find
(I)∆ACD
2∆ADC
(3)∆DAE
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Answer:
Angle ACB + Angle ACD = 180° (Linear Pair)
= 100 + ACD = 180
= Angle ACD = 80°
Angle ADC = 180 - (ACD + CAD)
[ASP of triangle]
= 180 - (80 + 50) = 180 - 130 = 50°
Angle BAC = 40° (Using ASP triangle)
Angle DAE = 180 - (DAC + BAC)
= 180 - (50 + 40) = 180 - 90 = 90°
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