Question

The vertices of a A PQR lie on a circle with centre O. SR is a tangent to the circle at the point R. If QR bisects theangle ORS, then what is the measure of RPQ?​

Answer 1

∠RPQ = 45° where ΔPQR lie on a circle with centre O , SR is Tangent & QR bisect ∠ ORS

Step-by-step explanation:

A ΔPQR lie on a circle with centre O

SR is a tangent to the circle at the point R

=> ∠ ORS = 90°

QR bisects the ∠ ORS

=> ∠ORQ = 90°/2 = 45°

in Δ ORQ

OR = OQ = Radius

=> ∠ORQ = ∠OQR

=> ∠OQR = 45°

∠ORQ + ∠OQR + ∠RQQ = 180°  ( sum of Angles of a Triangle)

=>  45° +  45° + ∠RQQ = 180°

=>  ∠RQQ = 90°

=> RQ chord makes an angle of 90° at center

=> ∠RPQ = (1/2)90° = 45°  (as  The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle )

Learn More :

Prove that equal chords of circle subtend equal angles at the centre

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Answer 2

∠RPQ=45°

Step-by-step explanation:

Consider the provided information.

SR is a tangent to the circle at the point R.

OR is the radius which is perpendicular to SR. Therefore ∠ORS=90°

It is given that QR bisects the angle ORS,

$\angle ORQ=\frac{1}{2}\angle ORS\\\angle ORQ=\frac{1}{2}\times 90^0\\\angle ORQ=45^0$

In ΔOQR

OQ=OR (Radius)

Thus, ΔOQR is isosceles triangle.

Therefore, ∠OQR=∠ORQ=45°

∠OQR+∠ORQ+∠QOR=180°

45°+45°+∠QOR=180°

∠QOR=180°-90°

∠QOR=90°

Theorem: Angle at the centre of a circle is twice the size of the angle at the circumference.

Therefore, $\angle RPQ=\frac{1}{2}\angle QOR =\frac{90^0}{2}=45^0$

Hence, ∠RPQ=45°

#Learn more

Prove that the angle subtended by an arc at the center is twice of the angle subtended by the same arc at the remaining part of the circle.

https://brainly.in/question/2841090