Question

FG ∥ CB, A ∈ FG, D ∈ AB, E ∈ AC

Find the value of x. Give reasons to justify your solutions!

Please help, 10 points will be rewarded.

Answer 1

The value of x is
`31^(\circ)`.

Explanation:

Please look at the figure attached to get more clear solution.

We have given:

FG||CB

And the line that cut the parallel line is transversal so, here BA is transversal

And alternate interior angles on transverse line are equal

So, ∠1=∠4

And ∠4=
`28^(\circ)`

Hence, ∠1=∠4=
`28^(\circ)`

And On FG the sum of angles will be
`180^(\circ)`

∠3+∠2+∠1=
`180^(\circ)`

`90^(\circ)`+∠2+
`28^(\circ)`=
`180^(\circ)`

Hence, ∠2=
`62^(\circ)`

Now, we know that the sum of interior angles is equal to the exterior angle:

Therefore, ∠2+∠5=∠6+∠7

`62^(\circ)+3x=x+4x`

On simplification we get:

`2x=62^(\circ)`

`x=31^(\circ)`

Hence, the value of x is
`31^(\circ)`.