Question

In the adjoining figure, AB || DE, find the value of x ​

Answer 1

Answer:

95°

Step-by-step explanation:

refer figure

a + b = 125

a + b + c = 180

c = 55°

Now,

c + d = 140

d = 140 - c = 140 - 55

d = 85°

Now,

b + c + d = 180

b + c = 180 - d = 180 - 85

b+c = 95°

∠x = b + c = 95°

Answer 2

Answer:

x = 95°

Step-by-step explanation:

Given:- AB || DE and ∠BAC = 125°,∠EDC = 140°

To find:- ∠ACD (x°)

Construction:- Let's draw a Line parallel to AB AB the point C and let the line be CF.

[Please do refer the above image]

Proof:-

We know that,

AB || DE (given) ----- 1

and,

CF || AB (by construction) ------- 2

Thus, from eq.1 and eq.2 we get,

CF || AB || DE

Then,

AC is a transversal to the Parallel lines AB and CF

∴ ∠BAC and ∠FCA are co-interior angles

Hence,

∠BAC + ∠FCA = 180°

But we know that, ∠BAC = 125°

Thus,

125° + ∠FCA = 180°

∠FCA = 180° - 125°

∠FCA = 55° ----- 3

Similarly,

CD is a transversal to the Parallel lines DE and CF

∴ ∠EDC and ∠FCD are co-interior angles

Hence,

∠EDC + ∠FCD = 180°

But we know that, ∠EDC = 140°

Thus,

140° + ∠FCD = 180°

∠FCD = 180° - 140°

∠FCD = 40° ----- 4

Now, if you observe closely,

∠ACD = ∠FCA + ∠FCD

From eq.3 and eq.4 we get,

∠ACD = 55° + 40°

∴ ∠ACD = 95°

Thus, ∠ACD = x = 95°

Hope it helped and you understood it........All the best