# Question Video: Using Cosine Role to Calculate an Unknown Angle of a Triangle

Calculate 𝑚∠𝑄 to the nearest degree.

03:02

### Video Transcript

Calculate the measure of angle 𝑄 to the nearest degree.

So, in this question, we’ve been given the lengths of all three sides of a triangle and we’re asked to calculate one of the angles. Now, this is exactly the setup required in order to use the law of cosines. So, if you were to look up the standard law of cosines, it would probably look something like this. 𝑎 squared is equal to 𝑏 squared plus 𝑐 squared minus two 𝑏𝑐 cos 𝐴.

If you’re looking to calculate the size of an angle, as we are in this question, then that can be rearranged in just a couple of steps to give you this formula here. Cos of 𝐴 is equal to 𝑏 squared plus 𝑐 squared minus 𝑎 squared all over two 𝑏𝑐. However, there are no 𝑎s, 𝑏s, and 𝑐s in this question. This question involves 𝑄s, 𝑅, and 𝑆s, so we need to think about how we can apply this version of the law of cosines within this question.

So, we need to look at the structure within this law of cosines. There are two sides that are always treated identically, 𝑏 and 𝑐 are both squared and added in the numerator, whereas 𝑎 squared but then subtracted. 𝑏 and 𝑐 also both appear in the denominator. So, these two sides that are treated identically are the two sides that enclose the angle we’re looking for. So, in this case, that would be the side of eight centimetres and 10 centimetres.

The third side, which is treated differently in the law of cosines, it only appears in the numerator and is subtracted rather than added. This is the side that is opposite the angle that we’re looking to calculate. So, I don’t need 𝑎s, 𝑏s, and 𝑐s. I can just write down the law of cosines by thinking about where the sides are in relation to this angle.

So, we have cos of 𝑄 is equal to, well, the two sides next to it, first of all, are eight and 10, so we’ll have 8 squared plus ten squared. Then, I need to subtract the square of the opposite side, so that will be minus 15 squared. Then, I need to divide this by two multiplied by the two sides that enclose this angle. So, this is two multiplied by eight multiplied by 10.

If I then evaluate all this, I have the cause of angle 𝑄 is equal to negative 61 over 160. In order to work out angle 𝑄, I need to use cosine inverse. So, 𝑄 is equal to cosine inverse of negative 61 over 160. This gives me a value of 112.411132. And I’ve been asked to give this to the nearest degree, so my final answer then is that angle 𝑄 is 112 degrees to the nearest degree. So, again, there were no 𝑎s, 𝑏s, or 𝑐s within our question. We just looked at the structure of the law of cosines in order to work out which value should be substituted where.