The lengths of a triangle's sides are consecutive integers. If the largest angle is twice the smallest angle, find the cosine of the smallest angle?

Question

This question is a part of my Trig class' homework, and I really can't think of how to solve it, like how to use the sine law and cosine law to find the answer. Thanks in advance for your help.

Answer 1

Let $\theta$ be the smallest angle with the side lengths $n-1, n, n+1$ units. It is therefore obvious that the side opposite $\theta$ must have length $n-1$ units and that opposite $2\theta$ must have length $n+1$. Then, by rule of sines, we have: $$\frac{\sin \theta}{n-1}=\frac{\sin 2\theta}{n+1}=\frac{2\sin \theta \cos \theta}{n+1} \implies \cos \theta = \frac{n+1}{2(n-1)} \tag{1}$$

Now, using rule of cosines, we must have: $$(n-1)^2 = n^2 + (n+1)^2-2n(n+1)\cos \theta \implies \cos \theta = \frac{n+4}{2(n+1)}\tag{2}$$

Thus, solving from $(1)$ and $(2)$ gives us: $n =5 \implies \cos \theta = \frac34$.