A geometry problem by Paola Ramírez

3y = 180º - 59º = 81º...(exterior angle property)→→

→y = 27º

→Now by angle sum property ,

2y + 59º + x = 180º

→x = 180º - 113º = 67º... Ans :-D

First, we consider $\triangle GIF$ . Let the green colored angle be $a$ . Since $IFB$ is a straight line and a straight line is $180^\circ$ ,

$a=180^\circ-140^\circ=40^\circ$

Now, we consider $\triangle EGF$ . Let the blue colored angle be $3y$ . We know that the sum of the interior angle of a triangle is $180^\circ$ , so

$3y=180-59-a=180-59-40=81^\circ$

It follows that,

$y=27^\circ$

Now, consider $\triangle GIF$ again. Let the purple colored angle be $c$ . Then

$c=180-y-a=180-27-40=113^\circ$

Let the yellow colored angle be $x$ . Since $HIF$ is a straight line, we have

$x=180-c=180-113=\boxed{67^\circ}$

$y=(180-59-40)/3=27$

Then $180-59-2(27)$ = $x=67$

3y=81 so y=27 now x=y+40 (exterior angle is equal to sum of opposite two interior angles) so x=27+40=67°

The supplement of 140 = 40. So 3y + 59 + 40 = 180, 3y = 81, y = 27. The supplement of x is 180 - x. So y + 40 + 180 - x = 180, x = y + 40 = 67. Ed Gray

A straight line is $180^\circ$ . Thus,

$a+140=180$ $\implies$ $a=40$

The sum of interior angles of triangle is $360^\circ$ . Thus,

$3y+59+40=180$ $\implies$ $y=27^\circ$

By the exterior angle theorem, we have

$x=y+a=27+40=$ $\boxed{67}$