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In the given figure, PQ is a tangent from an external point P to a circle with center O, and OP cuts the circle at T and QOR is a diameter. If $\angle POR={{130}^{\circ }}$ and S is a point on the circle, find $\angle 1+\angle 2$
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Last updated date: 15th Jun 2024
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Answer
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Hint: We firstly need to find the angles $\angle 1$ and $\angle 2$. As we know that: The angle subtended by an arc of a circle at its center is twice the angle it subtends anywhere on the circle's circumference. So, we can say that: $\angle ROT=2\angle RST$. So, we get a value of $\angle 2$. Now, we know that $\angle POR={{130}^{\circ }}$. We get the value of $\angle POQ$ by supplementary relation. Now, in $\Delta POQ$ , we have, $\angle PQO={{90}^{\circ }}$ . We have found the value of $\angle POQ$. So, by using the sum of all angles of triangles relation, get the value of $\angle 1$. Now, add the values of $\angle 1$ and $\angle 2$ to get $\angle 1+\angle 2$

Complete step-by-step solution
We know that: The angle subtended by an arc of a circle at its centre is twice of the angle it subtends anywhere on the circle's circumference. So, we get:
$\angle ROT=2\angle RST$
Also, $\angle ROT=\angle POR={{130}^{{}^\circ }}$
So, we get:
\[\begin{align}
  & \Rightarrow {{130}^{{}^\circ }}=2\angle RST \\
 & \Rightarrow \angle RST={{65}^{{}^\circ }}......(1) \\
\end{align}\]
Therefore, $\angle 2={{65}^{{}^\circ }}$
Now, we know that the sum of all the angles on a line is ${{180}^{\circ }}$ . So, we get:
$\begin{align}
  & \Rightarrow \angle ROT+\angle QOT={{180}^{{}^\circ }} \\
 & \Rightarrow {{130}^{{}^\circ }}+\angle QOT={{180}^{{}^\circ }} \\
 & \Rightarrow \angle QOT={{50}^{{}^\circ }}......(2) \\
\end{align}$
Now, in $\Delta POQ$ , we have,
$\angle PQO={{90}^{\circ }}$ (angle subtended by a tangent at a circle)
$\angle QOT={{50}^{{}^\circ }}$
So, we get:
$\begin{align}
  & \Rightarrow \angle QOT+\angle PQO+\angle OPQ={{180}^{{}^\circ }} \\
 & \Rightarrow {{50}^{{}^\circ }}+{{90}^{{}^\circ }}+\angle 1={{180}^{{}^\circ }} \\
 & \Rightarrow \angle 1={{180}^{{}^\circ }}-{{140}^{{}^\circ }} \\
 & \Rightarrow \angle 1={{40}^{{}^\circ }} \\
\end{align}$
Now, we need to find $\angle 1+\angle 2$
We have:
$\angle 1={{40}^{{}^\circ }}$
 $\angle 2={{65}^{{}^\circ }}$
So,
$\begin{align}
  & \angle 1+\angle 2={{40}^{{}^\circ }}+{{65}^{{}^\circ }} \\
 & ={{105}^{{}^\circ }}
\end{align}$

Note: Important theorems included in this question are:
1. The angle formed at the center of the circle by lines originating from two points on the circle's circumference is double the angle formed on the circumference of the circle by lines originating from the same points.
2. Tangent to a circle is always perpendicular to the radius.
3. the sum of angles of a triangle equals the straight angle, i.e. ${{180}^{\circ }}$
4. Angles on a straight line add up to ${{180}^{\circ }}$