Question

In the given figure, $SR\parallel QP$ and $\mathrm{\angle }RPQ={30}^{\circ }$. Find the value of $\mathrm{\angle }SQR$.

Answer 1

Hint: We use the theorem that tangents from the common point are equal and then calculate the values of $\mathrm{\angle }PRQ$ and $\mathrm{\angle }PQR$ by using angle sum property of a triangle. Then, we are given that $SR\parallel QP$, then apply the properties of parallel lines to find the angle $\mathrm{\angle }SRQ$. Next, we will find $\mathrm{\angle }RSQ$ by angles in the same segment theorem. At last, apply angle sum property to find the value of $\mathrm{\angle }SQR$

Complete step-by-step answer:
We are given the value of $\mathrm{\angle }RPQ={30}^{\circ }$ and $SR\parallel QP$
We have to find the value of $\mathrm{\angle }SQR$
Here, we can see that tangents $PQ$ and $PR$ are drawn from a common point $P$, then
$PQ=PR$
It is known that angles opposite to equal sides are equal in a triangle $PQR$
Therefore, $\mathrm{\angle }PRQ=\mathrm{\angle }PQR$
And sum of all the angles of a triangle is ${180}^{\circ }$
For triangle $PQR$, $\mathrm{\angle }PRQ+\mathrm{\angle }PQR+\mathrm{\angle }RPQ={180}^{\circ }$
On substituting the value $\mathrm{\angle }RPQ={30}^{\circ }$ and $\mathrm{\angle }PRQ=\mathrm{\angle }PQR$, we will get,
$$\begin{gathered} \mathrm{\angle }PQR+\mathrm{\angle }PQR+{30}^{\circ }={180}^{\circ } \\ \Rightarrow 2\mathrm{\angle }PQR={150}^{\circ } \\ \Rightarrow \mathrm{\angle }PQR={75}^{\circ } \end{gathered}$$
Therefore, we have
$\mathrm{\angle }PRQ=\mathrm{\angle }PQR={75}^{\circ }$
Now, $SR\parallel QP$
Then, $\mathrm{\angle }PQR=\mathrm{\angle }SRQ={75}^{\circ }$ as they are alternate interior angles.
Also, the angle between the chord and tangent is equal to the angle in the alternate segment.
Therefore, $\mathrm{\angle }PQR=\mathrm{\angle }RSQ={75}^{\circ }$
Hence, in triangle, $RSQ$, we have
$\mathrm{\angle }SRQ=\mathrm{\angle }RSQ={75}^{\circ }$
Therefore, $RSQ$ is also an isosceles triangle.
And the sum of all the angles of a triangle is ${180}^{\circ }$
$$\begin{gathered} \mathrm{\angle }SRQ+\mathrm{\angle }RSQ+\mathrm{\angle }SQR={180}^{\circ } \\ \Rightarrow {75}^{\circ }+{75}^{\circ }+\mathrm{\angle }SQR={180}^{\circ } \\ \Rightarrow {150}^{\circ }+\mathrm{\angle }SQR={180}^{\circ } \\ \Rightarrow \mathrm{\angle }SQR={30}^{\circ } \end{gathered}$$
Hence, the value of $\mathrm{\angle }SQR$ is ${30}^{\circ }$.

Note: Many students make mistakes by assuming that $RQ$ is perpendicular on $QP$ and $PR$. But, one has to take care $RQ$ is not the diameter as it is not passing from the centre. And the property states that the line from the centre is perpendicular to the tangent at the point of contact.