# Question Video: Forming and Solving a System of Linear Equations in Three Unknowns Involving Angles of a Triangle Mathematics

In the triangle 𝐴𝐵𝐶, one of the angles is the arithmetic mean of the other two. Find each angle of the triangle given the difference between the smaller and larger angles is 61°.

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### Video Transcript

In the triangle 𝐴𝐵𝐶, one of the angles is the arithmetic mean of the other two. Find each angle of the triangle given the difference between the smaller and larger angles is 61 degrees.

It’s sensible to begin by defining each of the angles in our triangle. We’ll say that each of the three angles in our triangle are equal to 𝑥 degrees, 𝑦 degrees, and 𝑧 degrees. Before even using any of the information in the question, we know that the sum of the angles is 180 degrees. So we can say that 𝑥 plus 𝑦 plus 𝑧 must be equal to 180. If we then say that 𝑥 is the smallest and 𝑧 is the largest, it follows that 𝑦 must be the arithmetic mean of 𝑥 and 𝑧. In other words, when we find the sum of 𝑥 and 𝑧 and then divide that by two, we end up with the third angle; that’s 𝑦. So 𝑥 plus 𝑧 over two equals 𝑦. Let’s neaten this up a little bit and multiply through by two.

When we do, we find that 𝑥 plus 𝑧 equals two 𝑦. The other piece of information we have is that the difference between the smaller and larger angles is 61 degrees. So 𝑧 minus 𝑥 must be equal to 61. And we’ve created a system of linear equations in three variables, 𝑥, 𝑦, and 𝑧. So how do we solve each of these equations? Well, ideally, we want to create an equation in just one variable. And so, we’re gonna begin by manipulating the first two equations I’ve labelled one and two. If we subtract two 𝑦 from both sides of our second equation, we get 𝑥 minus two 𝑦 plus 𝑧 equal zero. Let’s call this equation three. Notice that in equation one and equation three, we have exactly the same number of 𝑥s and 𝑧s. The only thing that changes is the number of 𝑦s we have.

So we’re going to subtract equation three from an equation one. This will have the effect of eliminating both 𝑥 and 𝑧. And we’ll just have an equation in terms of 𝑦. When we do, we’re left with 𝑦 minus negative two 𝑦 equals 180 minus zero. 𝑦 minus negative two 𝑦 is three 𝑦. And we can solve this equation for 𝑦 by dividing through by three. And when we do, we obtain 𝑦 to be equal to 60. So we’ve calculated the sides of the middle angle. We need to work out the size of angle 𝑥 and 𝑧. We’re going to take the value of 𝑦 and substitute it into equation two. And when we do, we’ll be left with two equations purely in terms of 𝑥 and 𝑧.

When we do, we get 𝑥 plus 𝑧 equals two times 60. So that’s 120. So we’ve got equation four and five. These are purely in terms of 𝑥 and 𝑧. Notice, though, that the signs of the 𝑥s are different in each equation. So we’re going to add these two equations, and that will eliminate the 𝑥. We could, alternatively, subtract the equations to eliminate the 𝑧. When we add the expressions on the left-hand side, we get 𝑧 minus 𝑥 plus 𝑥 plus 𝑧, which is simply two 𝑧. And on the right-hand side, we get 120 plus 61 which is 181.

We solved this equation by dividing through by two. And we see that 𝑧 is equal to 90.5. All that’s left is to calculate the size of angle 𝑥. It’s sensible to go back to either equation four or five. Now, I’m going to choose equation five because there are no negatives. We get 𝑥 plus 90.5 equals 120. And then, we solve for 𝑥 by subtracting 90.5 from both sides, to find that 𝑥 is equal to 29.5.

In ascending order, our angles are 29.5 degrees, 60 degrees, and 90.5 degrees. Now, we could absolutely check our working out by going back to equation one. We replace 𝑥 with 29.5, 𝑦 with 60, and 𝑧 with 90.5. Their sum is 180 as required. So we know we’ve done our working out correctly. The angles are 29.5, 60, and 90.5 degrees.