Question

Find the value of in the figure given below in which HF is parallel to AB

Answer 1

Step-by-step explanation:

HF parallel to AB and CD is a transversal line passing through the points F and D

angle HFC=angle ADF ( corresponding angles)

angle ADF = 28°

angle ADF +FDE+EDB= 180° (linear pairs)

28°+x+72°=180°

x+100°=180°

x=180°-100°

x=80°

I hope it's correct answer

Answer 2

The value of the angle x is 80°

Step-by-step explanation:

Given:

• HF║AB
• And also ∠EBD = 72° and ∠CFH = 28°
• From the figure, we get ADB is a line

To find:

The value of ∠x (i.e. ∠FDE)

So,

HF║AB (given)

CD is the transversal.

Now, ∠CFH = ∠FDA  [∵ corresponding angles are equal]

We got the value of ∠FDA = 28°

As ADB is a line, then

∠FDA + ∠FDE + ∠EBD = 180° [Linear pair]

⇒ 28° + ∠x + 72° = 180°

⇒ 100° + ∠x = 180°

⇒ ∠x = 180° - 100°

⇒ ∠x = 80°

Hence,

• ∠x = ∠FDE = 80°