- Home
- >
- Sss Triangle – Explanation & Examples

JUMP TO TOPIC

# SSS Triangle – Explanation & Examples

How do we solve for the unknown parts of a triangle when we only know the measure of all its **three sides**? It’s pretty simple. We will use the **SSS (side-side-side) triangle **combination to find the angles.

*SSS (side-side-side) triangle is basically a triangular combination when we know the measure of three sides of a triangle and need to determine the remaining parts (angles) of a triangle.*

After studying this lesson, we are expected to learn the concepts driven by the following questions and be able to address them.

- What is a SSS triangle?
- How to solve a SSS triangle?
- What is the combinational role of the Law of Cosines and the Law of Sines to solve a SSS triangle?

It is also the goal of this lesson to clear up any confusions you might have about the concept of SSS triangles. Let’s begin.

## What Is a SSS Triangle?

SSS (side-side-side) triangle is a triangular combination that we get when we know the measure of three sides of a triangle and need to determine the remaining parts (angles) of a triangle.

Consider a triangle $△ABC$ with the sides $a$, $b$, and $c$ facing the angles $\alpha$, $\beta$, and $\gamma$ respectively as shown in Figure 16-1. We can observe that we are given the **three sides** $a$, $b$**,** and $c$. Therefore, the below Figure 16-1 illustrates a triangular combination which is known as a **SSS triangle**.

## How To Solve a SSS Triangle

When we know the measure of three sides of a triangle, we can apply **three-step methods** to solve a SSS triangle: using the Law of Cosines, using the Law of Sines, and determining the measure of the third angle.

**Step 1 of 3**

- Use the Law of Cosines to measure the largest angle — an angle opposite the longest side.

**Step 2 of 3**

- Use the Law of Sines to determine either of the remaining acute angles.

**Step 3 of 3**

- Determine the measure of the third angle by subtracting the already measured angles in the previous steps from $180^{\circ }$.

**Example 1**

In triangle $△ABC$, $a = 10$, $b = 17$ and $c = 20$ cm. Solve the triangle.

Solution:

We are given three sides $a = 10$, $b = 17$, and $c = 20$. Thus, we will apply the three-step method to solve a SSS triangle.

**Step 1 of 3**

*Use the Law of Cosines to measure the largest angle — an angle opposite the longest side.*

The angle $\gamma$ is opposite the longest side $c = 20$. Thus, we need to find the angle $\gamma$.

Applying the law of cosines

$c^2\:=\:a^2\:+b^2\:-\:2ab\:\cos\:\gamma$

$\cos\:\gamma=\frac{\left(a^2\:+\:b^2\:-\:c^2\right)}{2ab}$

substituting $a = 10$, $b = 17$ and $c = 20$ in the formula

$\cos\:\gamma=\frac{\left(10^2\:+\:17^2\:-\:20^2\right)}{2\:(10)\:(17)}$

$\cos\:\gamma=\frac{\left(100\:+\:289\:-\:400\right)}{340}$

$\cos\:\gamma=-\frac{11}{340}$

$\gamma = \cos^{-1} (-0.0324)$

$\gamma = 91.857…^{\circ }$

$\gamma ≈ 92^{\circ }$ (angle to the nearest degree)

**Step 2 of 3**

*Use the Law of Sines to determine either of the remaining acute angles. *

Now it is very straightforward to determine the angle $\alpha$.

Applying the law of sines

$\frac{a}{\sin\:\alpha\:}=\:\frac{c}{\sin\:\gamma}$

$\sin\:\alpha=a\:\frac{\sin\:\gamma}{c}$

substituting $a = 10$, $c = 20$ and $\gamma = 91.857^{\circ }$

Please note that it is recommended to use the **unrounded** value of angle $\gamma = 91.857^{\circ }$ because the angle $\gamma = 92$ is rounded to the nearest degree.

$\sin\:\alpha=10\:\frac{\sin\:91.857^{\circ }}{20}$

$\sin\:\alpha=10\:\frac{0.999}{20}$

$\sin\:\alpha = 10 (0.04995)$

$\sin\:\alpha = 0.4995$

$\alpha = \sin^{-1} (0.4995)$

$\alpha = 29.967…^{\circ }$

$\alpha ≈ 30^{\circ }$ (angle to the nearest degree)

**Step 3 of 3**

** Determine the measure of the third angle by subtracting the already measured angles in the previous steps from 180º**.

$\beta = 180^{\circ }\: – \alpha\: – \gamma$

substituting $\alpha = 30^{\circ }$ and $\gamma = 92^{\circ }$

$\beta= 180^{\circ }\: -\: 30^{\circ }\: –\: 92^{\circ }$

$\beta = 58^{\circ }$

Thus, the solution for the given SSS triangle is:

$\alpha = 30^{\circ }$, $\beta = 58^{\circ }$ and $\gamma = 92^{\circ }$

### Determining the Largest Angle

If we know the cosine of an angle, we can easily determine whether the angle is acute or obtuse depending upon the following conditions:

*The condition for the acute Angle*

$\cos \theta > 0%%EDITORCONTENT%%nbsp; for $0^{\circ } < \theta < 90^{\circ }$** **

*The condition for the obtuse Angle*

$\cos \theta < 0%%EDITORCONTENT%%nbsp; for $90^{\circ } < \theta < 180^{\circ }$

Since the $\gamma ≈ 92^{\circ }$ was obtuse, we already figured out that the angles $\alpha$ and $\beta$ must be acute. This is the reason a single triangle **cannot** have **multiple obtuse** angles.

It is also very important to remember that if the largest angle is acute, then the remaining two angles would also be acute.

**Example 2**

In triangle $△ABC$, $a = 24$, $b = 18$ and $c = 29$ cm. Solve the triangle.

Solution:

We are given three sides $a = 24$, $b = 18$, and $c = 29$. Thus, we will apply the three-step method to solve a SSS triangle.

**Step 1 of 3**

The angle $\gamma$ is opposite the longest side $c = 29$. Thus, we need to find the $\gamma$.

Applying the law of cosines

$c^2\:=\:a^2\:+b^2\:-\:2ab\:\cos\:\gamma$

$\cos\:\gamma=\frac{\left(a^2\:+\:b^2\:-\:c^2\right)}{2ab}$

substituting $a = 24$, $b = 18$ and $c = 29$ in the formula

$\cos\:\gamma=\frac{\left(24^2\:+\:18^2\:-\:29^2\right)}{2\:(24)\:(18)}$

$\cos\:\gamma=\frac{\left(576\:+\:324\:-\:841\right)}{864}$

$\cos\:\gamma=\frac{59}{864}$

$\gamma = \cos^{-1} (0.0682)$

$\gamma = 86.084…^{\circ }$

$\gamma ≈ 86^{\circ }$ (angle to the nearest degree)

**Step 2 of 3**

Now it is very straightforward to determine angle. Thus, we will determine the angle $\beta$.

Applying the law of sines

$\frac{a}{\sin\:\alpha\:}=\:\frac{c}{\sin\:\gamma}$

$\sin\:\beta=b\:\frac{\sin\:\gamma}{c}$

substituting $b = 18$, $c = 29$ and $\gamma = 86.084^{\circ }$

$\sin\:\beta=18\:\frac{\sin\:86.084^{\circ }}{29}$

$\sin\:\beta=18\:\frac{0.998}{29}$

$\sin\:\beta = 18 (0.03441)$

$\sin\:\beta = 0.61938$

$\beta = \sin^{-1} (0.61938)$

$\beta = 38.271…^{\circ }$

$\beta ≈ 38^{\circ }$ (angle to the nearest degree)

**Step 3 of 3**

Subtract $m∠\beta = 38^{\circ }$ and $m∠\beta = 86^{\circ } from $180^{\circ }$.

$\alpha = 180^{\circ }\: – \beta\: – \gamma$

$\alpha = 180^{\circ }\: -\: 38^{\circ }\: –\: 86^{\circ }$

$\beta = 56^{\circ }$

Thus, the solution of the given SSS triangle is:

$\alpha = 56^{\circ }$, $\beta = 38^{\circ }$ and $\gamma = 86^{\circ }$

**Example 3**

The following diagram illustrates three ships $A$, $B$ and $C$ and the distance between them. Determine the bearing of ship $B$ from ship $A$.

Solution:

Looking at the diagram, it is clear that the triangle $△ABC$ represents a SSS case as we have:

$a = 10$, $b = 4.9$ and $c = 6.3$

We need to determine the angle $A$ to find the bearing of ship $B$ from ship $A$.

Applying the law of cosines

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

$\cos\:A=\frac{\left(b^2\:+\:c^2\:-\:a^2\right)}{2bc}$

substituting $a = 10$, $b = 4.9$ and $c = 6.3$ in the formula

$\cos\:A=\frac{\left(4.9^2\:+\:6.3^2\:-\:10^2\right)}{2\:(4.9)\:(6.3)}$

$\cos\:A=\frac{\left(24.01\:+\:36.69\:-\:100\right)}{61.74}$

$\cos\:A=\frac{-36.3}{61.74}$

$\cos\:A=-0.5879$

$A = \cos^{-1} (-0.5879)$

$A = -0.5879…^{\circ }$

$A = 126.008…^{\circ }$

$A ≈ 126^{\circ }$ (angle to the nearest degree)

Therefore, the bearing of ship $B$ from ship $A$ is $126^{\circ }$.