In the diagram, |SR| = |QR|. < SRP = 65^o and < RPQ = 48^o, find < PRQ

65^o

45^o

25^o

19^o

Correct answer is D

< RSQ = < RPQ = 48^o (angle in the same segment)

< SQR < RSQ (Base angle of an isosceles \(\bigtriangleup\))

< SQR = 480

< QRS + < RSQ + < RSQ = 180^o(sum of interior angles of a \(\bigtriangleup\))

i.e. < QRS + 48^o + 48^o = 180

< QRS = 180 - (48 + 48) = 180 - 96 = 84^o

but < PRQ + < PRS = < QRS

< PRQ = < QRS - < PRS - 84 - 65

= 19^o

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In the diagram, |SR| = |QR|. < SRP = 65o and < RPQ = 48o, find < PRQ

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In the diagram, |SR| = |QR|. < SRP = 65^o and < RPQ = 48^o, find < PRQ