Question

In the following figure, PQ and PR are tangents to the circle, with centre O. If $\angle QPR={60}^o$, calculate :

(i)$\angle QOR,$
(ii)$\angle OQR,$
(iii)$\angle QSR,$

Answer 1

(i) In the figure form ,
PQ=PR,
Hence it made an isosceles triangle
PQR= 180
2Q=180-60
2Q=120
Q=120/2
Q=60
And,
Angle OQP=90
In which RQP=60
So,
angle OQR=90-60=30
Hence, the angle is 30

$\angle QOR={180}^0-2\left({30}^0\right)$

$\angle QOR={120}^0$

(ii) $\angle OQR={30}^0$ from(i)

(iii) $\angle QSR=0.5\ast \angle QOR$

$\angle QSR={60}^0$