Q. 333.7( 3 Votes )

# In Fig. 10.95, PQ

Answer :

Given:

OPQ = 35°

Property 1: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

Property 2: The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

Property 3: The sum of all angles of a triangle = 180°.

By property 1,

QP = QR (tangent from Q)

By property 2, ∆OPQ is right-angled at OQP (i.e., OQP = 90°) and ORP is right-angled at ORP (i.e., ORP = 90°).

OQ QP

OR RP

Now,

OQP = ORP = 90° [Property 1]

QP = QR [Property 2]

OP = OP [Given]

OPQ OPR By SAS

Hence, OPQ = OPR = 35° By CPCTC

i.e. a = 35°

By property 3,

OQP + OPQ + QOP = 180°

90° + 35° + QOP = 180°

125° + QOP = 180°

QOP = 180° - 125°

QOP = 55°

i.e. b = 55°

Hence, a = 35° and b = 55°

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