Using the Law of Sines
In any triangle $\Delta ABC$, it will be true that $\frac{\sin{A}}{a} = \frac{\sin{B}}{b} = \frac{\sin{C}}{c}$
To use it, we must already know three out of four of the pieces of information. So we need to have one of the following situations:
• We know one angle and the length of its opposite side, and we know the angle opposite the side we want to find. (This is AAS from the congruence theorems of geometry. In case we have ASA, we can find the third angle and then use the Law of Sines.)
• We know one angle and the length of its opposite side, and we know the length of the side opposite the angle we want to find. (This is an ambiguous case and care must be taken in using the Law of Sines in this case: see at the end of the Law of Cosines example below.)
Notice that in both cases, we need to know one angle and the length of that angle’s opposite side. If we don’t have that, the Law of Sines cannot be used.
Example:
In $\Delta ABC$, $\angle A = 30^{\circ}$, $\angle B = 50^{\circ}$, and $a = 5$. Find the length of side $b$.
For this problem we will use the first two ratios in the Law of Sines:
$\frac{\sin{A}}{a} = \frac{\sin{B}}{b}$
Substituting in the given information:
$\frac{\sin{30^{\circ}}}{5} = \frac{\sin{50^{\circ}}}{b}$
Now solve for $b$ (do it algebraically before you compute any numbers!)
$b\sin{30^{\circ}} = 5\sin{50^{\circ}}$
$b = \frac{5\sin{50^{\circ}}}{\sin{30^{\circ}}}$
Now enter that last expression into your calculator: first check that you are in degree mode!
$b \approx 7.66$
Continuing the Example:
We can now go on to solve this triangle: the remaining unknowns are $c$ and $\angle C$. Remember, this is not a right triangle, so we cannot use the Pythagorean Theorem. But we can use the fact that the sum of all the angles in the triangle must be $180^{\circ}$ to find $\angle C = \180^{\circ} -(30^{\circ} + 50^{\circ} = 100^{\circ}$, and then use the Law of Sines again to find the length $c$. i There are two ways to do this, but since we had to round $b$, it is better to use $a$ and $\angle A$ in our proportion:
$\frac{\sin{A}}{a} = \frac{\sin{C}}{c}$
$\frac{\sin{30^{\circ}}}{5} = \frac{\sin{100^{\circ}}}{c}$
Solving this proportion as before, we get
$c = \frac{5\sin{100^{\circ}}}{\sin{30^{\circ}}}$
Enter this in the calculator to compute:
$c \approx 9.85$
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We can also use the Law os Sines to find an angle. That can be problematic, though. We will do that next time if time permits.
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Using the Law of Cosines:
For any triangle $\Delta ABC$, the following are true:
$a^{2} = b^{2} + c^{2} – 2bc\cos{A}$
$b^{2} = a^{2} + c^{2} – 2ac\cos{B}$
$c^{2} = a^{2} + b^{2} – 2ab\cos{C}$
[Notice that if $C$ is a right angle, so $\cos(C) = 0$, the last identity is just the Pythagorean Theorem.]
Using them:
The Law of Cosines is especially useful when we know the length of two sides and their included angle. (This is SAS from the congruence theorems of geometry.) It is also our only choice to use for finding an angle in the SSS case. However, care must be taken in that case!
Example:
If $a = 13$, $b = 12$ and $\angle C = 78^{\circ}}$, find the length of side $c$.
We use the third formula above:
$c^{2} = a^{2} + b^{2} – 2bc\cos{C}$
$c^{2} = 13^{2} + 12^{2} – 2(13)(12)\cos{78^{\circ}}$
Don’t compute anything yet, just solve for $c$:
$c = \sqrt{13^{2} + 12^{2} – 2(13)(12)\cos{78^{\circ}}}$
Only the positive square root is needed here because $c$ is a length.
Now enter that into your calculator to find $c$:
$c^{2} = 13^{2} + 12^{2} – 2(13)(12)\cos{78^{\circ}}$75
$c \approx 15.75$
Continuing this Example:
We can now go on to solve this triangle. Remember that it is always best to avoid using rounded numbers in our computations, and when forced to do so we should include some extra digits in the decimal part.
We now know all three sides of the triangle and the measure of $\angle C$. We need to find the measures of the other two angles. Notice that once we know one of them, the other will be easy to find.
Here is one thing we could do:
Using sides $a$ and $c$ and angle $C$, we could use the Law of Sines to find $\angle A$. Since we only have $c$ as a rounded number, we will do one of two things (depending on what device/program we are using to calculate):
• Compute $c$ to a few more places, let’s say to 6 places at least. (The more the better.)
• Don’t compute $c$ at all, but enter the expression using the square root into our formula. This is the best way, but if doing it on a hand-held calculator may be hard to enter.
I will show both ways.
We use this version of the Law of Sines:
$\frac{\sin{A}}{a} = \frac{\sin{C}}{c}$
$\frac{\sin{A}}{13} = \frac{\sin{78^{\circ}}}{c}$
Before we substitute for $c$ in that formula, we need a few more decimal digits, so go back and compute:
$c \approx 15.752192$ to six decimal places
Substitute this in and solve for $\sin{A}$:
$\frac{\sin{A}}{13} \approx \frac{\sin{78^{\circ}}}{15.752192}$
$\sin{A} \approx \frac{13\sin{78^{\circ}}}{15.752192}$
Don’t compute anything else yet! We want $\angle A$, not $\sin{A}$. So let’s compute
$\sin^{-1}\left(\frac{13\sin{78^{\circ}}}{15.752192}\right) \approx 52.83^{\circ}$
This may or may not give us $\angle A$. In this case, since the angle is positive, we can be assured that we did in fact get the correct angle. The problem arises when $\sin^{-1}$ gives a negative angle: then we will have to correct it to the actual angle in the triangle.
So $\angle A \approx 52.83^{\circ}$.
We can now use the fact that all the angles in the triangle add up to $180^{\circ}$ to find an approximation for angle $B$:
$\angle B \approx 180^{\circ} – \left(78^{\circ} + 52.83^{\circ}\right)$
$\angle B \approx 49.17^{\circ}$
We have now solved the triangle.
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Finding angle $B$ without approximating $c$:
We would do the same as above, but instead of substituting in the approximation $c \approx 15.752192$, we substitute in the formula $c = \sqrt{13^{2} + 12^{2} – 2(13)(12)\cos{78^{\circ}}}$:
$\sin{A} \approx \frac{13\sin{78^{\circ}}}{\sqrt{13^{2} + 12^{2} – 2(13)(12)\cos{78^{\circ}}}}$}$
$A \approx \sin^{-1}\left\frac{13\sin{78^{\circ}}}{\sqrt{13^{2} + 12^{2} – 2(13)(12)\cos{78^{\circ}}}}\right)$}$
Enter this in your calculator – easier to do if you are using Desmos scientific calculator or something like that! – and find that $\angle A \approx 52.83^{\circ}$ to two decimal places, just as in the other way.
As long as you include enough decimal places in the numbers you enter into your calculations, the final result will be the same after rounding. But it’s always better to avoid using rounded numbers in calculations if you can.
