What are ASA Triangles?

ASA Triangles are one of the first types of triangles that you’ll encounter when learning about triangle congruency in geometry. These triangles are easy to identify and exhibit interesting applications, so learning about them will help us in geometry and trigonometry as well!

In this article, we’ll show you what ASA stands for and how we can use this as a guide when identifying ASA triangles. We’ll also show you common geometric and trigonometric problems that involve the ASA triangles. By the end of our discussion, we want you to feel confident when working with ASA triangles.

What Are ASA Triangles?

ASA triangles are special types of angles defined by the side lengths and angle measures that are available. The abbreviation, ASA, stands for angle – side – angle. The order of the words matter because in an ASA triangle, the values that are given are two consecutive angles and the side sandwiched in between them.

What makes ASA triangles special is that with their given measures alone, we can find the three other measures of each ASA triangle. This also why when we’re given two ASA triangles, we can find their measures and confirm their congruency using appropriate geometric and trigonometric properties. 

Understanding ASA Triangles in the Context of Triangle Congruency

We’ll further appreciate the concept of ASA triangles when dealing with triangle congruency. When two ASA triangles show the same measures, we can conclude that the two triangles are congruent.

Take a look at the pair of triangles shown above – each triangle contains two consecutive angles and the included sides’ measures are given. Since all of these are equal for the two triangles, through the ASA postulate, we can conclude that the two triangles are congruent. This means that all three sides and all three angles of the two triangles are equal. 

We can also use the concept of ASA triangles to also find the unknown sides and angles of two congruent triangles. The example below shows you how to apply what you’ve just learned.

Problem 1

The two triangles shown below are said to be congruent through the ASA postulate. Determine the value of a, b, and c.

Since these two ASA triangles are congruent, their corresponding angles and sides will also be equivalent. Match up the congruent angles and side for these two ASA triangles to find the values for a, b, and c.

  • We only know one side for each ASA triangle, so a = 9.4 ft.
  • The angles, b\degree and 56 \degree, are equal, so we have b = 56.
  • Similarly, c\degree and 62 \degree, are also equal. Hence, we have c = 62.

Problem 2

Show that the two triangles, \Delta ABC and \Delta DEF , are congruent to each other.

We’ve established that through the ASA postulate, we can show that two ASA triangles are congruent. Recall that ASA triangles will have two consecutive angles and one included side. When their corresponding sides and angles are equal, the two triangles are congruent.

\begin{aligned}\angle C &= \angle F= 88\degree\\ \angle B &= \angle E= 36\degree\\\overline{BC} &= \overline{EF}=14\end{aligned}

We’ve shown that the corresponding consecutive angles are equal and the included sides for the two ASA triangles are also equal. This means that through the ASA postulate,  katex] \Delta ABC [/katex] and \Delta DEF are congruent.

These two problems show how important it is that we know how to identify ASA triangles as well as learn how to use them to solve geometric problems. You might get curious on how the rest of an ASA triangle’s angle and sides are determined, so the next section will address that. 

How To Find the Missing Angles and Sides of ASA Triangles?

As we have mentioned, we can find the rest of the angles and sides of an ASA triangle using geometric and trigonometric properties.

 It’s time for us to break down the steps to find the values of the missing angles and sides of an ASA triangle.  

  • The sum of the triangle’s interior angles is equal to 180\degree, so find the value of the remaining angle by subtracting the sum of the two consecutive angles from 180\degree.
  • Apply Law of Sines to find the length of each side. Given an ASA triangle, \Delta ABC, the measures of the missing side lengths can be calculated using the relationship  shown below. 
\begin{aligned}\boldsymbol{\dfrac{a}{\sin A} =\dfrac{b}{\sin B} =\dfrac{c}{\sin C} }\end{aligned}

The Law of Sines is especially helpful when working ASA triangles because we are immediately given two angles and one included side. Once we have the measure of the missing angle, we can use its ratio and the included side to find the rest of the triangle’s sides. Don’t worry, the best way to understand this is by working out on some examples. 

Problem 3

Calculate the missing sides and angle of the ASA triangle, \Delta ABC. Estimate your answers to two decimal places.

 We have the values of the angles: \angle A =40\degree and \angle C = 60\degree. Find the measure of \angle B using the fact that \angle A + \angle B + \angle C =180\degree.

\begin{aligned}\angle A + \angle B + \angle C &=180\degree\\40\degree + \angle B + 60\degree&= 180\degree\\\angle B &= 80\degree\end{aligned}

Now that we have a pair of angle and its opposite side, we can relate the ratios of the sides and the angles’ sine values using the Law of Sines. This means that we can equate \dfrac{\overline{AC}}{\sin B} and \dfrac{\overline{BC}}{\sin A} to find the length of \overline{BC}.

\begin{aligned}\dfrac{\overline{AC}}{\sin B} &= \dfrac{\overline{BC}}{\sin A}\\\dfrac{12}{\sin 80\degree} &= \dfrac{\overline{BC}}{\sin 40\degree}\\ \overline{BC}&= \dfrac{12\sin 40\degree}{\sin 80\degree}\\&\approx 7.83\end{aligned}

Apply a similar process to find the length of \overline{AB}.

\begin{aligned}\dfrac{\overline{AC}}{\sin B} &= \dfrac{\overline{AB}}{\sin C}\\\dfrac{12}{\sin 80\degree} &= \dfrac{\overline{AB}}{\sin 60\degree}\\ \overline{AB}&= \dfrac{12\sin 60\degree}{\sin 80\degree}\\&\approx 10.55\end{aligned}

The missing angle and sides of our ASA triangle, \Delta ABC, are as follows:

  • \overline{AB} \approx 10.55 ft
  • \overline{BC} \approx 7.83 ft
  • \angle B = 80 \degree ft

Are you getting the hang of it? Try out the next problem to further master the process of solving ASA triangles.

Problem 4

This time, why don’t we solve the triangle, PQR? Estimate your answers to the nearest one decimal place.

First, let’s find the measure of \angle Q as shown below.

\begin{aligned}\angle P + \angle Q +\angle R &= 180\degree\\\angle Q&= 180\degree - (55\degree + 65\degree)\\&= 60\degree\end{aligned}

Now that we have all a pair of opposite angle and sides, we can now use the Law of Sine to find the lengths of p and r

\begin{aligned}\dfrac{p}{\sin P} =\dfrac{q}{\sin Q} =\dfrac{r}{\sin R} \end{aligned}

Use these relationships to find the values of p and r. The table below summarizes the calculations for the missing sides’ lengths.

\boldsymbol{p}\boldsymbol{r}
\begin{aligned}\dfrac{q}{\sin Q} &= \dfrac{p}{\sin P}\\\dfrac{8.2}{\sin 60\degree} &= \dfrac{p}{\sin 55\degree}\\ p&= \dfrac{8.2 \sin 55\degree}{\sin 60\degree}\\&\approx 7.8\end{aligned} \begin{aligned}\dfrac{q}{\sin Q} &= \dfrac{r}{\sin R}\\\dfrac{8.2}{\sin 60\degree} &= \dfrac{r}{\sin 65\degree}\\ r&= \dfrac{8.2 \sin 65\degree}{\sin 60\degree}\\&\approx 8.6\end{aligned}

Hence, we have 60\degree, p \approx 7.8 cm, and 8.6 cm.

Finding the Area of an ASA Triangle

By applying more advanced trigonometric properties, we can find the area of an ASA triangle using the formulas.  

When \boldsymbol{a} is the given included sided.When \boldsymbol{b} is the given included sided.When \boldsymbol{c} is the given included sided.
\begin{aligned}\text{Area} &= \dfrac{a^2 \sin B\sin C}{2\sin A}\end{aligned} \begin{aligned}\text{Area} &= \dfrac{b^2 \sin A\sin C}{2\sin B}\end{aligned} \begin{aligned}\text{Area} &= \dfrac{c^2 \sin A\sin B}{2\sin C}\end{aligned}

We use the triangle shown below as reference:

Don’t worry, the areas may appear daunting at first, but with enough practice, you’ll get the hang of it as well!

Problem 5

Calculate area of the triangle, \Delta ABC, using two consecutive angles and one included angle provided for each item.

a. a = 10 ft

With a = 10 ft, the triangle becomes an ASA triangle with \angle B = 50\degree and \angle C = 60\degree as the two consecutive angles. Using the first formula, we can calculate the area of the ASA triangle shown below.

\begin{aligned}\text{Area} &= \dfrac{a^2 \sin B\sin C}{2\sin A}\\&= \dfrac{10^2 \sin 70\degree\sin 60\degree}{2\sin 50\degree}\\&\approx 53.1\end{aligned}

This means that when a = 10 ft, the area of \Delta ABC is equal to 53.1 ft2.

b. b = 12 cm

We’ll apply a similar process to calculate the triangle’s area given the new dimension: c = 12 cm. This time we use c^2 and 2\sin C in the numerator and denominator, respectively.

\begin{aligned}\text{Area} &=\dfrac{b^2 \sin A\sin C}{2\sin B}\\&= \dfrac{12^2 \sin 50\degree\sin 60\degree}{2\sin 70\degree}\\&\approx 50.8\end{aligned}

Hence, of \Delta ABC will have an area of 50.8 cm2 instead.

c. c = 20.4 inches

Now, what if we’re given c’s length instead? Let’s use the third formula and see how it differs from the first two.

\begin{aligned}\text{Area} &= \dfrac{c^2 \sin A\sin B}{2\sin C}\\&= \dfrac{20.4^2 \sin 50\degree\sin 70\degree}{2\sin 60\degree}\\&\approx 173.0\end{aligned}

This shows that when c = 20.4 inches, the area of our new ASA triangle is now approximately equal to 173.0 inches2.

These three examples show that angles alone can’t tell us how large a triangle is – this is why congruencies and relationships of triangles often depend on their angles and sides. Want to try out more examples? Try finding the unknown measures of the lengths for Problem 5a- c and confirm whether the areas will remain the same for each item.