precalculus - length of the sides of a triangle

Question

Folks,

I'm just about done with the precalculus part of my return to mathematics and have enjoyed relearning much that I had forgotten. I've successfully worked through many problems, but there are a few loose that I'm hoping I can get some help with.

I have no idea where to start with this one, but I'm sure it's something I forgot many years ago...

Two of the sides of a triangle have lengths $4$ cm and $6$ cm, and the angle between them is $120^\circ$. Calculate the length of the third side giving your answer in the form $m\sqrt{p}$ where $m$ and $p$ are integers and $p$ is prime.

The book gives the answer as $2\sqrt{19}$cm, but I don't know how to get there.

Thank You

Gary

Answer 1

Using the cosine rule, the third side is $$\sqrt{4^2+6^2-2\times4\times6cos(120)}$$. Note that $cos120=-0.5$, so we have $\sqrt{16+36+24}=\sqrt{76}=2\sqrt{19}$.

Answer 2

HINT: Do you remember the cosine rule?

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

This question is a direct application of this formula.

Answer 3

Personally, I like to solve this problem with a geometric approach using the Pythagoras theorem.

Start by drawing the triangle with one side as the base. Which side you choose is arbitrary but we need the known angle on it.

Now extend the base to form a right-angled triangle. We know the hypotenuse already and find the compliment angle ($180 - 120 = 60$). This is enough for us to fill in the other two sides.

Now we have a big triangle that is right-angled and we know two sides. The base is the sum of the original base and the intermediate triangle. Straightforward to find the hypotenuse using $c^2 = a^2 + b^2$.

This is the missing length of the original triangle and evaluates to $2 \sqrt{19}$.