In the diagram, |SR| = |QR|. < SRP = 65o and < RPQ = 48o, find < PRQ
65o
45o
25o
19o
Correct answer is D
< RSQ = < RPQ = 48o (angle in the same segment)
< SQR < RSQ (Base angle of an isosceles \(\bigtriangleup\))
< SQR = 480
< QRS + < RSQ + < RSQ = 180o(sum of interior angles of a \(\bigtriangleup\))
i.e. < QRS + 48o + 48o = 180
< QRS = 180 - (48 + 48) = 180 - 96 = 84o
but < PRQ + < PRS = < QRS
< PRQ = < QRS - < PRS - 84 - 65
= 19o
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