Free SAT Math Geometry and Trigonometry

SAT Math · Geometry and Trigonometry · free lesson

Right-Triangle Trigonometry (SOH-CAH-TOA)

The idea, then worked examples solved step by step, then practice questions whose numbers change every time you try them.

The idea

Three ratios of a right triangle

Stand at one of the two acute angles of a right triangle and call it $\theta$ . From there the three sides have names. The hypotenuse is the longest side, across from the right angle. The opposite side is across from $\theta$ . The adjacent side touches $\theta$ and is not the hypotenuse.

Definition.

Sine, cosine and tangent of the angle $\theta$ are ratios of two sides:

$\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}$ , $\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}$ , $\tan\theta = \frac{\text{opposite}}{\text{adjacent}}$ .

The word SOH-CAH-TOA holds all three: Sine is Opposite over Hypotenuse, Cosine is Adjacent over Hypotenuse, Tangent is Opposite over Adjacent.

Take a right triangle with legs 8 and 15. The angle $\theta$ is across from the leg 8. The hypotenuse comes from the Pythagorean theorem, and then each ratio is one division:

\begin{aligned} c &= \sqrt{a^{2} + b^{2}} \\ &= \sqrt{8^{2} + 15^{2}} \\ &= 17 \\ sine &= \frac{a}{c} \\ &= \frac{8}{17} \\ cosine &= \frac{b}{c} \\ &= \frac{15}{17} \\ tangent &= \frac{a}{b} \\ &= \frac{8}{15} \end{aligned}

Each ratio is a single number with no unit. Sine and cosine of an acute angle are always less than 1, because the hypotenuse is the longest side.

The other acute angle

Move to the other acute angle and the names opposite and adjacent trade places. The hypotenuse stays. In the triangle above, the second angle is across from the leg 15:

\begin{aligned} sine &= \frac{b}{c} \\ &= \frac{15}{17} \\ cosine &= \frac{a}{c} \\ &= \frac{8}{17} \\ tangent &= \frac{b}{a} \\ &= \frac{15}{8} \end{aligned}

The sine of one acute angle is the cosine of the other, and the two tangents are reciprocals. Before you write a ratio, find the angle the question names and label the sides from there.

The ratio belongs to the angle, not to the triangle

Right triangles with the same acute angle are similar: one is the other made larger or smaller. Scaling multiplies every side by the same number, so each ratio of two sides stays as it was. Drag the size slider and watch the sides change while the three ratios do not. Then drag the angle slider and watch all three ratios change.

Rule.

If two right triangles are similar, matching angles have the same sine, the same cosine and the same tangent. A ratio such as $\frac{3}{5}$ tells you the shape only: the sides could be 3 and 5, or 6 and 10, or 30 and 50.

A side from an angle and a side

Because the ratio is fixed by the angle, one angle and one side give the other sides. A right triangle has a hypotenuse of 12 and an angle of $35^\circ$ . The side opposite that angle is $x$ . Sine joins opposite and hypotenuse, so $\sin 35^\circ = \frac{x}{12}$ , and multiplying both sides by 12 gives $x = 12\sin 35^\circ$ .

\begin{aligned} x &= 12\sin(\frac{35\pi}{180}) \\ &= 6.882917236 \end{aligned}

The engine measures angles in radians, so it is given 35 degrees as $\frac{35\pi}{180}$ (the next topic explains radians). On your own calculator, set degree mode and type 12 sin(35). The side is about $6.88$ .

Warning.

When the unknown side is on the bottom of the ratio, you divide. If the side of 12 is opposite the $35^\circ$ angle and the hypotenuse is unknown, then $\sin 35^\circ = \frac{12}{h}$ and $h = \frac{12}{\sin 35^\circ}$ , not $12\sin 35^\circ$ . The test puts both expressions among the options.

Heights and distances

A height that cannot be measured is found from a distance along the ground and an angle. A person stands 50 feet from a flagpole and looks up at its top at an angle of $28^\circ$ above level ground. The ground is adjacent to the angle and the pole is opposite, so tangent joins them.

\begin{aligned} h &= 50\tan(\frac{28\pi}{180}) \\ &= 26.58547158 \end{aligned}

The pole is about $26.6$ feet tall, measured from the level of the eye.

Note.

On the SAT these questions come in a few shapes: find a ratio from two sides (often after finding the third side), choose the expression for a side, carry a ratio from one similar triangle to another, or use a ratio to find real lengths. In every one, the first step is the same: mark the angle, then name the three sides from it.

Worked examples

Each example is one of the kinds of question the test asks about the ratios of a right triangle. The numbers in the working are found by the engine.

Example.

1. A ratio from two sides. In triangle $ABC$ the angle at $C$ is a right angle, $AC = 24$ and $BC = 7$ . What is $\sin A$ ?

Sine needs the side opposite $A$ and the hypotenuse. The side opposite $A$ is $BC$ . The hypotenuse $AB$ is not given, so find it first.

\begin{aligned} c &= \sqrt{b^{2} + a^{2}} \\ &= \sqrt{24^{2} + 7^{2}} \\ &= 25 \\ sine &= \frac{a}{c} \\ &= \frac{7}{25} \end{aligned}

So $\sin A$ is $\frac{7}{25}$ . The reciprocal and the ratio $\frac{7}{24}$ , which is $\tan A$ , are the wrong options to expect.

Example.

2. A side that is a radical. A right triangle has a hypotenuse of 9 and one leg of 6. The angle $\theta$ is opposite the leg of 6. What is $\tan\theta$ ?

Tangent is opposite over adjacent. The adjacent leg is the missing one, and it is a leg, so subtract under the root.

\begin{aligned} b &= \sqrt{c^{2} - a^{2}} \\ &= \sqrt{9^{2} - 6^{2}} \\ &= 3\sqrt{5} \\ tangent &= \frac{a}{b} \\ &= \frac{6}{3\sqrt{5}} \\ &= \frac{2\sqrt{5}}{5} \end{aligned}

The engine clears the root from the bottom of the fraction, as the test's options do: $\tan\theta$ is $\frac{2\sqrt{5}}{5}$ . The sine needed no root at all:

\begin{aligned} sine &= \frac{a}{c} \\ &= \frac{6}{9} \\ &= \frac{2}{3} \end{aligned}

Example.

3. Which expression gives the side? A wire runs from the top of a pole to a point on level ground 14 feet from the foot of the pole. The wire meets the ground at an angle of $40^\circ$ . Which expression gives the length of the wire: $14\cos 40^\circ$ or $\frac{14}{\cos 40^\circ}$ ?

Seen from the $40^\circ$ angle, the 14 feet of ground is the adjacent side and the wire is the hypotenuse. Cosine joins those two: $\cos 40^\circ = \frac{14}{\text{wire}}$ . The wire is on the bottom, so it equals 14 divided by the cosine.

\begin{aligned} w &= \frac{14}{\cos(\frac{40\pi}{180})} \\ &= 18.27570205 \\ wrong &= 14\cos(\frac{40\pi}{180}) \\ &= 10.7246222 \end{aligned}

The first number is the wire. The second is shorter than 14 feet, which a hypotenuse can never be: that is a quick way to throw out $14\cos 40^\circ$ .

Example.

4. Similar triangles. Triangles $ABC$ and $DEF$ are similar right triangles, with right angles at $C$ and $F$ and with $A$ matching $D$ . In triangle $ABC$ , $BC = 5$ and $AB = 13$ . Each side of $DEF$ is 3 times the matching side of $ABC$ . What is $\sin D$ ?

Find the two sides of $DEF$ that the sine uses, then divide.

\begin{aligned} EF &= ka \\ &= 3 \times 5 \\ &= 15 \\ DE &= kc \\ &= 3 \times 13 \\ &= 39 \\ sinD &= \frac{EF}{DE} \\ &= \frac{15}{39} \\ &= \frac{5}{13} \\ sinA &= \frac{a}{c} \\ &= \frac{5}{13} \end{aligned}

The factor 3 cancels, so $\sin D$ equals $\sin A$ . You never need the sides of the second triangle: matching angles of similar triangles have equal ratios.

Example.

5. From a ratio to real lengths. In a right triangle, $\tan\theta = \frac{8}{15}$ and the leg adjacent to $\theta$ is 45. What is the perimeter of the triangle?

The fraction gives the shape, not the sides. The adjacent leg is 45, not 15, so every side is the same number of times larger than in an 8, 15 triangle.

\begin{aligned} k &= \frac{b}{15} \\ &= \frac{45}{15} \\ &= 3 \\ a &= 8k \\ &= 8 \times 3 \\ &= 24 \\ c &= \sqrt{a^{2} + b^{2}} \\ &= \sqrt{24^{2} + 45^{2}} \\ &= 51 \\ P &= a + b + c \\ &= 24 + 45 + 51 \\ &= 120 \end{aligned}

The perimeter is $120$ . Adding 8, 15 and 17 would give the perimeter of the small model triangle, which is the trap.

Example.

6. An altitude to the hypotenuse. Triangle $ABC$ has a right angle at $C$ . Point $D$ is on $AB$ , and $CD$ is perpendicular to $AB$ . If $AD = 4$ and $CD = 6$ , what is $\tan B$ ?

The side $DB$ is not given, so $\tan B$ cannot be read from triangle $CDB$ directly. Use angles instead. In the large triangle, $A$ and $B$ add to $90^\circ$ . In the small right triangle $ADC$ , the angle $A$ and the angle $ACD$ add to $90^\circ$ too. So the angle $ACD$ equals $B$ , and the tangent of $B$ can be read in triangle $ADC$ : opposite $AD$ , adjacent $CD$ .

\begin{aligned} tanB &= \frac{AD}{CD} \\ &= \frac{4}{6} \\ &= \frac{2}{3} \end{aligned}

Check with lengths. Triangles $ADC$ and $CDB$ are similar, so $\frac{DB}{CD} = \frac{CD}{AD}$ , which gives $DB$ , and then $\tan B$ is $CD$ over $DB$ :

\begin{aligned} DB &= \frac{6 \times 6}{4} \\ &= 9 \\ tanB &= \frac{CD}{DB} \\ &= \frac{6}{9} \\ &= \frac{2}{3} \end{aligned}

Both ways give $\frac{2}{3}$ .

Practice

Answer SAT-style questions on right-triangle trigonometry (soh-cah-toa). Each answer is checked, with the full solution, and every try brings new numbers. Practice needs a free account: it keeps your progress.

Practise right-triangle trigonometry (soh-cah-toa) with new numbers