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


Question

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

Options

A) 65o

B) 45o

C) 25o

D) 19o

The correct answer is D.

Explanation:

< RSQ = < RPQ = 48\(^{\circ}\) (angle in the same segment)
< SQR < RSQ (Base angle of an isosceles \(\bigtriangleup\))
< SQR = 480
< QRS + < RSQ + < RSQ = 180\(^{\circ}\)(sum of interior angles of a \(\bigtriangleup\))
i.e. < QRS + 48\(^{\circ}\) + 48\(^{\circ}\) = 180
< QRS = 180 - (48 + 48) = 180 - 96 = 84\(^{\circ}\)
but < PRQ + < PRS = < QRS
< PRQ = < QRS - < PRS - 84 - 65
= 19\(^{\circ}\)

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Dicussion (1)

  • < RSQ = < RPQ = 48\(^{\circ}\) (angle in the same segment)
    < SQR < RSQ (Base angle of an isosceles \(\bigtriangleup\))
    < SQR = 480
    < QRS + < RSQ + < RSQ = 180\(^{\circ}\)(sum of interior angles of a \(\bigtriangleup\))
    i.e. < QRS + 48\(^{\circ}\) + 48\(^{\circ}\) = 180
    < QRS = 180 - (48 + 48) = 180 - 96 = 84\(^{\circ}\)
    but < PRQ + < PRS = < QRS
    < PRQ = < QRS - < PRS - 84 - 65
    = 19\(^{\circ}\)

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