SAT skill guides / Geometry and Trigonometry

How to solve SAT lines, angles, and triangles questions

This skill covers angle relationships formed by intersecting and parallel lines, the angle and side properties of triangles, and similar and congruent triangles. On the Digital SAT these questions usually ask you to find an angle measure or a side length, often by setting up a short equation from a diagram. Geometry and Trigonometry is about 15% of the Math section, or 5 to 7 of the 44 questions. Those questions are split across four skills, so expect 1 or 2 questions on this skill per test.

What the questions look like

Angle chasing. Two parallel lines are cut by a transversal, or several lines meet at a point, and the angles are labeled with expressions like (3x+10)∘(3x + 10)^\circ. You find xx or one angle measure.

Triangle angle facts. You use the triangle angle sum, the exterior angle rule, or isosceles triangle properties to find a missing angle. Some questions extend this to the interior angles of a polygon.

Similar triangles. Two triangles share angles, often because a segment inside a triangle is parallel to one side. You set up a proportion to find a side length.

Step-by-step method

  1. Mark everything you know on the figure. Write given angles and lengths directly on it. Don't assume lines are parallel or angles are equal unless the question says so or the markings show it. Figures are not always drawn to scale.
  2. Apply angle relationships.
    • Angles on a straight line add to 180∘180^\circ.
    • Vertical angles are equal.
    • When parallel lines are cut by a transversal, corresponding angles and alternate interior angles are equal, and same-side interior angles add to 180∘180^\circ.
  3. Apply triangle facts.
    • The angles of a triangle add to 180∘180^\circ.
    • An exterior angle equals the sum of the two interior angles that are not next to it.
    • In an isosceles triangle, the angles opposite the equal sides are equal.
    • The interior angles of an nn-sided polygon add to (n−2)⋅180∘(n - 2) \cdot 180^\circ.
  4. For similar triangles, match vertices by angles. Write the similarity in order, such as △ADE∼△ABC\triangle ADE \sim \triangle ABC. Corresponding sides are in the same positions, so ADAB=DEBC=AEAC\frac{AD}{AB} = \frac{DE}{BC} = \frac{AE}{AC}.
  5. Write one equation and solve it.

Worked example

Example 1. Two parallel lines are cut by a transversal. Two same-side interior angles measure (3x+10)∘(3x + 10)^\circ and (5x−22)∘(5x - 22)^\circ. What is the value of xx?

Same-side interior angles formed by parallel lines are supplementary, so they add to 180∘180^\circ:

(3x+10)+(5x−22)=180(3x + 10) + (5x - 22) = 180
8x−12=1808x - 12 = 180
x=24x = 24

Check: the angles are 3(24)+10=82∘3(24) + 10 = 82^\circ and 5(24)−22=98∘5(24) - 22 = 98^\circ, which add to 180∘180^\circ. If you had set the expressions equal instead, you would get x=16x = 16, a likely wrong choice.

Example 2. In triangle ABCABC, point DD lies on AB‾\overline{AB} and point EE lies on AC‾\overline{AC}, with DE‾\overline{DE} parallel to BC‾\overline{BC}. If AD=6AD = 6, DB=4DB = 4, and DE=9DE = 9, what is BCBC?

Because DE‾∥BC‾\overline{DE} \parallel \overline{BC}, angle ADEADE equals angle ABCABC as corresponding angles, and the two triangles share angle AA. So △ADE∼△ABC\triangle ADE \sim \triangle ABC.

The side of the large triangle that matches ADAD is the whole segment AB=6+4=10AB = 6 + 4 = 10. Set up the proportion:

ADAB=DEBC⇒610=9BC\frac{AD}{AB} = \frac{DE}{BC} \quad \Rightarrow \quad \frac{6}{10} = \frac{9}{BC}
6⋅BC=90⇒BC=156 \cdot BC = 90 \quad \Rightarrow \quad BC = 15

Common traps

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