Question

In the given figure, PQ and PR are tangents to a circle with centre A. If $\angle$ QPA = ${27}^o$ then $\angle$ QAR equals

(a) ${63}^o$ (b) ${117}^o$ (c) ${126}^o$ (d) ${153}^o$

Answer 1

Here PQ and PR are tangents and AQ and AR are radii so
∠AQP= ∠ARP = ${90}^{\circ }$
Now, we know that
AP bisect angle QPR.
So, ∠QPR= 2∠QPA= 2×27 = ${54}^{\circ }$
In quadrilateral QARP,
90+90+54+∠QAR= 360
∠QAR= ${126}^{\circ }$

(c) 126° is correct answer