Question

D is a point on the side BC and P is a point on the side AC of ∆ABC. PD produced meets AB produced at Q. If angle A = 80°, angleABC = 60° and anglePDC = 15°,
find: (i) angleAQD
(ii) angleAPD.
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Answer 1

Answer:

consider ∆ BQD, ang QBD = 180-60 = 120

ang BDQ =15

Therefore ang BQD = AQD = 180-(120+15) = 45

Consider ∆ AQP , ang A =80, ang AQP = 45

Therefore ang APQ = 180 -(80+45) = 55

Here ∆ property that sum of three angles of a ∆ is 180 degree has to be used. Also vertically opposite angles (here 15 degree) are equal