Solve using Law of Cosines or Law of Sines

Question

I'm trying to solve these sets of problems please.

Determine the number of triangles with the given parts and solve each triangle (if possible).

1. $\alpha=39.6^\circ,c=18.4,a=3.7$
2. $\gamma=60^\circ,b=20,c=10\sqrt{3}$
3. $\beta=32.7^\circ,a=37.5,b=28.6$

It's asking to find the number of triangles using the parts they gave us. I assumed that it wants us to use law of sines or cosines but when I try my answer either goes over $180^\circ$ or the triangle isn't unsolvable. Can someone please explain to me how to solve these problems so I can know this once and for all?

Answer 1

The three problems you gave all have the following givens: an angle, a side opposite that angle, and another side. Say we are given $\alpha$, $a$, and $c$. To "solve the triangle" means to find the remaining angles and sides. For these set of givens, you may have zero, one, or two triangles.

You have zero triangles if the side opposite the given angle (in this case, $a$) is too short. This will happen if $a<c$ and $\alpha>90^\circ$.

You have two triangles if the given angle is acute and the side opposite it is less than the other given side. In this case, this will happen when $\alpha<90^\circ$ and $a<c$. (In the two triangles, one will have a short $b$, the other will have a long $b$.)

Otherwise, the givens describe a unique triangle.

1. Use a sine law to find the angle of the given side that doesn't have a given angle. In this case, $\frac{\sin\alpha}{a}=\frac{\sin\gamma}{c}$. Rearranging yields $\gamma=\sin^{-1}\left(\frac{c}{a}\sin\alpha\right)$.
2. Use the fact that the sum of the angles in a triangle is $180^\circ$ to find the third angle. In this case, $\beta=180^\circ-\alpha-\gamma$.
3. Use a sine law to find the third side. In this case, $b=a\frac{\sin\beta}{\sin\alpha}$.