Question

In the figure given below, angleSPQ=45 degree, anglePOT= 150 degree, and O is the centre of the circle. Find the measures of angleRQT, angleRTQ and anglePUT. I want the answer fast

Answer 1

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

Answer 2

Answer:

Step-by-step explanation:

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