Angles in a Snakey Segment

This is the easiest way:

I will add a line segment parallel with the line $AB, DE$ called $XC )$

Then, $\displaystyle \angle XCD=50°$ because $\displaystyle \angle XCD$ and $\angle CDE$ are alternate interior angles

Then $\displaystyle \angle BCX=70°$ because $\displaystyle \angle XCB+ \angle ABC=180°$ ( Consecutive interior angles )

Lastly, take $\displaystyle 50°+70°=\boxed{120°}$

$\angle CDE\quad and\quad \angle BPC\quad$ are alternate interior angles, so $\angle BPC$ = 50

Then $\angle PBC$ = 180 - 110 = 70

Then $\angle BCP$ = 180 - 70 - 50 = 60
Finally x = 180 - 60 = $\boxed{120}$

The is why I love math so much there are many different ways to solve problems like this. I personally drew a line perpendicular to line segment AB to create two triangles. The bottom triangle containing angle CDE and the newly created 90 degree angle allowed me to find the third angle of that triangle by subtracting those two angles from 180, since the sum of all interior angles of a triangle is 180 degrees. So 180-90-50=40, with this information I also know that the opposite angle from where this perpendicular line crosses line segment CD is also 40 degrees giving us one of the angle of the top triangle created. Now let's look at the other angle at point B, since I drew a perpendicular line from point B we can subtract 90 from 110 to discover the top angle of the top triangle, giving us 110-90=20. With this information we can now subtract the measurement of both of the new angles of the top triangle from 180 to give us the solution for what angle x is. This would be 180-40-20=120, so x=120.

The ans is 120, how ever the picture is a bit misleading. The angle does not look like 120, it looks like even less than 90 !

Problems are better if they have parallel solutions, :]

NCERT question Grade IX

Am I the only one who did it this way? Draw a segment from A to D. This makes ABCD into a quadrilateral, therefore the total angles of ABCD = 360 degrees. We know <ADE+<BAD=180-50=130, while <ABC=110. Therefore, <BCD=360-130-110=120

Trace a line parallel to AB that cross the point C. Let’s call it XC where X is the leftmost point.

∠XCB + ∠ABC = 180° or 180° - ∠ABC =∠XCB

∠XCB = 180° - 110° = 70°

∠XCD = ∠CDE = 50°

∠BCD = ∠XCD + ∠XCB = 50° + 70° = 120°

Be F the point that DC intercepts AB, Angle BFC is 50º. Angle CBF is 180º - 110º = 70º. So angle BCF is 180º - 50º - 70º = 60º. Thus x = 180º - BCF = 180º - 60} = 120º

draw a st. line passing through c parallel to eother ab or de then get the sum of 50 and 70 to get your ans..simple