Question

b) In the figure given below "O' is the centre of the circle. If QR = OP and LORP = 20°.
Find the value of 'x' give reasons.​

Answer 1

$\bold \red{ANSWER}$

QR=OP⟹QR=OQ (as OP=OQ are radii of circle)

∴∠ROQ=∠ORP=20°

Hence, using sum of interior angles theorem, ∠ORQ=180°−∠ROQ−∠ORP=140°

∴∠OQP=180° −∠OQR=40°

Also, OP=OQ as both are radii of the same circle.

∴∠OQP=∠OPQ=40° (isoceles triangle property)

∴∠POQ=180° −∠OQP−∠OPQ=100° (sum of interior angles of a triangle)

Now,

∠TOP+∠POQ+∠QOR=180° (angles on a straight line)

∴∠TOP=180°−100°−20°=60°