Answers

2016-02-13T22:47:29+05:30
Construction: TQ

angle TOQ + angle POT = 180 (Linear pairs)
=> angle TOQ = 180-150 = 30
In triangle OTQ,
=> OT = OQ
Therefore, OTQ is isosceles
=> angle OQT = angle OTQ (base angles of isosceles triangle) ...(1)

=> angles TOQ+OQT+OTQ = 180 (sum of interior angles)
=> 30+2OQT = 180 (From 1)
=> OQT = (180-30)/2 = 75
angle OQT + angle RQT = 180 (linear pairs)
=> angle RQT = 180-75 = 105

=> angle PUT = angle POT/2 = 150/2 = 75
(Theorem: Angle at circumferance will be half of angle at centre, when both are subtended by same arc (here it is arc PT))

angle PUT + angle PST = 180 (opposite sides of cyclic quadrilateral PUTS)
=> angle PST = 180-75 = 105

In triangle PSR,
angles PST+PRS+RPS = 180
=> 105+PRS+45 = 180
=> PRS = 180-150 = 30

Finally, in triangle RQT,
angles PRS+RQT+RTQ = 180
=> 30+105+RTQ = 180
=> RTQ = 180-135 = 45
5 5 5
Ah ok
He/she means to say that u have copied the answer from that 
But u have not...Good answer 
Hmm it's harder to find a particular question in the book for answers than to solve it though
Thanks man