Since the angle is called , it has vertex  - the middle letter is always the vertex - and it is the union of rays  and . Another name for  is  , since  is also on that ray, so the angle can be said to be the union of  and ; this makes  a valid name for the angle.

  and  are not valid, since the middle letter is not vertex .  is not valid, since  and  are on the same side of the angle.  is not valid; an angle can be named using only its vertex only if it is the only angle in the diagram with that vertex, and that is not the case here.

Coresponding angles can be found when a line crosses two parallel lines. Angles 10 and 14 are equal, because corresponding angles are equal. Angles 14 and 13 are supplementary because together they form a straight line. If angles 10 and 14 are equal, then angles 10 and 13 must be supplementary as well.

By properties of parallel lines A+B = 180^o, B = 45^o, C = A = 135^o, so 2*|B-C| = 2* |45-135| = 180^o

Since the angles are supplementary, their sum is 180 degrees.  Because they are in a ratio of 1:4, the following expression could be written:

Since we know opposite angles are equal, it follows that angle  and .  

Imagine a parallel line passing through point .  The imaginary line would make opposite angles with  & , the sum of which would equal .  Therefore, .

When the measure of an angle is added to the measure of its supplement, the result is always 180 degrees. Put differently, two angles are said to be supplementary if the sum of their measures is 180 degrees. For example, two angles whose measures are 50 degrees and 130 degrees are supplementary, because the sum of 50 and 130 degrees is 180 degrees. We can thus write the following equation:

Subtract 40 from both sides.

Add  to both sides.

The answer is .

When two parallel lines are intersected by another line, the sum of the measures of the interior angles on the same side of the line is 180°. Therefore, the sum of the angle that is labeled as 100° and angle y is 180°. As a result, angle y is 80°.

Another property of two parallel lines that are intersected by a third line is that the corresponding angles are congruent. So, the measurement of angle x is equal to the measurement of angle y, which is 80°.

The angles are alternate exterior angles and are, therefore, equal.

Let A represent the measure, in degrees, of angle A. By definition, the sum of the measures of A and its complement is 90 degrees. We can write the following equation to determine an expression for the measure of the complement of angle A.

A + measure of complement of A = 90

Subtract A from both sides.

measure of complement of A = 90 – A

Similarly, because the sum of the measures of angle A and its supplement is 180 degrees, we can represent the measure of the supplement of A as 180 – A.

The problem states that the measure of the supplement of A is 40 degrees larger than twice the measure of the complement of A. We can write this as 2(90-A) + 40.

Next, we must set the two expressions 180 – A and 2(90 – A) + 40 equal to one another and solve for A:

180 – A = 2(90 – A) + 40

Distribute the 2:

180 - A = 180 – 2A + 40

Add 2A to both sides:

180 + A = 180 + 40

Subtract 180 from both sides:

A = 40

Therefore the measure of angle A is 40 degrees. 

The question asks us to find the sum of the measures of the supplement and complement of A. The measure of the supplement of A is 180 – A = 180 – 40 = 140 degrees. Similarly, the measure of the complement of A is 90 – 40 = 50 degrees.

The sum of these two is 140 + 50 = 190 degrees.

Since A and B are parallel, and the triangle is isosceles, we can use the supplementary rule for the two angles,  and  which will sum up to . Setting up an algebraic equation for this, we get . Solving for , we get . With this, we can get either  (for the smaller angle) or  (for the larger angle - must then use supplementary rule again for inner smaller angle). Either way, we find that the inner angles at the top are 80 degrees each. Since the sum of the angles within a triangle must equal 180, we can set up the equation as

 degrees.