# If \[\angle RPQ={{45}^{\circ }}\], find \[\angle PRQ\]

A. \[{{60}^{\circ }}\]

B. \[{{90}^{\circ }}\]

C. \[{{120}^{\circ }}\]

D. \[{{150}^{\circ }}\]

Last updated date: 19th Mar 2023

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Answer

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Hint: Consider, \[\vartriangle PQR\], use the properties of the isosceles triangle and find the \[\angle PRQ\]. PR and QR are tangents from a point R meeting the circle at P and Q.

“Complete step-by-step answer:”

Given a figure, with a circle whose center is O. From the figure we can say that PR and QR are tangents to the circle from a point R. Three tangents from an external point to the circle are equal.

\[\therefore PR=QR\]

Now let us consider the isosceles triangle, \[\vartriangle PRQ\].

As it is an isosceles triangle, we know that two sides of an isosceles triangle are equal.

In \[\vartriangle PRQ\], we know that PR = QR.

Thus the angle RPQ is equal to angle PQR, because they are the base angles of an isosceles triangle.

We have been given, \[\angle RPQ={{45}^{\circ }}\].

\[\angle RPQ=\angle PQR={{45}^{\circ }}\].

The angle sum property of triangle, states that the sum of interior angles of a triangle is \[{{180}^{\circ }}\].

Hence, in the isosceles triangle PRQ, the sum of all the interior angles is \[{{180}^{\circ }}\].

\[\therefore \angle RPQ+\angle PQR+\angle PRQ={{180}^{\circ }}\]

We need to find the \[\angle PRQ\].

We already found out that, \[\angle RPQ=\angle PQR={{45}^{\circ }}\].

Thus substituting these values in the sum of triangle,

\[\begin{align}

& \angle RPQ+\angle PQR+\angle PRQ={{180}^{\circ }} \\

& {{45}^{\circ }}+{{45}^{\circ }}+\angle PRQ={{180}^{\circ }} \\

& \angle PRQ=180-45-45=180-90 \\

& \angle PRQ={{90}^{\circ }} \\

\end{align}\]

Hence, we got \[\angle PRQ={{90}^{\circ }}\], which makes \[\vartriangle PRQ\] right angled at R.

\[\therefore \]Option (b) is the correct answer.

Note: By seeing the figure, you may try to take \[\vartriangle POQ\] instead of \[\vartriangle PRQ\], understand the question that we only need to find \[\angle PRQ\]. So we should take, \[\vartriangle PRQ\]. As tangents from the external point of the circle are equal, the sides of the triangle are same, thus angles are same and we can get \[\angle PRQ\]easily.

“Complete step-by-step answer:”

Given a figure, with a circle whose center is O. From the figure we can say that PR and QR are tangents to the circle from a point R. Three tangents from an external point to the circle are equal.

\[\therefore PR=QR\]

Now let us consider the isosceles triangle, \[\vartriangle PRQ\].

As it is an isosceles triangle, we know that two sides of an isosceles triangle are equal.

In \[\vartriangle PRQ\], we know that PR = QR.

Thus the angle RPQ is equal to angle PQR, because they are the base angles of an isosceles triangle.

We have been given, \[\angle RPQ={{45}^{\circ }}\].

\[\angle RPQ=\angle PQR={{45}^{\circ }}\].

The angle sum property of triangle, states that the sum of interior angles of a triangle is \[{{180}^{\circ }}\].

Hence, in the isosceles triangle PRQ, the sum of all the interior angles is \[{{180}^{\circ }}\].

\[\therefore \angle RPQ+\angle PQR+\angle PRQ={{180}^{\circ }}\]

We need to find the \[\angle PRQ\].

We already found out that, \[\angle RPQ=\angle PQR={{45}^{\circ }}\].

Thus substituting these values in the sum of triangle,

\[\begin{align}

& \angle RPQ+\angle PQR+\angle PRQ={{180}^{\circ }} \\

& {{45}^{\circ }}+{{45}^{\circ }}+\angle PRQ={{180}^{\circ }} \\

& \angle PRQ=180-45-45=180-90 \\

& \angle PRQ={{90}^{\circ }} \\

\end{align}\]

Hence, we got \[\angle PRQ={{90}^{\circ }}\], which makes \[\vartriangle PRQ\] right angled at R.

\[\therefore \]Option (b) is the correct answer.

Note: By seeing the figure, you may try to take \[\vartriangle POQ\] instead of \[\vartriangle PRQ\], understand the question that we only need to find \[\angle PRQ\]. So we should take, \[\vartriangle PRQ\]. As tangents from the external point of the circle are equal, the sides of the triangle are same, thus angles are same and we can get \[\angle PRQ\]easily.

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