Question

find
(I)∆ACD
2∆ADC
(3)∆DAE​

Answer 1

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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Step-by-step explanation: