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?
Answers
∠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 )
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∠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,
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,
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.
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