Question

In the given figure, PQ is the diameter. \[\angle SQP\] is equal to

\[\left( a \right){{40}^{\circ }}\]
\[\left( b \right){{30}^{\circ }}\]
\[\left( c \right){{60}^{\circ }}\]
\[\left( d \right){{50}^{\circ }}\]

Answer 1

Hint: To solve this question, we will first use a theorem stated as, “If A, B and C are distinct points on a circle where the line AC is a diameter, then the angle \[\angle ABC={{90}^{\circ }}.\] Using this, we will calculate \[\angle PSQ\] and then use the fact that the opposite angles of a cyclic quadrilateral add up to \[{{180}^{\circ }}\] to get the measure of \[\angle SQP.\]

Complete step by step answer:
We are given the circle. Let us define R in it as below.

PQ is the diameter and \[\angle SRQ={{150}^{\circ }}\] is given.
Consider triangle PSQ first. We have a theorem that states, “If A, B and C are distinct points on a circle where the line AC is a diameter, then the angle \[\angle ABC={{90}^{\circ }}.\] Using this, we will calculate \[\angle PSQ.\] This means in the circle, AC is the diameter, then \[\angle ABC={{90}^{\circ }}.\]

Using this theory in our triangle and circle, we have, \[\angle PSQ={{90}^{\circ }}\] as PQ is the diameter of the circle and P, S and Q lie on the circle. We have now,

Now, considering the cyclic quadrilateral PSRQ, given \[\angle SRQ={{150}^{\circ }}.\] Then as the sum of the opposite angles of a cyclic quadrilateral is \[{{180}^{\circ }}.\]
\[\Rightarrow \angle SPQ+\angle SRQ={{180}^{\circ }}\]
\[\Rightarrow \angle SPQ+{{150}^{\circ }}={{180}^{\circ }}\]
\[\Rightarrow \angle SPQ={{180}^{\circ }}-{{150}^{\circ }}\]
\[\Rightarrow \angle SPQ={{30}^{\circ }}\]
Therefore, \[\angle SPQ={{30}^{\circ }}.\]
Now, finally considering triangle PSQ, the angle sum property of a triangle says that the sum of all the angles of a triangle is \[{{180}^{\circ }}.\] So, in triangle PSQ,
\[\Rightarrow \angle P+\angle S+\angle Q={{180}^{\circ }}\]
\[\Rightarrow {{30}^{\circ }}+{{90}^{\circ }}+\angle Q={{180}^{\circ }}\]
\[\Rightarrow \angle Q={{180}^{\circ }}-{{30}^{\circ }}-{{90}^{\circ }}\]
\[\Rightarrow \angle SQP={{180}^{\circ }}-{{120}^{\circ }}\]
\[\Rightarrow \angle SQP={{60}^{\circ }}\]

So the measure of \[\angle SQP={{60}^{\circ }}\] which is the required result.

Note: Cyclic quadrilaterals are those whose all sides lie on the circumference of a circle. Here in our question, the quadrilateral PSRQ is cyclic as all sides lie on the circumference of the circle. A key point to note here is that the measure of \[\angle PSQ={{90}^{\circ }}\] and not angle \[\angle QSR.\] Therefore, we cannot use triangle SRQ to calculate our angle Q.