**Finding Angles**

**Intersecting Lines**

1. Consecutive angles are supplementary: Angles 1 and 2 add to 180^{o}.

2. Vertical angles are congruent: Angles 1 and 3 are congruent.

EXAMPLE 1: If angle *BPC* is four times as large as angle *CPD*, then what is the measure of angle *APB*?

SOLUTION: If *x* is the measure of angle *CPD*, then 4*x* is the measure of angle *BPC*:

Angles *BPC* and *CPD* are supplementary, so 4*x* + *x* = 180. Therefore 5*x* = 180, so *x* = 36. That is, angle *CPD* measures 36^{o}. Angles *APB* and *CPD* are vertical angles and therefore congruent. So angle *APB* measures 36^{o}.

EXAMPLE 2: Find the values of *x* and *y*:

SOLUTION:

Using vertical angles, we know that 2*x* – 8 = –2(4 – *x*)

The right side simplifies to –8 + 2*x* or 2*x* – 8 so this equation is an identity:

2*x* – 8 = 2*x* – 8

Any number makes this true, so we cannot find *x* directly.

Again by vertical angles, 2*y* + 4 = 3*y* – 64. This equation is not an identity and we can solve for *y*:

4 = *y* – 64

68 = *y*

So we know *y* = 68, and therefore the two angles containing *y*measure 2(68) + 4 = 140^{o}.

Since the angle marked 2*x* – 8 is supplementary to this angle, we now know

2*x* – 8 = 140, so 2*x* = 148, and therefore *x* = 74.

**Parallel Lines**

When two parallel lines are cut by a transversal, there are 8 pairs of congruent angles and 8 pairs of supplementary angles. This can be understood by identifying a pair of corresponding angles, and using vertical angles and linear pairs. Corresponding angles can be thought of as angles that match when a picture of one angle at the intersection of one of the parallel lines with the transversal is moved to the other parallel line:

The following summarizes angle congruences for two parallel lines cut by a transversal:

All other pairs of angles (that is, pairs marked with an even and an odd number) are supplementary.

EXAMPLE 3: Find *x* and *y* if lines *a* and *b* are parallel, and determine if lines *c* and *d* are parallel.

SOLUTION: Since *a* and *b* are parallel, 2*x* + 13 = 3*x* – 24 (they are alternate interior angles). Solving gives *x* = 37.

These angles measure 2 ** ^{.}** 37 + 13 = 87 degrees, and the angle marked 3

Solving this equation gives *y* = 23.

Then 4*y* – 5 = 92 – 5 = 87, which again is supplementary to 3*y* + 24.

Since these are same-side interior angles for lines *c* and *d* cut by transversal *a*, lines *c* and *d* are parallel.

**Parallelograms**

A parallelogram is a quadrilateral in which opposite sides are parallel. Pairs of interior angles are either congruent or supplementary:

Opposite angles are congruent (angles 1 and 3, and angles 2 and 4).

Consecutive angles are supplementary (angles 1 and 2, angles 2 and 3, angles 3 and 4, and angles 1 and 4).

EXAMPLE 4: Could the following figure be a parallelogram?

SOLUTION: If it is a parallelogram, then 5*x* = 3*x* + 46 since opposite angles are congruent.

Solving gives *x* = 24, so 5*x* = 120.

If it is a parallelogram, then 5*y* = 180 – 120 = 60 since consecutive angles are supplementary, so *y* = 12.

Therefore 3*y* + 28 = 36 + 28 = 64. But the angles marked 5*y* and 3*y* + 28 should be congruent, which is not the case.

Therefore this cannot be a parallelogram.

**Triangles**

The main results about triangles are these:

1. The angles of a triangle add to 180^{o}.

2. An exterior angle of a triangle equals the sum of the two remote interior angles:

3. The base angles of an isosceles triangle are congruent, and conversely:

4. The side opposite the smallest angle in a triangle is the shortest side and the side opposite the largest angle is the longest side, and conversely.

EXAMPLE 5:

One angle of a triangle is twice as large as another angle, and the third angle is 20^{o} less than the largest angle. What are the measures of the angles in this triangle?

SOLUTION: We can let *x* and 2*x* denote the measures of two angles.

Then 2*x* – 20 is the measure of the third angle.

Since the angles add to 180^{o}, we can write the following equation:

(*x*) + (2*x*) + (2*x* – 20) = 180

Solving: 5*x* – 20 = 180, 5*x* = 200, *x* = 40

The angles are:

*x* = 40^{o}

2*x* = 80^{o}

2*x* – 20 = 60^{o}

EXAMPLE 6: Find the measure of angle *ABC*:

SOLUTION:

The external angle is the sum of its two remote interior angles, so we can write the following equation:

4*x* + 24 = 2*x* + 3*x*

Solving, we find that *x*= 24.

Therefore the exterior angle measures 4 ** ^{.}** 24 + 24 = 120

Angle *ABC* is supplementary to this, so it measures 180 – 120 = 60^{o}.

EXAMPLE 7: Which side is longest in the following triangle?

SOLUTION: First we solve for *x*:

4*x* + 14 + 5*x* + 4 + 6*x* – 3 = 180

15*x* + 15 = 180

15*x* = 165

*x* = 11

Now we find the angles:

angle *A*: 4*x* + 14 = 58^{o}

angle *B*: 5*x* + 4 = 59^{o}

angle *C*: 6*x* – 3 = 63^{o}

Since angle *C* is the largest angle, side *AB* is the longest side.

**Polygons**

A polygon of *n* sides can be partitioned into *n* – 2 non-overlapping triangles with vertices in common with the polygon. For example, a quadrilateral has 4 sides and can be split into 2 triangles, and a triangle has 3 sides and consists of 1 triangle.

Since the angles of a triangle add to 180 degrees, the angles of an *n*-sided polygon must add to 180(*n* – 2) degrees.

EXAMPLE 8: Find the measure of angle *A* in the following pentagon:

SOLUTION:

A pentagon has 5 sides, so can be partitioned into 3 triangles:

Therefore its angles add to 3 times 180^{o}, so

(2*x* + 3) + (3*x*) + (*x* + 3) + (3*x* – 5) + 2*x* = 3 ** ^{.}** 180

11*x* + 1 = 540

11*x* = 539

*x* = 49

So angle *A* measures 2 ** ^{.}** 49 + 3 = 101

**Return to Lesson 3**