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

Answer
475.2k+ views
Hint: We use the theorem that tangents from the common point are equal and then calculate the values of $\angle PRQ$ and $\angle PQR$ 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 $\angle SRQ$. Next, we will find $\angle RSQ$ by angles in the same segment theorem. At last, apply angle sum property to find the value of $\angle SQR$
Complete step-by-step answer:
We are given the value of $\angle RPQ = {30^ \circ }$ and $SR\parallel QP$
We have to find the value of $\angle SQR$
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, $\angle PRQ = \angle PQR$
And sum of all the angles of a triangle is ${180^ \circ }$
For triangle $PQR$, $\angle PRQ + \angle PQR + \angle RPQ = {180^ \circ }$
On substituting the value $\angle RPQ = {30^ \circ }$ and $\angle PRQ = \angle PQR$, we will get,
$
\angle PQR + \angle PQR + {30^ \circ } = {180^ \circ } \\
\Rightarrow 2\angle PQR = {150^ \circ } \\
\Rightarrow \angle PQR = {75^ \circ } \\
$
Therefore, we have
$\angle PRQ = \angle PQR = {75^ \circ }$
Now, $SR\parallel QP$
Then, $\angle PQR = \angle SRQ = {75^ \circ }$ as they are alternate interior angles.
Also, the angle between the chord and tangent is equal to the angle in the alternate segment.
Therefore, $\angle PQR = \angle RSQ = {75^ \circ }$
Hence, in triangle, $RSQ$, we have
$\angle SRQ = \angle RSQ = {75^ \circ }$
Therefore, $RSQ$ is also an isosceles triangle.
And the sum of all the angles of a triangle is ${180^ \circ }$
$
\angle SRQ + \angle RSQ + \angle SQR = {180^ \circ } \\
\Rightarrow {75^ \circ } + {75^ \circ } + \angle SQR = {180^ \circ } \\
\Rightarrow {150^ \circ } + \angle SQR = {180^ \circ } \\
\Rightarrow \angle SQR = {30^ \circ } \\
$
Hence, the value of $\angle SQR$ 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.
Complete step-by-step answer:
We are given the value of $\angle RPQ = {30^ \circ }$ and $SR\parallel QP$
We have to find the value of $\angle SQR$
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, $\angle PRQ = \angle PQR$
And sum of all the angles of a triangle is ${180^ \circ }$
For triangle $PQR$, $\angle PRQ + \angle PQR + \angle RPQ = {180^ \circ }$
On substituting the value $\angle RPQ = {30^ \circ }$ and $\angle PRQ = \angle PQR$, we will get,
$
\angle PQR + \angle PQR + {30^ \circ } = {180^ \circ } \\
\Rightarrow 2\angle PQR = {150^ \circ } \\
\Rightarrow \angle PQR = {75^ \circ } \\
$
Therefore, we have
$\angle PRQ = \angle PQR = {75^ \circ }$
Now, $SR\parallel QP$
Then, $\angle PQR = \angle SRQ = {75^ \circ }$ as they are alternate interior angles.
Also, the angle between the chord and tangent is equal to the angle in the alternate segment.
Therefore, $\angle PQR = \angle RSQ = {75^ \circ }$
Hence, in triangle, $RSQ$, we have
$\angle SRQ = \angle RSQ = {75^ \circ }$
Therefore, $RSQ$ is also an isosceles triangle.
And the sum of all the angles of a triangle is ${180^ \circ }$
$
\angle SRQ + \angle RSQ + \angle SQR = {180^ \circ } \\
\Rightarrow {75^ \circ } + {75^ \circ } + \angle SQR = {180^ \circ } \\
\Rightarrow {150^ \circ } + \angle SQR = {180^ \circ } \\
\Rightarrow \angle SQR = {30^ \circ } \\
$
Hence, the value of $\angle SQR$ 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.
Recently Updated Pages
Express the following as a fraction and simplify a class 7 maths CBSE

The length and width of a rectangle are in ratio of class 7 maths CBSE

The ratio of the income to the expenditure of a family class 7 maths CBSE

How do you write 025 million in scientific notatio class 7 maths CBSE

How do you convert 295 meters per second to kilometers class 7 maths CBSE

Write the following in Roman numerals 25819 class 7 maths CBSE

Trending doubts
Full Form of IASDMIPSIFSIRSPOLICE class 7 social science CBSE

Fill in the blanks with appropriate modals a Drivers class 7 english CBSE

What are the controls affecting the climate of Ind class 7 social science CBSE

The southernmost point of the Indian mainland is known class 7 social studies CBSE

What were the major teachings of Baba Guru Nanak class 7 social science CBSE

What was the approximate time period of the Indus Valley class 7 social science CBSE
