Question

centre of circle as shown in figure, find $\angle RQT$ and $\angle RTQ$

Answer 1

REF.Image

Angle POT +Angle TOQ = ${180}^{\circ }$

ngle TOQ = 40

OT=OQ (Radii)

Angle OTQ = Angle OQT

Consider triangle TOQ

By angle sum property,

Angle TOQ + Angle OTQ + Angle OQT = ${180}^{\circ }$

2. Angle OTQ= 140

OTQ = OQT = 70

Angle OQT + Angle RQT = 180

Therefore, Angle RQT=110.

Consider cyclic Quadrilateral PQTS.

Angle P + Angle T = ${180}^{\circ }$

Angle T = ${135}^{\circ }$

Angle SQT + Angle RTQ = ${180}^{\circ }$

Therefore, angle RTQ = ${180}^{\circ }$