Question

plz answer fast its urgent...​

Answer 1

(i) <AQD = <APF = 50° (FP and EQ are║ and both lie on the same plane)

<AQD = <CDQ = 50° (one of the property of triangle)

<CDQ + <CDG + <EDG = 180°

<EDG = 180° - (50 + 40)

(ii) <EDG = 90°

<EDG = <PDQ = 90° (vertically opp. angles)

<DPQ = 180° - <AQD - <PDQ

<DPQ = 50°

<DPQ + <DPF + <APF = 180° (st. line)

<DPF = 180° - 100°

(iii) <DPF = 80°

therefore,

(i) <AQD = 50°

(ii) <EDG = 90°

(iii) <DPF = 80°

Answer 2

Answer:

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