10 of 100: Parallel Puzzle

Here orange angle is 60 degree and blue angle is 65 degree.

So x=(180-60-65)= 55 degree

You can also skip computing for the third interior angle if you know Euclid's Elements I.32 (exterior angle = sum of opposite interior angles)

Let $M$ be a point of the top line, so that $AMC\angle=90°$ . Since the two line is parallel, $BAM\angle=(180°-25°)-90°=65°$ . We also know that $BCM\angle=180°-30°=150°$ . In a quadrilateral the amount of the angles is $360°$ , so $ABC\angle=360°-65°-90°-150°=\boxed{55°}$ .

Using line to connect both line:

Y + Z + 30 + 25 = 180

And X + Y + Z = 180

Hence X = 55

The "right" line.

if we make a triangle with the 30 degree and 25 degree. the outer angles are 60 degree and 65 degree.

so,180-(60+65)=55. so the answer is 55 degree.

add 25 degrees + 30

So x=(180-60-65)= 55 degree

Starting from the top line:

Each blue angle represents a clockwise rotation of the previous line, and the red angle is a counter clockwise rotation.

In order for the original line to end up parallel to itself, the amounts of clockwise and counter clockwise rotation have to be equal (or in other cases add to a multiple of 180º).

Thus, X = 25º + 30º = 55º

I just thought intuitively about it, if we slide the middle point vertically up we will end up with that shape from here we can solve it easily as x = 55

I did it by constructing a straight line creating two right triangles. Since every triangle = 180° , and the angles about a straight line are 180° , by subtraction, one gets 55°

The slope of the line produced by the 25° angle is $tan(25°)$ , while the slope of the line produced by the 30° angle is - $tan(30°)$ . The angle between two lines of slope u and v is $tan(m) = \frac{u - v}{1 + u \times v}\$ , therefore our angle of choice is $X = arctan( \frac{tan(25°) + tan(30°)}{1 - tan(25°) \times tan(30°)} )$ which computes to 55°.

$LaTeX$ Draw an auxiliary line parallel to the given two passing through "the point corresponding to the X angle". The angle gets divided in two parts whose value we can get by thinking about alternate interior angles (highlighted in the picture)

I drew a line perpendicular to the parallel lines at the intersection of the two lines that form 25 degrees. This created an irregular quadrilateral, which obviously has 360 degrees as the sum of internal angles. 90-25=65 degrees is the bottom angle. 90 degrees is the angle in the top left hand corner (perpendicular lines). 180-30=150 degrees as the top right angle. 65+90+150=305 and 360-305=55 degrees.

Sum of alternates.

Parallelity is why.

Fifty-Five is sum.

extending either existing line segment through point x to the second parallel line gives solution quickly

Add 30 and 25 together.