How to solve SAT right triangle and trigonometry questions

Right triangle questions on the Digital SAT ask you to find missing sides and angles with the Pythagorean theorem, special right triangles, and the trig ratios sine, cosine, and tangent. A few also test the link between sine and cosine of complementary angles and the conversion between degrees and radians. 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. The reference sheet gives the Pythagorean theorem and both special right triangles.

What the questions look like

Missing side. You find a side length using the Pythagorean theorem, a Pythagorean triple, or a $45^\circ$-$45^\circ$-$90^\circ$ or $30^\circ$-$60^\circ$-$90^\circ$ triangle. Answers are often exact values with square roots.

Trig ratios. You are given two sides and asked for $\sin$, $\cos$, or $\tan$ of an angle, or given a ratio such as $\cos P = \frac{8}{17}$ and a side, and asked for another side.

Complementary angles and radians. You use $\sin x^\circ = \cos(90 - x)^\circ$, or convert an angle between degrees and radians.

Step-by-step method

1. Find the right angle and the hypotenuse. The hypotenuse is the side across from the right angle and is always the longest side.
2. Label the legs relative to the angle you care about. The opposite leg is across from that angle. The adjacent leg touches it and is not the hypotenuse.
3. Use SOH CAH TOA.
   • $\sin \theta = \dfrac{\text{opposite}}{\text{hypotenuse}}$
   • $\cos \theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}$
   • $\tan \theta = \dfrac{\text{opposite}}{\text{adjacent}}$
4. Look for shortcuts before computing.
   • Common triples: $3, 4, 5$ and $5, 12, 13$ and $8, 15, 17$ and $7, 24, 25$, plus their multiples.
   • $45^\circ$-$45^\circ$-$90^\circ$: sides $x$, $x$, $x\sqrt{2}$.
   • $30^\circ$-$60^\circ$-$90^\circ$: sides $x$ opposite $30^\circ$, $x\sqrt{3}$ opposite $60^\circ$, and $2x$ for the hypotenuse.
5. Use complements. The two acute angles of a right triangle add to $90^\circ$, so the sine of one equals the cosine of the other. If $\sin a^\circ = \cos b^\circ$ for acute angles, then $a + b = 90$.
6. Convert angle units when needed. Multiply degrees by $\frac{\pi}{180}$ to get radians, and multiply radians by $\frac{180}{\pi}$ to get degrees.

Worked example

Example 1. In right triangle $PQR$, angle $Q$ is the right angle, $\cos P = \frac{8}{17}$, and $PR = 34$. What is the length of $\overline{QR}$, and what is $\sin R$?

Since angle $Q$ is the right angle, $PR$ is the hypotenuse. For angle $P$, the adjacent leg is $PQ$:

$$\cos P = \frac{PQ}{PR} = \frac{8}{17} \quad \Rightarrow \quad PQ = \frac{8}{17}(34) = 16$$

Use the Pythagorean theorem for $QR$:

$$QR = \sqrt{34^2 - 16^2} = \sqrt{1156 - 256} = \sqrt{900} = 30$$

This is the $8, 15, 17$ triple doubled. For angle $R$, the opposite leg is $PQ$, so

$$\sin R = \frac{PQ}{PR} = \frac{16}{34} = \frac{8}{17}$$

Notice that $\sin R = \cos P$. Angles $P$ and $R$ are complementary, so this always happens.

Example 2. A ladder leans against a vertical wall and makes a $60^\circ$ angle with the level ground. The top of the ladder touches the wall 9 feet above the ground. How long is the ladder?

The wall, the ground, and the ladder form a $30^\circ$-$60^\circ$-$90^\circ$ triangle. The 9-foot height is across from the $60^\circ$ angle, so it is the $x\sqrt{3}$ side:

$$x\sqrt{3} = 9 \quad \Rightarrow \quad x = \frac{9}{\sqrt{3}} = 3\sqrt{3}$$

The ladder is the hypotenuse, $2x = 6\sqrt{3}$ feet, which is about 10.39 feet.

Common traps

• Mixing up opposite and adjacent. These labels depend on which angle you are using. In Example 1, $PQ$ is adjacent to angle $P$ but opposite angle $R$. Relabel every time the angle changes.
• Swapping the $30^\circ$-$60^\circ$-$90^\circ$ sides. The shortest side is across from $30^\circ$, and the $x\sqrt{3}$ side is across from $60^\circ$. Putting 9 on the short side in Example 2 gives a ladder of 18 feet, a likely wrong choice.
• Using the wrong calculator mode. If you compute $\sin 30$ in radian mode, you get about $-0.988$ instead of $0.5$. Set degree mode for degree problems.
• Assuming a right angle. Only use SOH CAH TOA and the Pythagorean theorem when the triangle has a marked or stated right angle.
• Getting the complement equation wrong. If $\sin(2k + 10)^\circ = \cos(3k - 5)^\circ$, the angles add to $90^\circ$: $5k + 5 = 90$, so $k = 17$. Setting the two angles equal is the trap.

Desmos tip

Desmos starts in radian mode. Open the settings with the wrench icon and select Degrees before evaluating trig functions in a degree problem. Then $\sin(30)$ returns 0.5, and inverse functions such as $\cos^{-1}(8/17)$ return an angle in degrees, about $61.93^\circ$. Switch back to radians if a question gives angles in terms of $\pi$.