Use sine, cosine, and tangent

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Difficulty: Intermediate
Estimated time: 30 minutes
Domains: Geometry and Trigonometry
Techniques: Trigonometric-ratiosSide-rolesInverse-trigonometryComplementary-anglesReference-angles

What you’ll learn

Name the opposite side, adjacent side, and hypotenuse from any acute angle.
Pick sine, cosine, or tangent from the sides you know and need.
Write one ratio equation, then let Desmos finish the arithmetic in Degrees mode when the angle isn’t special.
Use $\sin A=\cos(90^\circ-A)$ for complementary angles.
Find sine, cosine, and tangent from a point’s coordinates or from a standard angle like $150^\circ$ .
Tell when an exact special-triangle ratio is quicker than the calculator.

Prerequisites

Why this matters on the SAT

Choose the ratio that connects what you know to what you need

An SAT question might give you one acute angle and one side of a right triangle, then ask for another side. The reference sheet won’t help: it doesn’t list sine, cosine, or tangent. So you need to know which side is which, and which ratio links them. Here’s a typical one.

Solution to the example

Stand at $\angle A$ and look at the sides. $AC=12$ touches the angle, so it’s adjacent, and $AB=x$ is the hypotenuse. Cosine is the ratio that uses those two:

\cos37^\circ=\frac{\text{adjacent}}{\text{hypotenuse}} =\frac{12}{x}.

Multiply both sides by $x$ , then divide by $\cos37^\circ$ :

x=\frac{12}{\cos37^\circ}.

The answer is D. Once you picked cosine, the algebra took one line. Picking the ratio was the real work.

Choice B comes from multiplying instead of dividing, and you can rule it out fast. $\cos37^\circ$ is less than $1$ , so $12\cos37^\circ$ is shorter than $12$ . But the hypotenuse is the longest side of a right triangle.

SAT example

Triangle

ABC

has a right angle at

C

\angle A=37^\circ

, side

AC=12

, and side

AB

labeled

x

In right triangle $ABC$ , $\angle C=90^\circ$ , $\angle A=37^\circ$ , and $AC=12$ . Which choice represents the length $x$ of hypotenuse $\overline{AB}$ ?

A
$12\sin37^\circ$
B
$12\cos37^\circ$
C
$\dfrac{12}{\tan37^\circ}$
D
$\dfrac{12}{\cos37^\circ}$

Name the sides from the angle

The hypotenuse never changes. It’s always the side across from the $90^\circ$ angle.

The other two names depend on which acute angle you’re working from:

The opposite side is across from your angle.
The adjacent side is the leg that touches your angle.

Switch angles and the two legs trade names. The hypotenuse stays the hypotenuse.

So before you write any ratio, mark the angle you’re working from. The same side can be adjacent to one acute angle and opposite the other.

Check your understanding:

In the right panel, two sides touch $\angle B$ . Why is the vertical leg the adjacent side, and not the slanted one?

Common mistake:

Calling every side that touches your angle adjacent. The hypotenuse touches both acute angles too. To avoid the mix-up, find the hypotenuse first, across from the right angle. The other side that touches your angle is adjacent.

Choose sine, cosine, or tangent

Each of the three ratios compares two sides of a right triangle:

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

The memory aid SOHCAHTOA packs all three into one word:

SOH: sine is opposite over hypotenuse.
CAH: cosine is adjacent over hypotenuse.
TOA: tangent is opposite over adjacent.

It only helps once you’ve named the sides, so name them first.

To pick a ratio, find the one that uses both the side you know and the side you need. In the SAT example, you knew the adjacent side and needed the hypotenuse, and only cosine uses both.

Sometimes the question asks for the ratio itself, like $\tan A$ , and gives you two sides. Then you’re done once you write that fraction and simplify it. A trig value is a number, not a side length.

Keep the thinking separate from the arithmetic:

Reason first. Name the sides and pick the ratio by hand.
Let Desmos do the arithmetic when it helps. If the angle isn’t special and you need a decimal, rearrange once and have Desmos evaluate the whole expression in Degrees mode.
Check the answer. Round only as the question asks, include units, and make sure you found the side or angle it asked for.

Check your understanding:

From angle $\theta$ , you know the opposite side and need the adjacent side. Which ratio should you use, and why?

Example: Find a cable length

Worked example

A straight cable runs from

P

on the ground to a point

18

meters up the tower and makes a

41^\circ

angle with the ground.

A vertical tower stands on level ground. A straight cable runs from point $P$ on the ground to an attachment point $18$ meters above the ground. The cable makes a $41^\circ$ angle with the ground at $P$ . What is the length of the cable, to the nearest tenth of a meter?

Step 1

Find the right angle

The tower is vertical and the ground is level, so they meet at a right angle. With the cable, that makes a right triangle.

Step 2

Name the sides from the angle

From the $41^\circ$ angle at $P$ :

the tower’s $18$ meters is across from the angle, so it’s opposite;
the cable $c$ is across from the right angle, so it’s the hypotenuse.

Sine is the ratio that uses opposite and hypotenuse.

Step 3

Write the ratio equation

Put in the angle and the side:

\sin41^\circ=\frac{18}{c}.

Write this out before you rearrange anything. It’s the line that shows you picked the right ratio.

Step 4

Select Degrees, then calculate

Multiply both sides by $c$ , then divide by $\sin41^\circ$ :

c=\frac{18}{\sin41^\circ}.

The $41^\circ$ has a degree symbol, so Desmos has to be in Degrees before you trust its answer:

Open Graph Settings.
Select Degrees.
Evaluate the whole expression, 18/sin(41), in one go. It’s already typed into the calculator below.
Keep every decimal Desmos shows until you round at the end.

In Degrees, Desmos gives

c\approx27.4365.

To the nearest tenth, the cable is

\boxed{27.4\text{ meters}}.

A quick check: it’s in meters, as asked, and it’s longer than the $18$ -meter tower, as a hypotenuse has to be.

Calculator loads as you approach

Degrees isn’t selected for you, so open Graph Settings and select it first. Then this should read about 27.4365. Try changing the angle or the height to see how the cable changes.

Use degree mode, and find missing angles

Why does the mode matter so much? Desmos can measure angles in degrees or in radians, a different unit. In radians, it reads sin(41) as $41$ radians, a completely different angle, and gives a different answer.

The mode matters just as much when you run a ratio backward. When you know two sides and need an angle, use an inverse trig function. Say the opposite and adjacent sides are $7$ and $15$ . Then

\tan\theta=\frac{7}{15}.

Inverse tangent turns that ratio back into the angle:

\theta=\tan^{-1}\left(\frac{7}{15}\right)\approx25.0^\circ.

Here $\tan^{-1}$ means “the angle whose tangent is $\frac7{15}$ .” The $-1$ looks like an exponent, but it doesn’t mean $\frac{1}{\tan\theta}$ .

In Desmos, select Degrees and enter arctan(7/15). This time the mode sets the unit of the answer: Degrees gives about $25.0$ , while radians would give about $0.44$ .

Calculator loads as you approach

Select Degrees first. Then this should read about 25.0 degrees.

Common mistake:

Typing a degree angle while Desmos is in radians. For the cable, radians turn 18/sin(41) into about $-113.5$ , a negative length. Wrong-mode answers don’t always look this strange, though, and right ones can look unfamiliar. So set the mode from the question before you type, not from the answer afterward: a degree symbol means Degrees.

Check your understanding:

From angle $\phi$ , the opposite side is $9$ and the hypotenuse is $20$ . What equation gives $\phi$ , and which calculator mode do you need for an answer in degrees?

Use complementary angles

The two acute angles in a right triangle add to $90^\circ$ . Two angles that add to $90^\circ$ are called complementary.

Look back at the side-roles figure. The side opposite $\angle A$ is the side adjacent to $\angle B$ , and the hypotenuse is the same for both. So

\sin A =\frac{\text{opposite to }A}{\text{hypotenuse}} =\frac{\text{adjacent to }B}{\text{hypotenuse}} =\cos B.

Since $B=90^\circ-A$ ,

\boxed{\sin A=\cos(90^\circ-A)}.

It works the other way too:

\cos A=\sin(90^\circ-A).

This is called a cofunction relationship: the sine of one acute angle equals the cosine of its complement. You don’t need any side lengths to use it.

Check your understanding:

Acute angles $R$ and $S$ are complementary, and $\sin R=\frac{\sqrt{21}}5$ . What is $\cos S$ ?

Common mistake:

Swapping sine for cosine without checking the angles. The swap only works when the two angles add to $90^\circ$ . For example, $\sin30^\circ=\frac12$ but $\cos30^\circ=\frac{\sqrt3}2$ , because $30^\circ+30^\circ$ isn’t $90^\circ$ . Check the sum first, then switch sine to cosine, or cosine to sine, and keep the value the same.

Find trig values from coordinates

Sometimes the SAT gives you a point instead of a triangle. Take $P(-3,4)$ in the figure below. Draw a line from the origin out to $P$ , then drop straight down from $P$ to the $x$ -axis. That makes a right triangle with legs $3$ and $4$ .

The triangle gives the sizes

3

4

, and

5

. The quadrant gives the signs.

The long side is $P$ ’s distance from the origin, called $r$ :

r=\sqrt{(-3)^2+4^2}=5.

Now build the ratios from the coordinates themselves, signs and all: sine is $y$ over $r$ , cosine is $x$ over $r$ , and tangent is $y$ over $x$ . The $x$ -coordinate is negative, so cosine and tangent come out negative:

\sin\theta=\frac45, \qquad \cos\theta=-\frac35, \qquad \tan\theta=-\frac43.

Here are the names for what you just used. The angle $\theta$ starts on the positive $x$ -axis, which puts it in standard position, and it ends at the ray through $P$ , its terminal ray. The triangle is the reference triangle, and its acute angle $\alpha$ at the origin, between the terminal ray and the $x$ -axis, is the reference angle.

The same rules work for any point $P(x,y)$ on the terminal ray except the origin:

r=\sqrt{x^2+y^2}, \qquad \sin\theta=\frac{y}{r}, \qquad \cos\theta=\frac{x}{r}.

Tangent is $\frac{y}{x}$ as long as $x\ne0$ . When $x=0$ , the ray lies on the $y$ -axis and tangent is undefined, because $\frac{y}{x}$ would mean dividing by zero.

Whenever the terminal ray is off the axes, you get a triangle like the one for $P(-3,4)$ . Its sides give the size of each ratio, always positive, and the coordinates give the signs. When the ray lies on an axis, called a quadrantal angle, there’s no triangle to draw, so use $\frac{y}{r}$ and $\frac{x}{r}$ directly for sine and cosine.

The reference angle also gives exact values for a standard angle like $150^\circ$ , with no unit circle to memorize. The terminal ray of $150^\circ$ stops $30^\circ$ short of the negative $x$ -axis, so the reference angle is

180^\circ-150^\circ=30^\circ.

The $30^\circ$ special triangle gives the sizes:

\left|\sin150^\circ\right|=\frac12 \quad\text{and}\quad \left|\cos150^\circ\right|=\frac{\sqrt3}{2}.

The terminal ray is in quadrant II, where $y$ is positive and $x$ is negative. So

\sin150^\circ=\frac12 \quad\text{and}\quad \cos150^\circ=-\frac{\sqrt3}{2}.

Stick with these exact values. A calculator would give decimals, which take longer and are less exact.

Check your understanding:

An angle’s terminal ray passes through $(5,-12)$ . Without finding the angle, what are the signs of sine, cosine, and tangent?

Choose the shortest method that works

Let what the question gives you pick the method:

With one side and an angle that isn’t special, like the cable’s $41^\circ$ , and a decimal side to find, set up the ratio and let Desmos finish it in Degrees.
With two sides and a missing angle, use an inverse, like $\tan^{-1}\left(\frac7{15}\right)$ , in the question’s angle unit.
With two sides and a missing side, the Pythagorean theorem is usually quicker than trig.
With a $30^\circ$ - $60^\circ$ - $90^\circ$ or $45^\circ$ - $45^\circ$ - $90^\circ$ triangle, use its exact side ratios when they lead straight to the answer.
With a point or a standard angle like $150^\circ$ , use $y/r$ , $x/r$ , or $y/x$ with the quadrant’s signs.

Say a right triangle has a $30^\circ$ angle and hypotenuse $18$ . The side opposite $30^\circ$ is half the hypotenuse, so it’s $9$ . Typing $18\sin30^\circ$ into Desmos gets $9$ too, but the special triangle gets there faster.

Try it yourself:

Pick a method for each one before you calculate anything: (1) legs $8$ and $15$ , find the hypotenuse; (2) an angle of $43^\circ$ and adjacent side $10$ , find the opposite side; (3) a $45^\circ$ - $45^\circ$ - $90^\circ$ triangle with leg $7$ , find the exact hypotenuse.

Answer: (1) The Pythagorean theorem, since you have two sides and need the third. (2) Tangent, because it links opposite and adjacent. (3) The exact $45^\circ$ - $45^\circ$ - $90^\circ$ ratio, which gives $7\sqrt2$ .

Practice problems

Mark the angle, name the sides, and pick a method before you calculate anything.

Scale up a tangent ratio

Practice problem

In right triangle $ABC$ , $\angle C=90^\circ$ , $\tan A=\frac{8}{15}$ , and $AC=45$ . What is the length of hypotenuse $\overline{AB}$ ?

Calculator loads as you approach

This one works out exactly by hand: scale the ratio, then use the Pythagorean theorem.

Find a height in degree mode

Practice problem

From a point on level ground $14$ meters from the base of a vertical tree, the angle of elevation to the top of the tree, meaning the angle up from the ground to the top, is $52^\circ$ . To the nearest tenth, what is the height of the tree, in meters?

Your answer

Calculator loads as you approach

Set up tangent by hand, then let Desmos finish in Degrees.

Use coordinates and quadrant signs

Practice problem

The terminal ray of angle $\theta$ in standard position passes through the point $(-8,15)$ . Which choice gives the ordered pair $\left(\sin\theta,\cos\theta\right)$ ?

Calculator loads as you approach

No decimals needed here: find

r

, then use the signs of the coordinates.

Finish the lesson

3 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

Find the hypotenuse first, across from the right angle. Then name opposite and adjacent from your angle.
SOHCAHTOA: pick the ratio that uses the side you know and the side you need.
Set up the ratio by hand. For a decimal with an angle that isn’t special, let Desmos evaluate the whole expression in Degrees.
Inverse trig turns two sides into an angle, in the question’s angle unit.
Complementary angles swap sine and cosine: $\sin A=\cos(90^\circ-A)$ and $\cos A=\sin(90^\circ-A)$ .
From a point $(x,y)$ , $\sin\theta=\frac yr$ and $\cos\theta=\frac xr$ , and $\tan\theta=\frac yx$ , which is undefined when $x=0$ . Off the axes, the triangle gives the size and the quadrant gives the sign. On an axis, use the coordinate formulas directly.
When a $30^\circ$ , $45^\circ$ , or $60^\circ$ triangle takes you straight to the answer, use its exact ratios.
Finish by rereading the question: give what it asks for, with units, rounded only as it says.

Next lesson

Use similarity inside right triangles

Use the similar triangles created by an altitude to a right triangle’s hypotenuse.

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316 SAT questions use what this lesson teaches. Practice a few in a study session at the difficulty you choose.

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