#### xaktly | Trigonometry

Non right triangle trig.

### The trigonometry of non-right triangles

So far, we've only dealt with right triangles, but trigonometry can be easily applied to non-right triangles because any non-right triangle can be divided by an altitude* into two right triangles.

Roll over or tap the triangle to see what that means →

Remember that an altitude is a line segment that has one endpoint at a vertex of a triangle intersects the opposite side (or an extension of it outside the triangle) at a right angle. See triangles.

### Customary labeling of non-right triangles

This labeling scheme is commonly used for non-right triangles. Capital letters are angles and the corresponding lower-case letters go with the side opposite the angle: side a (with length of a units) is across from angle A (with a measure of A degrees or radians), and so on.

X

### derivation

In mathematics, when we derive a formula, we basically invent it using one or more simpler, already-known principles. A derivation is essentially a proof that the formula we're creating works and can be relied upon.

Consider the triangle below. if we find the sines of angle A and angle C using their corresponding right triangles, we notice that they both contain the altitude, x.

The sine equations are

$$sin(A) = \frac{x}{c} \; \; \text{ and } \; \; sin(C) = \frac{x}{a}$$

We can rearrange those by solving each for x (multiply by c on both sides of the left equation, and by a on both sides of the right):

$$x = c\cdot sin(A) \; \; \text{ and } \; \; x = a\cdot sin(C)$$

Now the transitive property says that if both c·sin(A) and a·sin(C) are equal to x, then they must be equal to each other:

$$c \, sin(A) = a \, sin(C)$$

We usually divide both sides by ac to get the easy-to-remember expression of the law of sines:

$$\frac{sin(A)}{a} = \frac{sin(C)}{c}$$

We could do the same derivation with the other two altitudes, drawn from angles A and C to come up with similar relations for the other angle pairs. We call these together the law of sines. It's in the green box below.

The law of sines can be used to find the measure of an angle or a side of a non-right triangle if we know:

• two sides and an angle not between them or
• two angles and a side not between them.

#### Law of Sines

$$\frac{sin(A)}{a} = \frac{sin(B)}{b} = \frac{sin(C)}{c}$$

Although three fractions are related here with two equal signs, we only use them in pairs in practice.

### Example 1

Find all of the missing measurements of this triangle:

Solution: The missing angle is easy, it's just

$$180˚ - 76˚ - 34˚ = 70˚$$

Now set up one of the law of sines proportions and solve for the missing piece, in this case the length of the lower side:

\begin{align} \frac{sin(76˚)}{x} &= \frac{sin(34˚)}{11} \\[5pt] x &= \frac{11 \cdot sin(76˚)}{sin(34˚)} \\[5pt] &= 19.1 \; \text{in.} \end{align}

Then do the same for the other missing side. It's best to use the original known angle and side so that round-off errors or mistakes don't add up.

\begin{align} \frac{sin(70˚)}{y} &= \frac{sin(34˚)}{11} \\[5pt] y &= \frac{11 \cdot sin(70˚)}{sin(34˚)} \\[5pt] &= 18.5 \; \text{in.} \end{align}

#### Pro tip:

I've skipped a couple of steps in the algebra above. Remember that when the variable for which you're trying to solve is in a denominator, your first task is clear: Get it out of the denominator, usually by multiplying by that variable on both sides. Then the rest of the algebra to isolate that variable from everything else attached to it is fairly clear.

### Example 2

Find all of the missing measurements of this triangle:

Solution: This time we have to find our first missing angle using the LOS, then we can use the sum of the angles of a triangle to find the third. First, set up one law of sines proportion. We'll be solving for a missing angle, so we'll have to calculate an inverse sine:

\begin{align} \frac{sin(46˚)}{23} &= \frac{sin(Y˚)}{30} \\[5pt] sin(Y˚) &= \frac{30 \, sin(46˚)}{23} \\[5pt] Y &= sin^{-1}\left( \frac{30 \, sin(46˚)}{23} \right) \\[5pt] Y &= 69.8˚ \end{align}

Now it's easy to calculate the third angle:

$$180˚ - 46˚ - 69.8˚ = 64.2˚$$

Then apply the law of sines again for the missing side. We have two choices, we can solve

\begin{align} \frac{sin(46˚)}{23} &= \frac{sin(64.2˚)}{x} \; \; \color{#E90F89}{\text{ or }}\\[5pt] \frac{sin(69.8˚)}{30} &= \frac{sin(64.2˚)}{x} \end{align}

$$x = 28.8 \; cm$$

### Caution: Law of sines ambiguity

We have to be careful about the law of sines, because it can give ambiguous solutions when we use the inverse sine function [f(x) = sin-1(x)]. Here's how it works. For a given triangle in which only one side is known, two possible triangles can be formed. Take a look:

The magenta sides mark out an obtuse triangle (small one on the left) and an acute triangle (outer triangle) using the same combination of two sides, where the third (bottom) side can be one of two lengths because it isn't initially known. The two possible angles are always related: their sum is 180˚.

How do we handle this? Let's do an example.

First, we solve for angle A using the LOS:

$$\frac{sin(30˚)}{7} = \frac{sin(A)}{10}$$

We can rearrange and use the inverse sine function to get the angle:

$$A = sin^{-1}\left( \frac{10 \, sin(30˚)}{7} \right) = 45.6˚$$

Now if there is an ambiguity, its measure will be 180˚ minus the angle we determined:

$$180˚ - 45.6˚ = 134.4˚$$

Now if that angle, added to the original angle (30˚) is less than 180˚, such a triangle can exist, and we have an ambiguous case. Here are the two possible triangles in this example:

In such a case, we have to know a little more about our triangle. Is it acute or obtuse? Is a given diagram drawn to scale (as these are)? If so, that might be enough to resolve the ambiguity.

In some cases, the quantity (180˚ minus our calculated angle) added back to the calculated angle will be greater than 180˚, and such a triangle cannot exist, so the solution is unique.

Make sure to be aware of this as you work problems. Check for ambiguities unless you're clear about the results from other clues in the problem.

### The law of cosines

The law of cosines is used when we know:

• the lengths of two sides of a triangle and the measure of the angle between them, OR

• the lengths of all three sides of a triangle, but no angle measures.

The law of cosines is computationally a little more complicated to use than the law of sines, but fortunately, it only needs to be used once. After the law of cosines is applied to a triangle, the resulting information will always make it possible to use the law of sines to calculate further properties of the triangle.

### Derivation of the law of cosines

Consider another non-right triangle, labeled as shown with side lengths x and y. We can derive a useful law containing only the cosine function.

First use the Pythagorean theorem to derive two equations for each of the right triangles:

\begin{align} c^2 &= y^2 + x^2 \; \; \text{ and} \\[5pt] a^2 &= (b - y)^2 + x^2 \end{align}

Notice that each contains and x2, so we can eliminate x2 between the two using the transitive property:

$$c^2 - y^2 = a^2 - (b - y)^2$$

Then expand the binomial (b - y)2 to get the equation below, and note that the y2 cancel:

$$c^2 - y^2 = a^2 - b^2 + 2by - y^2$$

Now we still have a y hanging around, but we can get rid of it using the cosine solution, notice that

$$cos(A) = \frac{y}{c}, \; \text{ so } \; y = c \, cos(A)$$

Substituting c·cos(A) for y, we get

$$a^2 = b^2 + c^2 - 2 bc \, cos(A)$$

which is the law of cosines

The law of cosines can be used to find the measure of an angle or a side of a non-right triangle if we know:

• two sides and the angle between them or
• three sides and no angles.

We could again do the same derivation using the other two altitudes of our triangle, to yield three versions of the law of cosines for any triangle. They are listed in the box below.

#### Law of Cosines

The Law of Cosines is just the Pythagorean relationship with a correction factor, e.g. -2bc·cos(A), to account for the fact that the triangle is not a right triangle. We can write three versions of the LOC, one for every angle/opposite side pair:

\begin{align} a^2 = b^2 + c^2 - 2 bc \, cos(A) \\[5pt] b^2 = a^2 + c^2 - 2 ac \, cos(B) \\[5pt] c^2 = a^2 + b^2 - 2 ab \, cos(C) \end{align}

#### Pro tip

You'll only need to use the law of cosines once if you need it at all. After finding the missing angle or side with the LOC, the law of sines will work for the rest.

### Example 3

Find all of the missing measurements of this triangle:

Solution: Set up the law of cosines using the only set of angles and sides for which it is possible in this case:

\begin{align} a^2 &= 8^2 + 4^2 - 2(8)(4) \, cos(51˚) \\[5pt] a^2 &= 39.72 \; m \\[5pt] a &= 6.3 \; m \end{align}

Now using the new side, find one of the missing angles using the law of sines:

\begin{align} \frac{sin(51˚)}{6.3} &= \frac{sin(B)}{8} \\[5pt] sin(B) &= \frac{8 \, sin(51˚)}{6.3} \\[5pt] B &= sin^{-1} \left( \frac{8 \, sin(51˚)}{6.3} \right) \\[5pt] &= 80.7˚ \end{align}

And then the third angle is

$$180˚ - 51˚ - 80.7˚ = 48.3˚$$

In general, try to use the law of sines first. It's easier and less prone to errors (but be careful of the ambiguity when solving for non-acute angles). But in this case, that wasn't possible, so the law of cosines was necessary.

### Example 4

Find all of the missing measurements of this triangle:

Solution: Set up the law of cosines to solve for either one of the angles:

$$3.5^2 = 8^2 + 8.5^2 - 2(8)(8.5) \, cos(A)$$

Rearrange to solve for A. You'll need an inverse cosine to get the angle.

\begin{align} cos(A) &= \frac{3.5^2 - 8^2 - 8.8^2}{-2(8)(8.5)} \\[5pt] A &= cos^{-1} \left( \frac{3.5^2 - 8^2 - 8.8^2}{-2(8)(8.5)} \right) \\[5pt] &= 24.2˚ \end{align}

Use the law of sines to find a second angle

\begin{align} \frac{sin(24.2˚)}{3.5} &= \frac{sin(B)}{8} \\[5pt] sin(B) &= \frac{8 \, sin(24.2˚)}{3.5} \\[5pt] B &= sin^{-1} \left( \frac{8 \, sin(24.2˚)}{3.5} \right) \\[5pt] &= 69.5˚ \end{align}

Finally, just calculate the third angle:

$$180˚ - 24.2˚ - 29.9˚ = 125.9˚$$

### Practice problems

Find the measures of all missing sides and angles of these triangles:

(Hint: Always use the LOS first if you can. It's simpler. When that fails, use the LOC, but then you can usually use other means to fill in the rest of the measurements using the LOS or the Pythagorean theorem.)

### Video examples

#### 1. Derivation of the law of sines

The law of sines is an important relationship between the angle measures and side lengths of non-right triangles. In this video you'll see where it comes from.