Question

Р
16.
In the figure AB//QR |BAQ = 142°, and ABP = 100° find
(i) APB (ii) |AQR QRP
100°
A
B
142°
Q
R​

Answer 1

Answer:

Abp=

142°=100°+abp(exterior angle theorem)

abp=142°-100°

abp=42°

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