Right-triangle trigonometry

SOH—CAH—TOA

Shhhhh the ... mathematicians don't like this too much, but for most of you, about 80% of all you'll ever need to know about trig. is in that mnemonic: SOH-CAH-TOA.

Most of the trigonometry you'll ever need to do involves right triangles, though we'll generalize this to any triangle, and do a lot of other interesting things later. Take a look at the right triangles on the right so we can be on the same page with our terms.

Right triangles have a right angle, of course, and a hypotenuse across from that. We'll define an angle, θ (the Greek letter "theta"), and our choice of θ will determine the labels of the other two sides. The opposite side (o) is opposite angle θ, and the side that is a part of theta will be called the adjacent side (a).

The trig functions have limits

The trigonometric functions can't take on just any values — they have limits. Look at this right triangle, for example:

In this triangle, the angle is very small, making the opposite side small. That means that the sine of $\theta$ will be small and the cosine of $\theta$ will be close to 1.

$$ \begin{align} sin(\theta) &= \frac{o}{h} \; \rightarrow \frac{0}{h} \; \rightarrow 0 \\[5pt] cos(\theta) &= \frac{a}{h} \; \rightarrow \frac{a}{h} \; \rightarrow 1 \\[5pt] tan(\theta) &= \frac{o}{a} \; \rightarrow \frac{0}{a\approx h} \; \rightarrow 0 \end{align}$$

Notice that as the angle $\theta$ gets close to zero, the sine ratio gets closer to zero, and is exactly zero when $\theta = 0$. Likewise, as $\theta$ gets close to zero, the lengths of the adjacent side and the hypotenuse approach being the same length, so the cosine ratio approaches 1. Finally, as $\theta$ approaches zero, the tangent ratio, opposite divided by adjacent, also approaches zero. These are hard limits on our three trig functions as $\theta \; \rightarrow 0$.

Example 1

Calculate the measures of the missing sides and angles of a right triangle.

Solution: The three trig. functions allow us to determine everything there is to know about a right triangle starting with very little initial information. In the triangle below, only one side and one angle (and the right angle) are known.

The angle (θ = 25˚) opens onto the opposite side and the adjacent side is part of that angle. The hypotenuse is the unique side of a right triangle; it is not a part of the right angle.

That's plenty of information, though, to find all of the missing information.

To begin, just write down one of the trigonometric relationships — let's choose sine — and rearrange it to find the missing quantity:

Trig. functions & your calculator

Almost any calculator or calculator application (e.g. on your phone) will be able to calculate sine, cosine and tangent functions, and a few other trig. functions that we'll cover in due time. Calculators make the trig. functions act like "black boxes" – you just key in your angle and you get its sine, cosine or tangent.

You'll need to be careful about the units. Your calculator can use two units of angle measure: Degrees and Radians. Make sure you're working with the right units. Learn more about angle measurements here.

Example 2

Calculate all of the missing angles and sides of this triangle. This time the hypotenuse is one of the missing sides.

Solution: We begin by writing down one of the trigonometric functions. The 18 cm side of this triangle is the side adjacent to the 34˚ angle, so let's use cosine. Here we'll write out the definition of cosine and plug in the side and angle we have, then multiply both sides by h and divide both by cos(34˚).

$$ \begin{align} cos(34˚) = \frac{18}{h} \color{#E90F89}{\longrightarrow} h &= \frac{18}{cos(34˚)} \\[5pt] h &= 21.7 \; cm \end{align}$$

We can find the length of the opposite side, o, using the tangent function in a similar way:

$$ \begin{align} tan(34˚) = \frac{o}{18} \color{#E90F89}{\longrightarrow} o &= 18 \cdot tan(34˚) \\[5pt] o &= 12.1 \; cm \end{align}$$

Practice problems 1

Try to find all angle and side measurements of the three triangles below using SOH-CAH-TOA trig. Roll over the problems to see the answers.

Example 3

In the scenario below, a surveyor can calculate the height of a building by measuring the angle of elevation from her position to the top of the building, as long as the distance to the building (100 ft. in this case) is known. Calculate the height of the building.

Solution: The surveyor's lines-of-sight (dashed red lines) form a right triangle: The straight-line distance to the building, the height of the building (minus the 5 ft. from the ground to the surveyor's instrument) and the hypotenuse. To find the length of the opposite side, which we'll call o, we use the tangent function:

$$tan(19.3˚) = \frac{o}{100}$$

Here I'm calling o the height of the building above our 5 ft. mark. We'll have to add that 5 ft. to the total before we finish. We can rearrange to solve for o, multiplying by 100 ft. on both sides:

$$o = 100 \cdot tan(19.3˚) = 35$$

So the height of the building is that 35 ft. plus the 5 ft. height of the instrument, or 40 ft.

Interesting fact: The tangent function is approximately equal to the line y = x between, say, ±π/6. Here are overlapping graphs to show this:

Practice problems 2

For problems 1-4, calculate the measure of any missing angles and the lengths of any missing sides of the triangle.

1.  

   

   Solution

   The missing angle is $x = 180 - 90 - 39 = 51^{\circ}$.

   The length of the hypotenuse (h) can be obtained using

   $$ \begin{align} sin(39^{\circ}) &= \frac{2}{h} \\[5pt] h &= \frac{2}{sin(39^{\circ})} = 3.2 \, \text{cm} \end{align}$$

   The adjacent side is

   $$ \begin{align} tan(39^{\circ}) &= \frac{2}{a} \\[5pt] a &= \frac{2}{tan(39^{\circ})} = 2.5 \, \text{cm} \end{align}$$

2.  

   

   Solution

   The missing angle is $x = 180 - 90 - 15 = 75^{\circ}$.

   The length of the opposite side (o) can be obtained using

   $$ \begin{align} sin(15^{\circ}) &= \frac{o}{12} \\[5pt] o &= 12 \cdot sin(15^{\circ}) = 3.1 \, \text{m} \end{align}$$

   We can get the adjacent side like this:

   $$ \begin{align} cos(15^{\circ}) &= \frac{a}{12} \\[5pt] a &= 12 \cdot cos(15^{\circ}) = 11.6 \, \text{m} \end{align}$$

5. A surveyor knows the depth of the canyon and can measure the angle shown. Calculate the distance from one side of the canyon to the other.

   

   Solution

   
   

   Here's the basic geometry of the problem:

   

   
   

   Our task is to find the length of the adjacent side of that triangle, which suggests the tangent:

   $$ \begin{align} tan(14.3^{\circ}) &= \frac{250}{a} \\[5pt] a &= \frac{250}{tan(14.3^{\circ})} = 981 \, \text{m} \end{align}$$

   The canyon is about 981 m across.




6. Roof trusses are what hold up the roof and keep the walls parallel in some houses. They distribute any forces (mostly the downward forces of gravity, but also wind forces) evenly among all of the wooden members ("chords") that make them up. Using the angles and measurements given, determine the lengths A and B shown in the figure.

   

   Solution

   
   

   Here's the basic geometry of the problem:

   

   To find angle $\theta$ we subtract: $180 - 55 = 2\theta = 125^{\circ}$, so $\theta = 62.5^{\circ}$

   
   

   We can use the sine function to find the $B$ length:

   $$ \begin{align} sin(62.5^{\circ}) &= \frac{4}{B} \\[5pt] B &= \frac{4}{sin(62.5^{\circ})} = 4.51 \, \text{ft.} \end{align}$$

   To find $A/2$ we can either use the Pythagorean theorem or the tangent function. Given that we're here to learn trig, let's use the tangent.

   $$ \begin{align} tan(62.5^{\circ}) &= \frac{4}{A/2} \\[5pt] \frac{A}{2} &= \frac{4}{tan(62.5^{\circ})} \\[5pt] A &= 2 \cdot \frac{4}{tan(62.5^{\circ})} \\[5pt] &= 4.16 \, \text{ft.} \end{align}$$

The small-angle approximation

Here's an interesting problem and an opportunity to show how approximations can be good and save a lot of time. Consider a boat out on the ocean, a distance d from a lighthouse of known height – in this case 25 m. We'll explore two ways of calculating the distance to the lighthouse. Our measurement from the boat is the angle of elevation between the horizontal (sea surface) and the top of the lighthouse, 2˚.

First, let's do this in the straightforward way using the tangent function. First let's convert the angle to radians, because we'll use it that way later:

$$2˚ \left( \frac{\pi \, \text{rad}}{180˚} \right) = 0.03491 \, \text{rad}$$

The tangent of the angle is

$$tan(0.03491 \, \text{rad}) = \frac{25 \, m}{d},$$

so the distance is

$$d = \frac{25 \, m}{tan(0.03491 \, \text{rad})} = 715.836 \, m$$

Now let's do that in a different way. Notice that the circle of radius d is shown, and that, to a good approximation, the part of that circle that crosses through the lighthouse is almost a line. That is, if we take a small enough part of any curve, it's nearly a line.

Video examples

1. The special triangles: 45-45-90 and 30-60-90

You'll make your journey through trigonometry much easier if you memorize everything needed to derive the trig functions of 30˚, 45˚ and 60˚ angles. Here's how to do it.

Minutes of your life: 3:42