AOPS Pre-Algebra Chapter 10.1

Concepts

Points, Segments, Lines, Rays, Angles

Depiction of a point. Normally labeled with capital letters:

Line segment or segment: straight path from one point to another.

Line: a segment that continues forever in both directions.

Ray: Segment that continues forever past only one endpoint:

Angles: made by two intersecting lines or rays that share an origin:

Types of Angles

Right angles are 90 degrees and depicted with a little box.

Acute angles are less than 90 degrees and obtuse angles are greater than 90 degrees but less than 180°.

Two angles that add to 180° are supplementary. Two angles that add to 90° are complementary.

Vertical angles are opposite each other and have the same measure.

Congruent angles have the same measure.

In-Chapter Problems

✅ Problem 10.1

I used a Mac protractor app. I got 90° for $\angle AOB$, 80° for $\angle CXD$, 37° for $\angle DXE$. and 117° for $\angle CXE$.

They basically agreed but had 36° for $\angle DXE$

✅ Problem 10.2

a) The minute hand would be pointing to 12 at the top of the clock. The hour hand would be pointing to 5. 5 is 5/12 of a rotation through the clock. 360° are in a full rotation. So the hour hand and the minute hand have an angle of $\frac{5}{12} \times 360°$ or 150° between them in the smaller angle.

b) The minute hand will be at the position indicating 24 minutes, which is a bit before 5 on the clock. picture above. It will be 4/10 through a rotation around the clock. So, starting from the top of the clock, it will have rotated $\frac{4}{10} \times 360°$ or 144°.

The hour hand will be past the 5 since it will keep gradually moving as the hour passes. There are 360 degrees of rotation in the clock. There are 10 increments of 6 minutes in an hour and 12 hours which are represented on the clock. So one way to think about things is that the hour hand moves $\frac{1}{120}$ of a rotation around the clock for every six minutes that passes. $\frac{1}{120} \times 360° = 3°$. So the hour hand moves 3 degrees for every 6 minutes that passes. So it moves 150 degrees for the 5 hours from 12pm to 5pm, and then it moves an additional 12 degrees for the 24 minutes from 5:00pm to 5:24pm. So it moves a total of 162°.

So the measure of the smaller angle between the hour and minute hands of a clock at 5:24pm is 162° - 144° or 18°.

✅ Problem 10.3

The sum of the angles has to be 360°, so $x = 360° - (64° + 80° + 143°)$

$x = 73$.

✅ Problem 10.4

180°.

They explain this a bit:

✅ Problem 10.5

$\angle SOP$ is 37°. $\overline{ROS}$ is a straight line where the measure of $\angle ROS$ is 180°. So $\angle POR = 180 - 37°$ or 143°.

✅ Problem 10.6

$\angle WPY$ has to be supplementary (sum to 180°) to $\angle YPX$ since they're both on a straight line. So $\angle WPY$ has to be 119°.

$\angle WPZ$ has to be supplementary to $\angle WPY$ since they're both on a straight line. So $\angle WPZ$ has to be 61°.

By this analysis we can see what the idea of vertical angles is based on.

✅ Problem 10.7

Because $\angle WPZ$ and $\angle YPX$ are supplementary to the same angle, $\angle WPY$, so they must have the same value in order to both sum up to 180° with $\angle WPY$.

✅ Problem 10.8

a & b)

If x is the bigger angle, which is how they seem to want to set it up, then we have this:

In that case we add x and $\frac{x}{3}$, and get $\frac{4x}{3} = 180°$, which results in $x = 135°$ and $\frac{x}{3} = 45°$

If x is the smaller angle, which I find a more intuitive setup, then we have this:

And so $4x = 180°$ and $x = 45°$ and $3x = 135°$.

So in either case we correctly figure out that the big angle is 135° and the small angle is 45°.

NOTE: They actually wanted to go with the 3x / x setup. I read the part of part b that says "Let $x$ be the measure of the first angle mentioned in the problem." as referring to "The measure of one angle formed by two intersecting lines is three times the measure of another angle formed...". So the first angle mentioned in the problem was the big one, and it seemed like they wanted that to be $x$.

c) the opposite angles of each angle will be the same. So

Exercises

✅ 10.1.1

62°.

✅ 10.1.2

$\frac{360°){5} = 72°$

✅ 10.1.3

I zoomed in and got 43° using my computer protractor. They got 42°. Close enough!

✅ 10.1.4

$2x + 10 = 180°$

$2x = 170°$

$x = 85°$

✅ 10.1.5

Let's say the smaller angle is x. The ratio of the measures of the smaller and larger angles being 1:5 means that the total combined value of the angles $\angle ABX$ and $\angle CBX$ is $x + 5x = 6x$, so $6x = 90°$.

$x = 15°$. So the smaller angle's measure is 15°.

✅ 10.1.6

The supplement of an angle is the the angle you combine the first angle with to get 180°.

Let's call the angle that's 15 more than double the supplement x and the supplement y.

The problem thus gives us the following information:

$x = 2y + 15$

$x + y = 180°$

So we can substitute:

$(2y + 15) + y = 180°$

$3y = 165°$

$y = 55$

and $x = 125°$. So the answer is 125°. 125° is 15° more than twice its supplement (which is 55°) so it all checks out.

They solved it a bit differently. They just made the supplement x.

✅ 10.1.7

342°. $360° - 18°$.

✅ 10.1.8

125 clerts.

(90° is $\frac{1}{4} \times 360°$. 125 clerts is $\frac{1}{4} \times 500 clerts$.

✅ 10.1.9

$\overline{FC}$ is a straight line. $\angle COB$ and $\angle BOF$ need to add up to 180°. To have a 7:2 ratio of the angles, $\angle COB$ needs to be 140° and $\angle BOF$ needs to be 40°.

$\overline{OD}$ splits $\angle COE$ into two equal angles. $\angle BOF$ is opposite $\angle COE$ and needs to have the same measure as $\angle BOF$. We know that $\angle BOF$ has a measure of 40° so $\angle COE$ also has a measure of 40°. $\angle COD$ and $\angle DOE$ are thus 20° each. And therefore we have our answer which is that $\angle COD$ is 20°.

✅10.1.10

Some facts we know:

$\angle ABP = \angle PBQ = \angle CBQ$

$\angle ABM = \angle MBC$

$\angle MBQ = 28°$

Let's say we want to figure out $\angle MBP$. We know that $\angle ABM$ is one half of $\angle ABC$, and we know that $\angle ABP$ is one third of $\angle ABC$. We can see from the picture that $\angle ABM$ is made up of $\angle ABP$ plus $\angle PBM$. So we can subtract $\angle ABP$ from $\angle ABM$ to get $\angle PBM$, which is 1/6 of $\angle ABC$.

$$\angle ABM - \angle ABP = \angle PBM$$

We can use the same reasoning for $\angle MBQ$. $\angle CBM$ is one half of $\angle ABC$ and $\angle CBQ$ is one third of $\angle ABC$ so we can subtract them and get $\angle MBQ$ which is 1/6 of $\angle ABC$.

We know that $\angle PBM$ and $\angle MBQ$ are the same proportion of $\angle ABC$, so they have to be equal to each other. And we know that the value of $\angle MBQ$ is 28°. So $\angle PBM$ also has to be 28°. So the total of $\angle PBQ$, which is one third of $\angle ABC$, is 56°.

To get the answer, we need to find $\angle CBP$. To get that, we need to figure out the value of $\angle CBQ$. We already know that. We have the value of $\angle PBQ$, which is 56°. $\angle PBQ$ is equal to $\angle CBQ$. So $\angle CBQ$ is 56° and $\angle CBP$ is 112°.

How they explained it: