Video Transcript
Capital 𝐴 capital 𝐵 capital 𝐶 is a triangle with sides capital 𝐴 capital 𝐵 is equal to lowercase 𝑐, capital 𝐴 capital 𝐶 is equal to lowercase 𝑏, and capital 𝐵 capital 𝐶 is equal to lowercase 𝑎. If the perimeter of triangle capital 𝐴 capital 𝐵 capital 𝐶 is equal to lowercase 𝑝, which of the given relations can we use to write the perimeter in terms of the sine of the angles using the sin rule? Is the answer option (A) lowercase 𝑝 is equal to the sin of angle capital 𝐴 plus the sin of angle capital 𝐵 plus the sin of angle capital 𝐶 all divided by the sin of angle capital 𝐴? Is the answer option (B) lowercase 𝑝 is equal to lowercase 𝑎 multiplied by the sin of capital 𝐴 plus the sin of capital 𝐵 plus the sin of capital 𝐶 all divided by the sin of capital 𝐴? Is it option (C) the perimeter is equal to lowercase 𝑏 multiplied by the sin of capital 𝐴 plus the sin of capital 𝐵 plus the sin of capital 𝐶 all over the sin of capital 𝐴? Is the answer option (D) the perimeter is equal to lowercase 𝑐 multiplied by the sin of capital 𝐴 plus the sin of capital 𝐵 plus the sin of capital 𝐶 all over the sin of capital 𝐴? Or is the answer option (E) the perimeter is equal to lowercase 𝑎 multiplied by the products of the sin of capital 𝐴, the sin of capital 𝐵, and the sin of capital 𝐶 over the sin of capital 𝐴?
In this question, we’re given some information about a triangle. So, one thing we can start by doing is sketching a picture of this triangle. First, we label the vertices of our triangle capital 𝐴, capital 𝐵, and capital 𝐶. Next, we can label the lengths of the sides of our triangle. It’s also worth pointing out here capital 𝐴, capital 𝐵, and capital 𝐶 not only represent the vertices of our triangle, they also represent the interior angle at this vertex of our triangle. But this isn’t much information to go on. This could, in fact, be any possible triangle.
We need to determine an expression for the perimeter of this triangle. Remember, that’s going to be the sum of all of the lengths which make up our triangle. So, we’ll start off with the following equation: the perimeter is the sum of the side lengths of our triangle. So, lowercase 𝑝 is going to be equal to lowercase 𝑎 plus lowercase 𝑏 plus lowercase 𝑐. Remember, the question wants us to express the perimeter in terms of the sine of the angles of our triangle, and we’re told to do this by using the sine rule.
So, let’s start by recalling the sine rule or the law of sines. One version of this tells us if lowercase 𝑎, 𝑏, and 𝑐 are the side lengths of a triangle opposite the angles capital 𝐴, capital 𝐵, and capital 𝐶, respectively, then lowercase 𝑎 over sin capital 𝐴 is equal to lowercase 𝑏 over sin capital 𝐵 is equal to lowercase 𝑐 divided by sin of capital 𝐶. And since we’re told in the question capital 𝐴𝐵𝐶 represents a triangle, the law of sines must apply for our triangle.
There’s actually a lot of different ways we can use the law of sines to rewrite our perimeter. We need to determine which one we need to use. And to do this, we’re going to need to look at our answers. We can see something all of our answers have in common. In the denominator of our fraction, we have the sin of angle capital 𝐴. This means when we rewrite our perimeter, we need to have the sin of capital 𝐴 in our denominator. Therefore, when we’re rewriting the equation for our perimeter, we want to rewrite lowercase 𝑎, lowercase 𝑏, and lowercase 𝑐 in terms of the sin of capital 𝐴.
There’s actually a few different ways of doing this. Let’s start with the first two parts of this equation. This means that lowercase 𝑎 over sin capital 𝐴 is equal to lowercase 𝑏 divided by sin capital 𝐵. We want to use this to find an expression for lowercase 𝑏 in terms of sin of capital 𝐴. Well, we can just rearrange this equation for lowercase 𝑏. We just need to multiply through by the sin of capital 𝐵. We get lowercase 𝑏 is equal to lowercase 𝑎 times the sin of capital 𝐵 divided by the sin of capital 𝐴.
And importantly, the sin of capital 𝐴 is in the denominator of this expression. We can substitute this into our equation for the perimeter 𝑝. Of course, we’re going to want to do exactly the same for lowercase 𝑐. However, this time we’re going to need to use the fact that lowercase 𝑎 over sin of capital 𝐴 is equal to lowercase 𝑐 divided by the sin of capital 𝐶.
And just as we did before, we want to use this equation to find an expression for our length lowercase 𝑐. And to do this, we’re going to multiply through by the sin of capital 𝐶. This gives us that lowercase 𝑐 is equal to lowercase 𝑎 multiplied by the sin of capital 𝐶 all divided by the sin of capital 𝐴. And once again, we’re going to substitute this into our equation for the perimeter lowercase 𝑝.
But now we seem to run into a problem. We can’t do exactly the same to find an expression for lower case 𝑎. Instead, we just need to notice we just want to write lowercase 𝑎 to have a denominator of the sin of capital 𝐴. And one way of doing this is to directly write lowercase 𝑎 as a fraction in this form. We just write it as lowercase 𝑎 multiplied by the sin of capital 𝐴 divided by the sin of capital 𝐴. Therefore, we found the following expression for the perimeter of our triangle lowercase 𝑝.
And before we continue, there is one interesting thing worth pointing out about the sin of capital 𝐴. Remember in the question we’re told that capital 𝐴 capital 𝐵 capital 𝐶 makes up a triangle. In particular, this means that the measure of angle capital 𝐴 is not equal to zero, and it’s not equal to 180 degrees. It’s some angle between these two values. And this is useful to know because it means the sin of capital 𝐴 is not equal to zero. In other words, we’re not dividing by zero in this expression.
Now, in fact, we could just leave our answer like this. However, we’re going to take out the shared factor of side lowercase 𝑎 from this expression. Then, we’ll write this as one fraction all divided by the sin of capital 𝐴. Doing this gives us the perimeter of our triangle lowercase 𝑝 is equal to lowercase 𝑎 multiplied by the sin of capital 𝐴 plus the sin of capital 𝐵 plus the sin of capital 𝐶 all divided by the sin of capital 𝐴, which we can see was our option (B).
Therefore, in this question, we were able to use the sin rule to find an expression for the perimeter of any given triangle given only one side length and all of the angles of our triangle. We were able to show the perimeter 𝑝 of this triangle is equal to lowercase 𝑎 multiplied by the sin of capital 𝐴 plus the sin of capital 𝐵 plus the sin of capital 𝐶 all divided by the sin of angle capital 𝐴.
