circ }}$.
Note: One may note that this is the congruence method to solve this question. We can also solve this question by using the theorem “length of tangents drawn from a given point are equal”. By using this theorem we can directly find out angle TOS and then substitute its value in the formula of sum of internal angles of a triangle is ${{180}^{\circ }}$.