Mathematics
If PQ is a tangent to the circle at R; calculate :
(i) ∠PRS,
(ii) ∠ROT.
Given, O is the center of the circle and angle TRQ = 30°.
Answer
(i) Since, ST passes through O, so ST is the diameter of the circle.
We know that,
Angle in a semi-circle is a right angle.
∴ ∠SRT = 90°.
Since, PQ is a straight line.
∴ ∠PRS + ∠SRT + ∠TRQ = 180°
⇒ ∠PRS + 90° + 30° = 180°
⇒ ∠PRS + 120° = 180°
⇒ ∠PRS = 180° - 120°
⇒ ∠PRS = 60°.
Hence, ∠PRS = 60°.
(ii) We know that,
The angle between a tangent and chord through the point of contact is equal to an angle in the alternate segment.
∠TSR = ∠TRQ = 30°.
Since, angle subtended by a segment at the center is double the angle suspended at the circumference.
∠ROT = 2∠TSR = 2 × 30° = 60°.
Hence, ∠ROT = 60°.
Related Questions
In the figure (i) given below, PQ is a tangent to the circle at A, DB is a diameter, ∠ADB = 30° and ∠CBD = 60°, calculate
(i) ∠QAB
(ii) ∠PAD
(iii) ∠CDB.
In the given figure, diameter AB and chord CD of a circle meet at P. PT is a tangent to the circle at T. CD = 7.8 cm, PD = 5 cm, PB = 4 cm. Find :
(i) AB
(ii) the length of tangent PT.
In the figure, ABCD is a cyclic quadrilateral with BC = CD. TC is tangent to the circle at point C and DC is produced to point G. If ∠BCG = 108° and O is the center of the circle, find :
(i) angle BCT
(ii) angle DOC
Two circles with centers O and O' are drawn to intersect each other at points A and B. Center O of one circle lies on the circumference of the other circle and CD is drawn tangent to the circle with center O' at A. Prove that OA bisects angle BAC.
