Question

In the figure given alongside find angle acd, angle adc, angle dae

Answer 1

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°

Answer 2

∠ACD = 80°, ∠ADC = 50°  and ∠DAE = 90°  

Given: From given figure In triangle ABC  

∠ABC = 40° and  ∠ ACB = 100° and ∠CAD = 50°

To find: The measures of ∠ ACD, ∠ ADC, ∠ DAE

Solution: From given figure,

⇒ ∠ ACB + ∠ ACD = 180°  [ ∵ ∠ACB, ∠ACD are Linear pair ]

⇒ 100° + ∠ ACD = 180°   [ ∵ ∠ ACB = 100° ]  

⇒ ∠ ACD = 180° - 100° = 80°

Therefore, ∠ACD = 80°

As we know sum of angles in a triangle = 180°

From Triangle, ACD

⇒ ∠ ACD + ∠ ADC +∠CAD = 180°

⇒   80°+ ∠ADC + 50° = 180°

⇒ ∠ADC + 130° = 180°

⇒ ∠ADC = 180° - 130° = 50°

Therefore, ∠ADC = 50°

From triangle ABC

⇒ ∠ABC + ∠BCA + ∠BAC = 180°

⇒ 40° + 100° + ∠CAB = 180°

⇒ ∠BAC + 140° = 180°

⇒ ∠BAC = 40°

From figure, ∠BAC + ∠CAD + ∠DAE = 180°  [ BAE is a straight line ]

⇒ 40° + 50° + ∠DAE = 180°

⇒ ∠DAE + 90° = 180°

⇒  ∠DAE = 180° - 90° = 90°

Therefore, ∠DAE = 90°

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