Question

please find x and y in given figure​

Answer 1

Answer:

y+3y+108=180

4y=180-108

y=18

sorry maybe its easy but I couldn't find the value of x

::: ( I think ADC is an isosceles triangle)

Answer 2

Answer :-

x = 54° and y = 18°.

Solution :-

∠CAE + ∠BAC + ∠BAD = 180° (Linear Pair)

108 + 3y + y = 180

108 + 4y = 180

4y = 180 - 108

4y = 72

y = 72/4

y = 18

We know that

Exterior angle = Sum of two interior angles of a triangle

∠CAE = ∠ACD + ∠ADC

108° = x + ∠ADC

108 - x = ∠ADC --(1)

108 - ∠ADC = x --(2)

Subtracting (1) and (2)

108 - x - (108 - ∠ADC) = ∠ADC - x

108 - x - 108 + ∠ADC = ∠ADC - x

- x + ∠ADC = ∠ADC - x

- x + ∠ADC = - x + ∠ADC

- x + x = ∠ADC - ∠ADC

0 = 0

This means ∠ADC = x

In ΔADC

∠ADC + ∠DAC+ ∠ACD = 180° (Angle sum property of a triangle)

∠ADC + 3y + y + x= 180

∠ADC + 4y + x = 180

∠ADC + 4(18) + x = 180

∠ADC + 72 + x = 180

x + 72 + x = 180 (Since ∠ADC = x)

2x = 180 - 72

2x = 108

x = 108/2

x = 54°

Therefore x = 54° and y = 18°.