10.1 Non-Right Triangles: Law of Sines

Generally, it's good practice to set up what you know and what you do not. I like to make a chart with the sides and angles of the triangle.

Because we know the angles in a triangle must add up to 180, we know $B=180-25-113=42$ and this importantly gives us an angle that is opposite a side we already know. So we can use the law of sines.

Again we set up our chart for what we know and do not know.

Using the sum of the angles in the triangle we find $A=180-26-120=34$

This gives us an angle opposite of a side that we know so we can use the law of sines.

We'll set up our chart for what we know and don't know, but we're going to have a second chart as well for the secondary triangle that could exist in a SSA problem. We'll label them with subscripts to tell them apart. Note that right now, both tables look the same because they only contain given information.

Now we focus on trying to find $C_1$ using the law of sines.

So we've found $C_1$! But we used the inverse sine function which has a limited range. It will only give us answers from $-90^\circ \leq \theta \leq 90^\circ$. There is another angle in quadrant II that has the same sine value as our $C_1$, and we need to check if it makes a valid triangle. $C_2=180-C_1=180-69.4=110.6$

Using the fact that the angles in a triangle add up to 180, we find that both of these triangles are valid.

Almost done. We just need to find those missing sides and the law of sines is perfect for it. As a rule of thumb for math problems in general, when you need to use numbers to calculate something you should use what you've been given. So let's use as much of the given information as possible when making these choices. That means using $\frac{\sin(A_1)}{a_1}$ for our known ratio instead of $\frac{\sin(C_1)}{c_1}$, because $C_1$ is something we've calculated and incidentally rounded.

And there we have it, two solutions!

Set up the knowns and unknowns.

Use law of sines to find a missing angle.

Remember to check the complement angle because inverse sine only gave you one of the possible angles. $C_2=180-C_1=180-56.8=123.2$

Use the sum of the angles being 180 to find the missing angles.

$$180-31-56.8=92.2$$

$$180-123.2-31=25.8$$

Both of these are valid triangles. Use the law of sines to find the missing sides.

We begin as we always do: expect 2 triangles with SSA because we have to use the inverse sine function.

Use the law of sines to find the missing angle.

$$ \frac{\sin(50^\circ)}{10} = \frac{\sin B_1}{9} \implies \sin B_1 = \frac{9\sin(50^\circ)}{10} \implies B_1 = \sin^{-1}\left(\frac{9\sin(50^\circ)}{10}\right)\approx 43.6^\circ $$

This means our complementary angle is $B_2=180^\circ - 43.6^\circ = 136.4^\circ$

However, when we try to find that third angle we run into an issue.

$$C_1=180-50-43.6=86.4$$ $$C_2=180-50-136.4=-6.4$$

$C_1$ makes sense, but $C_2$ shouldn't be negative. The other two angles in that triangle already add up to over $180^\circ$! So our second triangle here is not a valid triangle at all. We just have the one solution.

$$ \frac{\sin(50^\circ)}{10} = \frac{\sin(86.4^\circ)}{c_1} \implies c_1 = \frac{10\sin(86.4^\circ)}{\sin(50^\circ)} \approx 13.0 $$

Because this is SSA, we still start by expecting two possible triangles.

Use the law of sines to find $B_1$.

$$ \frac{\sin(125^\circ)}{80} = \frac{\sin(B_1)}{65} \implies \sin(B_1)=\frac{65\sin(125^\circ)}{80} \implies B_1=\sin^{-1}\left(\frac{65\sin(125^\circ)}{80}\right)\approx 41.7^\circ $$

Now check the complementary angle:

$$B_2=180^\circ-41.7^\circ=138.3^\circ$$

Find the third angles.

$$C_1=180-125-41.7=13.3$$ $$C_2=180-125-138.3=-83.3$$

$C_2$ is negative, so triangle 2 is not valid. We only have one solution.

$$ \frac{\sin(125^\circ)}{80}=\frac{\sin(13.3^\circ)}{c_1} \implies c_1=\frac{80\sin(13.3^\circ)}{\sin(125^\circ)}\approx 22.5 $$

Set up what we know and what we need.

Use the law of sines to try to find $B$.

$$ \frac{\sin(A)}{a}=\frac{\sin(B)}{b} \implies \frac{\sin(75^\circ)}{50}=\frac{\sin(B)}{70} \implies \sin(B)=\frac{70\sin(75^\circ)}{50}\approx 1.35 $$

At this point we hit a domain error for inverse sine, because sine values must satisfy $-1\leq \sin(\theta)\leq 1$. Since $1.35>1$, there is no real angle $B$ that works.

Therefore, no triangle exists with these measurements.

Set up what we know and what we need.

Use the law of sines to try to find $B$.

$$ \frac{\sin(A)}{a}=\frac{\sin(B)}{b} \implies \frac{\sin(120^\circ)}{1}=\frac{\sin(B)}{4} \implies \sin(B)=\frac{4\sin(120^\circ)}{1}\approx 3.46 $$

This gives a domain error for inverse sine. Since sine values must satisfy $-1\leq \sin(\theta)\leq 1$, the value $3.46$ is impossible.

Therefore, no triangle exists with these measurements.

Let triangle $ABC$ model the situation with $a=BC=100$, $B=65^\circ$, and $C=80^\circ$. We are asked to find $CA$, which is side $b$.

First find the missing angle.

$$A=180-65-80=35$$

Now use the law of sines to solve for $b$.

$$ \frac{\sin(A)}{a}=\frac{\sin(B)}{b} \implies \frac{\sin(35^\circ)}{100}=\frac{\sin(65^\circ)}{b} \implies b=\frac{100\sin(65^\circ)}{\sin(35^\circ)}\approx 158.0 $$

So the distance across the canyon from point $C$ to point $A$ is about $158.0$ feet.

Let $A$ and $B$ be the two mileposts with $AB=8$ km, and let $C$ be the plane. Using alternate interior angles, the angles of depression are equal to the angles of elevation at the ground points, so take $\angle A=37^\circ$ and $\angle B=44^\circ$. We want $CA$.

Find the third angle.

$$C=180-37-44=99$$

Since $CA$ is side $b$, use the law of sines.

$$ \frac{\sin(B)}{b}=\frac{\sin(C)}{c} \implies \frac{\sin(44^\circ)}{b}=\frac{\sin(99^\circ)}{8} \implies b=\frac{8\sin(44^\circ)}{\sin(99^\circ)}\approx 5.6 $$

So the plane is about $5.6$ km from point $A$.