# SAT Lines and Angles

To successfully solve problems about lines and angles on the SAT, you need to know:

- angles on the SAT are always measured in degrees
- the types of angles
- angle arithmetic
- the definition of complementary and supplementary angles
- the definition of vertical angles
- the definition of an angle bisector
- the definition of a midpoint
- the segment addition postulate
- the angle addition postulate

## Examples for Lines

In the figure above, if \(AB=x-1, BC=2x+1, CD=x,\) and \(AD=12,\) what is the length of \(\overline{BD}?\)

(A) \(\ \ 2\)

(B) \(\ \ 3\)

(C) \(\ \ 7\)

(D) \(\ \ 9\)

(E) \(\ \ 10\)

Correct Answer: E

Solution:By the segment addition postulate, we have:

\[\begin{array}{r c l} AB + BC + CD &=& 12\\ x-1 + 2x + 1 + x &=& 12\\ 4x &=& 12\\ x &=&3\\ \end{array}\]

We are looking for \(BD.\)

\[BD = BC + CD = 2x+1 + x = 3x + 1 = 3\cdot 3 + 1 = 9 + 1 =10.\]

Incorrect Choices:

(A)

Tip: Read the entire question carefully.

Tip: If a diagram is drawn to scale, trust it.

If you solve for \(AB\) instead of \(BD,\) you will get this wrong answer. Note that we can eliminate this answer since the diagram is drawn to scale, and since \(BD\) seems to be greater than half of \(AD,\) or greater than 6.

(B)

Tip: Read the entire question carefully.

Tip: If a diagram is drawn to scale, trust it.

If you solve for \(x\) instead of \(BD,\) you will get this wrong answer. Note that we can eliminate this answer since the diagram is drawn to scale, and since \(BD\) seems to be greater than half of \(AD,\) or greater than 6.

(C)

Tip: Read the entire question carefully.

If you solve for \(BC\) instead of \(BD,\) you will get this wrong answer.

(D)

Tip: Read the entire question carefully.

If you solve for \(AC,\) instead of \(BD,\) you will get this wrong answer.

In the diagram above, points \(A\) and \(B\) have coordinates \(a\) and \(b,\) such that \(b>a.\) If \(M\) is the midpoint of segment \(\overline{AB},\) and \(M\) has the coordinate \(x,\) all of the following are true EXCEPT:

(A) \(\ \ AM = \frac{1}{2}AB\)

(B) \(\ \ MB = b-x\)

(C) \(\ \ \overline{AM} \cong \overline{MB}\)

(D) \(\ \ 2x = a+b\)

(E) \(\ \ x = 2b-a\)

## Examples for Angles

The ratio of \(m\angle AOB\) to \(m\angle BOC\) to \(m\angle COD\) is 1 : 3 : 2. What is the measure of \(\angle BOD?\)

(A) \(\ \ 30\)

(B) \(\ \ 90\)

(C) \(\ \ 100\)

(D) \(\ \ 120\)

(E) \(\ \ 150\)

Correct Answer: E

Solution 1:

Tip: Angles on a line sum to \(180^\circ.\)

If \(\angle AOD\) were divided into 1 + 3 + 2 = 6 parts, then \(\angle BOC\) would equal to 3 out of 6 parts, and \(\angle COD\) would equal to 2 out of 6 parts.Since \(m\angle AOD = 180^\circ,\) it follows that

\(m\angle BOC = \frac{3}{6} \cdot 180^\circ = 90^\circ\) and \(m\angle COD = \frac{2}{6} \cdot 180^\circ = 60^\circ.\)

Therefore \(m\angle BOD = m\angle BOC + m\angle COD = 90^\circ + 60^\circ = 150^\circ.\)

Solution 2:

Tip: Angles on a line sum to \(180^\circ.\)

Let \(m\angle AOB = x.\) Then, according to the given ratio, \(m \angle BOC = 3x\) and \(m\angle COD = 2x.\)\[\begin{array}{r c l} m \angle AOB + m \angle BOD + m \angle DOC &=& 180^\circ\\ x + 3x + 2x &=& 180^\circ\\ 6x &=& 180^\circ\\ x &=& 30^\circ\\ \end{array}\]

Therefore \(m\angle BOD = m\angle BOC + m\angle COD = 3x + 2x = 5x = 5\cdot 30^\circ = 150^\circ.\)

Solution 3:

Tip: If a diagram is drawn to scale, trust it.

It is evident from the diagram that \(\angle BOD\) is obtuse. Therefore, we can eliminate options (A) and (B).

We draw segment \(\overline{OE},\) perpendicular to \(\overline{AD}\) and we divide \(\angle AOE \) in two \(45^\circ\) angles. Because \(m\angle BOD > 90^\circ+45^\circ=135^\circ,\) we eliminate choices (C) and (D), and select choice (E) as the correct answer.

Incorrect Choices:

(A)

Tip: Read the entire question carefully.

This is \(m\angle AOB,\) not \(m \angle BOD.\)

(B)

Tip: Read the entire question carefully.

This is \(m\angle BOC,\) not \(m \angle BOD.\)

(C)

This answer choice is just meant to confuse you.

(D)

Tip: Read the entire question carefully.

This is \(m\angle AOC,\) not \(m \angle BOD.\)

## Review

If you thought these examples difficult and you need to review the material, these links will help:

## SAT Tips for Lines and Angles

- Angles at a point sum to \(360^\circ.\)
- Angles on a line sum to \(180^\circ.\)
- \(\angle A\) and \(\angle B\) are complementary if \(m\angle A + m\angle B=90^\circ.\)
- \(\angle A\) and \(\angle B\) are supplementary if \(m\angle A + m\angle B=180^\circ.\)
- Vertical angles are congruent.
- The angle bisector divides an angle in half.
- The midpoint of a segment divides it in half.
- If a diagram is drawn to scale, trust it.
- SAT General Tips

**Cite as:**SAT Lines and Angles.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/sat-lines-and-angles/