Solving Triangles

In this tutorial, we’ll learn all about solving triangles. So let’s begin by looking at what it actually means to solve a triangle.

There are six values describing the six parts of a triangle — three sides and three internal angles.

Now, if we know any three of those six values, we can find the other three (with one exception that we’ll look at later). And that’s what solving triangles is all about.

To solve a triangle is to find its unknown (or required) side lengths and angles using what is known about the triangle.

Solving Triangles — Important Concepts

Before we dive into the different types of problems and examples and start solving them, here are the important concepts, formulas, and relations that are essential to being able to solve such problems.

So first things first. When naming sides and angles, a side is represented by the small case version of the letter for the opposite angle.

For example, the angle opposite to the side $\hspace{0.2em} AB \hspace{0.2em}$ is angle $\hspace{0.2em} C \hspace{0.2em}$ . So it (or its length) is represented by $\hspace{0.2em} c \hspace{0.2em}$ .

Sum of Angles

The sum of the three internal angles of a triangle is $\hspace{0.2em} 180 \degree$ .

A + B + C = 180 \degree

Law of Cosines

The law of cosines gives the relationship between the side lengths of a triangle and the cosine of any of its angles. It says —

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

Re-framing the formula for other sides, we have

b^2 = a^2 + c^2 - 2ac \, \cos B

c^2 = a^2 + b^2 - 2ab \, \cos C

For cases where we need to find angles using the cosine rule, the three formulas can be rearranged as —

\begin{align*} \cos A \, = \, \frac{b^2 + c^2 - a^2}{2bc} \\[1.5em] \cos B \, = \, \frac{a^2 + c^2 - b^2}{2ac} \\[1.5em] \cos C \, = \, \frac{a^2 + b^2 - c^2}{2ab} \end{align*}

Law of Sines

The law of sines states that the ratio of side length to the sine of opposite angle is the same for all sides in a triangle.

\frac{a}{\sin A} \, = \, \frac{b}{\sin B} \, = \, \frac{c}{\sin C}

Important

When we find an angle using its sine value, there are two possibilities. The angle can be acute or obtuse (angle greater than $\hspace{0.2em} 90 \degree$ ).

For example,

\begin{align*} &\sin \theta \, = \, \frac{1}{2} \\[1.3em] \Rightarrow \hspace{0.5em} &\theta \, = \, \sin ^{-1} \theta \\[1.3em] \Rightarrow \hspace{0.5em} &\theta \, = \,30 \degree \,\, \text{ or } \,\, \theta \, = \, 150 \degree \end{align*}

So which angle do we choose? Here are a couple of points to keep in mind.

A triangle can have a maximum of one obtuse angle. Otherwise, the sum would exceed $\hspace{0.2em} 180 \degree$ .
The obtuse angle (if present) would be the angle opposite the largest side (itself being the largest angle).

If it sounds confusing, don’t worry. Wait till we get to the examples.

The Different Types

Depending on which three of the six values are known to us, the problems under the topic of solving triangles can be divided into five categories.

1. Side-Side-Side (SSS)

In this type, all three sides are known to us.

To solve, start by finding one angle using the cosine rule. Then find the two remaining angles using the sine rule.

2. Side-Angle-Side (SAS)

Here, two sides and the included angle (the angle contained between them) are known to us.

To solve, first find the third side using the law of cosines. Then find the two unknown angles using the law of sines.

3. Side-Side-Angle (SSA)

In this case, two sides and one angle (not included) are known.

(angle sum rule). Finally, find the missing side using the sine rule.

4. Angle-Side-Angle (ASA)

Here, two angles and the included side (the side shared by the two angles) are known to us.

To solve, start by finding the third side using the angle sum property. Then find the two unknown sides using the sine rule.

5. Angle-Angle-Side (AAS)

In this type, two angles and one side (not included) are known.

The steps to solve this are the same as those for the ASA type. The first step is to find the third angle using the angle sum property. Then you find the two unknown sides using the sine rule.

6. Angle-Angle-Angle (AAA)

In this case, the three angles are known to us.

You can’t solve this type. Here's why.

A triangle is not uniquely determined by its three angles. For the same three angles, you can have infinitely many triangles. They will have the same shape but different sizes (similar triangles).

For example, here are two triangles with the same internal angles (so similar by AA criteria) but with different size.

So how would one know which triangle it is without knowing at least one side-length?

Solving Triangles — Examples

Alright, now it’s time to use what we have learned so far and solve examples covering the different types discussed above.

1. SSS

Example

The lengths of the three sides of a triangle are $\hspace{0.2em} 6 \hspace{0.2em}$ , $\hspace{0.2em} 3 \hspace{0.2em}$ , and $\hspace{0.2em} 4 \hspace{0.2em}$ . Find the measures of its angles.

Solution

Step 0. We start by drawing a rough sketch of the triangle and labeling the information given in the question. It’s not necessary but often makes things easier and helps avoid silly mistakes.

Step 1. Use the Cosine Rule to find the largest angle

When we know all the side lengths, we can use the Cosine Rule to find any of the angles.

However, I strongly recommend you find the largest angle first — the angle opposite to the longest side. Why?

Because if there is an obtuse angle $\hspace{0.2em} (>90 \degree) \hspace{0.2em}$ in the triangle, it has to be this angle. So once this is out of the way, we can use the sine rule to find the other angles without having to worry about the obtuse options.

Here the largest angle would be $\hspace{0.2em} A \hspace{0.2em}$ . So using the formula for $\hspace{0.2em} \cos A \hspace{0.2em}$ , we have

\begin{align*} \cos A \, &= \, \frac{a^2 - b^2 - c^2}{-2bc} \\[1.3em] &= \, \frac{6^2 - 3^2 - 4^2}{-2 \cdot 3 \cdot 4} \\[1.3em] &\approx \, -0.4583 \end{align*}

Taking cos inverse on both sides.

\begin{align*} A \, &\approx \, \cos ^{-1} (-0.4583) \\[1em] &\approx \, 117.28 \degree \end{align*}

Step 2. Use the Sine Rule for one of the remaining angles

Now that we know the three sides and one angle, we can use the Sine Rule to find any of the remaining two angles. Let's go for $\hspace{0.2em} B \hspace{0.2em}$ .

\begin{align*} \frac{a}{\sin A} \, &= \, \frac{b}{\sin B} \\[1.75em] \frac{6}{\sin 117.28 \degree} \, &\approx \, \frac{3}{\sin B} \\[1.75em] \sin B &\approx 0.4444 \end{align*}

Taking sine inverse on both sides

B \approx 26.38 \degree

Step 3. Use the Angle Sum Property to find the third angle

So,

\begin{align*} A + B + C \, &= \, 180 \degree \\[1em] 117.28 \degree + 26.38 \degree + C \, &= \, 180 \degree \\[1em] C \, &= \, 36.34 \degree \end{align*}

And we have solved the triangle.

Note — We could have found the third angle ( $\hspace{0.2em} C \hspace{0.2em}$ ) using the sine rule, as we did for angle $\hspace{0.2em} B \hspace{0.2em}$ . I used the angles sum property just because the calculation was easier. You can also use one to find the angle and the other to validate your answer(s).

2. SAS

Example

Two sides and the included angle in a triangle measure $\hspace{0.2em} 12 \hspace{0.2em}$ , $\hspace{0.2em} 14 \hspace{0.2em}$ , and $\hspace{0.2em} 110 \degree \hspace{0.2em}$ . Solve the triangle for the missing side and angles.

Solution

Again, we start with a rough sketch for the problem.

Step 1. Use the Cosine Rule to find the missing side

When we have two sides and the included angle, the law of cosines allows us to find the third side.

Applying the cosine rule for side $\hspace{0.2em} b \hspace{0.2em}$ ,

\begin{align*} b^2 \, &= \, a^2 + c^2 - 2ac \, \cos B \\[1em] &= \, 14^2 + 12^2 - 2 \cdot 14 \cdot 12 \, \cos 110 \degree \\[1em] &\approx \, 454.92 \end{align*}

Taking square root on both sides,

\begin{align*} b \, &\approx \, \sqrt{454.92} \\[1em] &\approx \, 21.33 \end{align*}

Step 2. Use the Sine Rule for the remaining angles

At this point, we know three sides and one angle. So, we can find the unknown angles using the sine rule.

\begin{align*} \frac{a}{\sin A} \, &= \, \hspace{0.85em} \frac{b}{\sin B} &= \, \frac{c}{\sin C} \\[1.5em] \frac{14}{\sin A} \, &\approx \, \hspace{0.35em} \frac{21.33}{\sin 110 \degree} &\approx \, \frac{12}{\sin C} \end{align*}

When using the sine rule to find an angle, go for the smaller angle (the one opposite the smaller side) if possible. Here’s why.

The smaller angle cannot be obtuse and so you can confidently accept the acute angle solution when taking the inverse of sine.

Here though, we already have an obtuse angle in the triangle. So, no matter which of the unknown angles we calculate first, we can safely ignore the obtuse options when taking sine inverse.

Alright, now lets split the equation and solve for $\hspace{0.2em} C \hspace{0.2em}$ .

\begin{align*} \frac{21.33}{\sin 110 \degree} \, &\approx \, \frac{12}{\sin C} \\[1.5em] \sin C \, &\approx \, 0.5286 \\[1.5em] C \, &\approx \, 31.91 \degree \end{align*}

And finally, as we did in the last example, let's calculate the last angle using the angle sum property.

\begin{align*} A + B + C \, &= \, 180 \degree \\[1em] A + 110 \degree + 31.91 \degree \, &= \, 180 \degree \\[1em] A \, &= \, 38.09 \degree \end{align*}

Solved.

3. SSA

When it comes to problems of type SSA, it’s useful to divide them into two sub-types. Let me explain with an example of each type.

3a. SsA

Example

Solve the triangle below for the missing sides and angles.

Solution

In this example, we don’t need to draw anything as the question itself provides the figure.

Now the question gives us two sides and one angle (duh!) but —

The known angle is adjacent to the smaller of the two known sides and hence opposite to the larger side. That’s what makes this sub-type SsA (lowercase/small “s” is placed adjacent to “A”).

Step 1. Use the Sine Rule to find the missing angle opposite to one of the known sides

Here, we know the sides $\hspace{0.2em} b \hspace{0.2em}$ and $\hspace{0.2em} c \hspace{0.2em}$ and the angle $\hspace{0.2em} B \hspace{0.2em}$ . So we need to find angle $\hspace{0.2em} C \hspace{0.2em}$ .

Applying the sine rule to $\hspace{0.2em} B \hspace{0.2em}$ and $\hspace{0.2em} C \hspace{0.2em}$ .

\begin{align*} \frac{b}{\sin B} \, &= \, \frac{c}{\sin C} \\[1.5em] \frac{28}{\sin 55 \degree} \, &= \, \frac{20}{\sin C} \\[1.5em] \sin C \, &\approx \, 0.5851 \end{align*}

Taking sine inverse on both sides

C \, \approx \, 35.81 \degree

We know $\hspace{0.2em} C \hspace{0.2em}$ is acute because the opposite side, $\hspace{0.2em} c \hspace{0.2em}$ , is certainly not the largest $\hspace{0.2em} (c < b) \hspace{0.2em}$ .

Step 2. Find the third angle using the angle sum property

\begin{align*} A \, &= \, 180 \degree - B - C \\[1em] A \, &\approx \, 180 \degree - 55 \degree - 35.81 \degree \\[1em] &\approx \, 89.19 \degree \end{align*}

Step 3. Find the missing side using the sine rule

\begin{align*} \frac{a}{\sin A} \, &= \, \frac{b}{\sin B} \\[1.5em] \frac{a}{\sin 89.19 \degree} \, &\approx \, \frac{28}{\sin 55 \degree} \\[1.5em] &\approx \, 34.17 \end{align*}

That’s it.

3b. sSS

Example

In $\hspace{0.2em} \triangle ABC \hspace{0.2em}$ , $\hspace{0.2em} b = 15 \hspace{0.2em}$ , $\hspace{0.2em} c = 21 \hspace{0.2em}$ , and $\hspace{0.2em} B = 42 \degree \hspace{0.2em}$ . Solve the triangle.

Solution

Here’s the drawing to get started.

This is an example of the other sub-type where the known angle is adjacent to the larger of the two known sides and hence opposite to the smaller side. That’s what makes this sub-type sSA (uppercase “S” is placed adjacent to “A”).

This case is often termed as the “ambiguous case” because, as you’ll see, it has two potential solutions. And unless we have more information about the problem, we can’t pick one over the other. So both solutions must be worked out.

Step 1. Use the Sine Rule to find the missing angle opposite to one of the known sides

\begin{align*} \frac{b}{\sin B} \, &= \, \frac{c}{\sin C} \\[1.5em] \frac{15}{\sin 42 \degree} \, &= \, \frac{21}{\sin C} \\[1.5em] \sin C \, &\approx \, 0.9368 \end{align*}

Unlike in the last example, here, we need to consider both acute and obtuse values for $\hspace{0.2em} C$ . That’s because side $\hspace{0.2em} c \, \, (AB) \hspace{0.2em}$ could be the largest side. It is larger than $\hspace{0.2em} b \hspace{0.2em}$ , and we don’t know $\hspace{0.2em} a \hspace{0.2em}$ yet.

Acute Solution for $\hspace{0.2em} C \hspace{0.2em}$

\begin{align*} C \, &\approx \, \sin ^{-1} 0.9368 \\[1em] C \, &\approx \, 69.52 \degree \end{align*}

Obtuse Solution for $\hspace{0.2em} C \hspace{0.2em}$

\begin{align*} C \, &\approx \, 180 \degree - 69.52 \degree \\[1em] C \, &\approx \, 110.48 \degree \end{align*}

Note — When using the sine rule, you can get the obtuse solution by subtracting the acute solution from $\hspace{0.2em} 180 \degree$ .

Next, we'll need to work out the remainder of the solution for each of the two cases.

First, let's go ahead assuming $\hspace{0.2em} C \hspace{0.2em}$ is acute.

Step 2. Find the third angle using the angle sum property

\begin{align*} A \, &= \, 180 \degree - B - C \\[1em] &\approx \, 180 \degree - 42 \degree - 69.52 \degree \\[1em] &\approx \, 68.48 \degree \end{align*}

Step 3. Find the missing side using the sine rule

\begin{align*} \frac{a}{\sin A} \, &= \, \frac{b}{\sin B} \\[1.5em] \frac{a}{\sin 68.48 \degree} \, &\approx \, \frac{15}{\sin 42 \degree} \\[1.5em] a \, &\approx \, 20.85 \end{align*}

Step 4. Repeat steps 2 and 3 assuming $\hspace{0.2em} C \hspace{0.2em}$ is obtuse.

Again, first, we find angle $\hspace{0.2em} A \hspace{0.2em}$ .

\begin{align*} A \, &= \, 180 \degree - B - C \\[1em] &\approx \, 180 \degree - 42 \degree - 110.48 \degree \\[1em] &\approx \, 27.52 \degree \end{align*}

And then side $\hspace{0.2em} a \hspace{0.2em}$

\begin{align*} \frac{a}{\sin A} \, &= \, \frac{b}{\sin B} \\[1.5em] \frac{a}{\sin 27.52 \degree} \, &\approx \, \frac{15}{\sin 42 \degree} \\[1.5em] a \, &\approx \, 10.36 \end{align*}

Great, we have found all the missing values for both possible solutions. To make things clearer, here are their drawings.

The two solutions (calculated values in lighter color)

Note - Before we move ahead, I want to make it clear you don’t always have to worry about multiple solutions. Of the six cases we discuss in this tutorial, this is the only one of that kind.

4. ASA

Example

Solve the triangle below.

Solution

Step 1. Find the third angle using the angle sum property

\begin{align*} A \, &= \, 180 \degree - B - C \\[1em] &= \, 180 \degree - 65 \degree - 30 \degree \\[1em] &= \, 85 \degree \end{align*}

Step 2. Find the missing side using the sine rule

\begin{align*} &\frac{a}{\sin A} \, = \, \frac{b}{\sin B} \, = \, \frac{c}{\sin C} \\[1.5em] &\frac{10}{\sin 85 \degree} \, = \, \frac{b}{\sin 85 \degree} \, = \, \frac{c}{\sin 30 \degree} \end{align*}

You know what to do now — split it into two equations and solve for $\hspace{0.2em} b \hspace{0.2em}$ and $\hspace{0.2em} c \hspace{0.2em}$ .

Solving for $\hspace{0.2em} b \hspace{0.2em}$

\begin{align*} \frac{10}{\sin 85 \degree} \, &= \, \frac{b}{\sin 85 \degree} \\[1.5em] b \, &\approx \, 9.10 \end{align*}

Solving for $\hspace{0.2em} c \hspace{0.2em}$

\begin{align*} \frac{10}{\sin 85 \degree} \, &= \, \frac{c}{\sin \sin 30 \degree} \\[1.5em] c \, &\approx \, 5.02 \end{align*}

Triangle solved.

5. AAS

Example

In $\hspace{0.2em} \triangle ABC \hspace{0.2em}$ , $\hspace{0.2em} c = 14 \hspace{0.2em}$ , $\hspace{0.2em} B = 45 \degree \hspace{0.2em}$ , and $\hspace{0.2em} C = 55 \degree \hspace{0.2em}$ . Solve the triangle.

Solution

Step 1. Here, we know two angles. So, we find the third angle using the angle sum property

\begin{align*} A \, &= \, 180 \degree - B - C \\[1em] &= \, 180 \degree - 45 \degree - 55 \degree \\[1em] &= \, 80 \degree \end{align*}

Step 2. Find the missing sides using the sine rule

\begin{align*} &\frac{a}{\sin A} \, = \, \frac{b}{\sin B} \, = \, \frac{c}{\sin C} \\[1.5em] &\frac{a}{\sin 80 \degree} \, = \, \frac{14}{\sin 45 \degree} \, = \, \frac{c}{\sin 55 \degree} \end{align*}

Solving for $\hspace{0.2em} a \hspace{0.2em}$

\begin{align*} \frac{a}{\sin 80 \degree} \, &= \, \frac{14}{\sin 45 \degree} \\[1.5em] a \, &\approx \, 19.50 \end{align*}

Solving for $\hspace{0.2em} c \hspace{0.2em}$

\begin{align*} \frac{14}{\sin 45 \degree} \, &= \, \frac{c}{\sin 55 \degree} \\[1.5em] c \, &\approx \, 16.22 \end{align*}

And with that, we come to the end of this tutorial on solving triangles. Until next time.