The Law of Cosines is a way of solving for the missing angles and side lengths for any oblique (non-right angled) triangle. This method uses the formula:
This formula is used whenever we have 2 side lengths and an angle and need to the missing side length. It can also be rearranged as such:
This formula is used whenever we have all 3 side lengths and need to determine an angle(s).
Find the missing side length of the following triangle:
As we have \(2\) side lengths and an angle, we can use the side length formula, \(\textcolor{blue}{c}^2 = \textcolor{red}{a}^2 + \textcolor{green}{b}^2 - 2\textcolor{red}{a}\textcolor{green}{b}\textcolor{royalblue}{\cos(\text{C})}\), to determine the missing side length:
\(\textcolor{red}{d}^2 = \textcolor{green}{e}^2 + \textcolor{blue}{f}^2 - 2\textcolor{green}{e}\textcolor{blue}{f}\textcolor{crimson}{\cos \text{D}}\)
\(\textcolor{red}{d}^2 = (\textcolor{green}{11})^2 + (\textcolor{blue}{14})^2 - 2(\textcolor{green}{11})(\textcolor{blue}{14})\textcolor{crimson}{\cos(82°)}\)
\(\textcolor{red}{d}^2 = 121 + 196 - (308)(0.139)\)
\(\textcolor{red}{d}^2 = 317 - 42.87\)
\(\sqrt{\textcolor{red}{d}^2} = \sqrt{274.13}\)
\(\textcolor{red}{d = 16.55\;[\text{cm}]}\)
Therefore, we can determine the length of \(\boldsymbol{\textcolor{red}{d = 16.55 [\textbf{cm}]}}\).
First, we can sketch the triangle based on the given information to help visualize the problem:
As we have \(2\) side lengths and an angle, we can use the formula \(\textcolor{blue}{c}^2 = \textcolor{red}{a}^2 + \textcolor{green}{b}^2 - 2\textcolor{red}{a}\textcolor{green}{b}\textcolor{royalblue}{\cos \text{C}}\) to determine the length of \(\textcolor{red}{m}\):
\(\textcolor{red}{m}^2 = c^2 + b^2 - 2\textcolor{green}{c}\textcolor{blue}{b}\textcolor{crimson}{\cos \text{M}}\)
\(\textcolor{red}{m}^2 = (\textcolor{green}{18})^2 + (\textcolor{blue}{21})^2 - 2(\textcolor{green}{18})(\textcolor{blue}{21})\textcolor{crimson}{\cos(61°)}\)
\(\textcolor{red}{m}^2 = 324 + 441 - (378)(0.485)\)
\(\textcolor{red}{m}^2 = 765 - 183.26\)
\(\sqrt{\textcolor{red}{m}^2} = \sqrt{398.48}\)
\(\textcolor{red}{m = 19.96\;[\text{cm}]}\)
Using the length of \(\textcolor{red}{m}\), we can determine the measurements of the missing angles. We can begin by using the Law of Sines to find \(\textcolor{royalblue}{\angle \text{B}}\):
\(\cfrac{\textcolor{royalblue}{\sin \text{B}}}{\textcolor{blue}{b}} = \cfrac{\textcolor{crimson}{\sin M}}{\textcolor{red}{m}}\)
\(\cfrac{\textcolor{royalblue}{\sin \text{B}}}{\textcolor{blue}{21}} = \cfrac{\textcolor{crimson}{\sin 61°}}{\textcolor{red}{19.96}}\)
\((21)\left(\cfrac{\textcolor{royalblue}{\sin \text{B}}}{\textcolor{blue}{21}}\right) = (21)(0.043818622)\)
\(\textcolor{royalblue}{\angle \text{B}} = \sin⁻¹(0.920191074)\)
\(\textcolor{royalblue}{\angle \text{B} = 66.95°}\)
We can find the final angle \(\textcolor{seagreen}{\angle \text{C}}\) by subtracting the other 2 angles, \(\textcolor{crimson}{\angle \text{M}}\) and \(\textcolor{royalblue}{\angle \text{B}}\) from \(180°\):
\(\textcolor{seagreen}{\angle \text{C}} = 180° - (\textcolor{crimson}{\angle \text{M}} + \textcolor{royalblue}{\angle \text{B}})\)
\(\textcolor{seagreen}{\angle \text{C}} = 180° - (61° + 66.95°)\)
\(\textcolor{seagreen}{\angle \text{C}} = 180° - 127.95°\)
\(\textcolor{seagreen}{\angle \text{C} = 52.05°}\)
Therefore, we can determine that \(\boldsymbol{\textcolor{red}{m = 19.96 [\textbf{cm}]}}\), \(\boldsymbol{\textcolor{royalblue}{\angle \text{B} = 66.95°}}\), and \(\boldsymbol{\textcolor{seagreen}{\angle \text{C} = 52.05°}}\).
Solve for the indicated angle, to the nearest whole degree:
As we have 3 side lengths, we can use the angle formula, \(\textcolor{royalblue}{\cos \text{C}} = \cfrac{\textcolor{red}{a}^2 + \textcolor{green}{b}^2 - \textcolor{blue}{c}^2}{2\textcolor{red}{a}\textcolor{green}{b}}\), to solve for the missing angle:
\(\textcolor{royalblue}{\cos \text{P}} = \cfrac{\textcolor{red}{z}^2 + \textcolor{green}{j}^2 - \textcolor{blue}{p}^2}{2\textcolor{red}{z}\textcolor{green}{j}}\)
\(\textcolor{royalblue}{\cos\text{P}} = \cfrac{(\textcolor{red}{5.1})^2 + (\textcolor{green}{3.8})^2 - (\textcolor{blue}{4.5})^2}{2(\textcolor{red}{5.1})(\textcolor{green}{3.8})}\)
\(\textcolor{royalblue}{\cos\text{P}} = \cfrac{26.01 + 14.44 - 20.25}{38.76}\)
\(\textcolor{royalblue}{\cos \text{P}} = \cfrac{20.2}{38.76}\)
\(\textcolor{royalblue}{\angle \text{P}} = \cos⁻¹(0.52115583)\)
\(\textcolor{royalblue}{\angle \text{P}} = 58.59° \approx 59°\)
Therefore, we can determine that the value of \(\textcolor{royalblue}{\angle P}\) to the nearest whole degree is \(\boldsymbol{59°}\).
First, we can sketch the triangle based on the given information to help visualize the problem:
As we have 3 side lengths, we can use the angle formula to solve for the missing angle:
We can begin with finding \(\textcolor{crimson}{\angle \text{M}}\):
\(\textcolor{crimson}{\cos \text{M}} = \textcolor{green}{r}^2 + \cfrac{\textcolor{blue}{v}^2 - \textcolor{red}{m}^2}{2\textcolor{green}{r}\textcolor{blue}{v}}\)
\(\textcolor{crimson}{\cos \text{M}} = \cfrac{ (\textcolor{green}{3.5})^2 + (\textcolor{blue}{4.0})^2 - (\textcolor{red}{3.2})^2}{2(\textcolor{green}{3.5})(\textcolor{blue}{4.0})}\)
\(\textcolor{crimson}{\cos \text{M}} = \cfrac{16 + 12.25 - 10.24}{28}\)
\(\cos M = \cfrac{18.01}{28}\)
\(\textcolor{crimson}{\angle \text{M}} = \cos^{-1}(0.643214285)\)
\(\textcolor{crimson}{\angle \text{M} = 49.97° \approx 50°}\)
We can find \(\textcolor{royalblue}{\angle \text{V}}\) using the same process:
\(\textcolor{royalblue}{\cos \text{V}} = \cfrac{\textcolor{red}{m}^2 + \textcolor{green}{r}^2 - \textcolor{blue}{v}^2}{2mr}\)
\(\textcolor{royalblue}{\cos \text{V}} = \cfrac{(\textcolor{red}{3.2})^2 + (\textcolor{green}{3.5})^2 - (\textcolor{blue}{4.0})^2}{2(\textcolor{red}{3.2})(\textcolor{green}{3.5})}\)
\(\textcolor{royalblue}{\cos \text{V}} = \cfrac{12.25 + 10.24 - 16}{22.4}\)
\(\textcolor{royalblue}{\cos \text{V}} = \cfrac{6.49}{22.4}\)
\(\textcolor{royalblue}{\angle \text{V}} = \cos⁻¹(0.289732142)\)
\(\textcolor{royalblue}{\angle \text{V} = 73.16° \approx 73°}\)
We can find \(\textcolor{seagreen}{\angle \text{R}}\) by subtracting the other 2 angles, \(\textcolor{crimson}{\angle \text{M}}\) and \(\textcolor{royalblue}{\angle \text{V}}\), from \(180°\):
\(\textcolor{seagreen}{\angle \text{R}} = 180° - (\textcolor{crimson}{\angle M} + \textcolor{royalblue}{\angle V})\)
\(\textcolor{seagreen}{\angle \text{R}} = 180° - (\textcolor{crimson}{50°} + \textcolor{royalblue}{73°})\)
\(\textcolor{seagreen}{\angle \text{R}} = 180° - 123°\)
\(\textcolor{seagreen}{\angle \text{R} = 57°}\)
Therefore, we can determine that \(\boldsymbol{\textcolor{crimson}{\angle \text{M} = 50°}}\), \(\boldsymbol{\textcolor{royalblue}{\angle \text{V} = 73°}}\), and \(\boldsymbol{\textcolor{green}{\angle \text{R} = 57°}}\).
Since we have 2 side lengths and an angle, we can determine the bridge length using the formula \(\textcolor{blue}{c}^2 = \textcolor{red}{a}^2 + \textcolor{green}{b}^2 - 2\textcolor{red}{a}\textcolor{green}{b}\textcolor{royalblue}{\cos \text{C}}\). In this instance, the bridge will be represented as \(c\) with the other \(2\) side lengths represented as \(a\) and \(b\):
\(\textcolor{blue}{c}^2 = (\textcolor{red}{35})^2 + (\textcolor{green}{63})^2 - 2(\textcolor{red}{35})(\textcolor{green}{63})\textcolor{royalblue}{\cos(67°)}\)
\(\textcolor{blue}{c}^2 = 1225 + 3969 - 1723.124277\)
\(\sqrt{\textcolor{blue}{c}^2} = \sqrt{3470.875723}\)
\(\textcolor{blue}{c = 58.9 \approx 59\; [\text{m}]}\)
Therefore, we can determine that the length of the bridge to the nearest meter is \(\textbf{59 [m]}\).
Laurissa is designing a reflecting pool, in the shape of a triangle, for her backyard.
i. As we have all 3 side lengths already determined, we can solve for the interior angles by using the following formula:
We can start by solving for \(\textcolor{crimson}{\angle \text{X}}\):
\(\textcolor{crimson}{\cos \text{X}} = \cfrac{\textcolor{green}{y}^2 + \textcolor{blue}{z}^2 - \textcolor{red}{x}^2}{2\textcolor{green}{y}\textcolor{blue}{z}}\)
\(\textcolor{crimson}{\cos \text{X}} = \cfrac{(\textcolor{green}{4.0})^2 + (\textcolor{blue}{3.8})^2 - (\textcolor{red}{3.8})^2}{2(\textcolor{green}{4.0})(\textcolor{blue}{3.8})}\)
\(\textcolor{crimson}{\cos \text{X}} = \cfrac{14.44 + 16 - 14.44}{30.4}\)
\(\textcolor{crimson}{\cos \text{X}} = \cfrac{16}{30.4}\)
\(\textcolor{crimson}{\angle \text{X}} = \cos⁻¹(0.526315789)\)
\(\textcolor{crimson}{\angle \text{X} = 58.24° \approx 58°}\)
As \(\textcolor{red}{x}\) and \(\textcolor{blue}{z}\) share the same length, we can determine that \(\textcolor{crimson}{\angle \text{X}} = \textcolor{royalblue}{\angle \text{Z}}\).
We can find \(\textcolor{seagreen}{\angle \text{Y}}\) by subtracting the other 2 angles, \(\textcolor{crimson}{\angle \text{X}}\) and \(\textcolor{royalblue}{\angle \text{Z}}\) from \(180°\):
\(\textcolor{seagreen}{\angle \text{Y}} = 180° - (\textcolor{crimson}{\angle \text{X}} + \textcolor{royalblue}{\angle \text{Z}})\)
\(\textcolor{seagreen}{\angle \text{Y}} = 180° - (\textcolor{crimson}{58°} + \textcolor{royalblue}{58°})\)
\(\textcolor{seagreen}{\angle \text{Y}} = 180° - 116°\)
\(\textcolor{seagreen}{\angle \text{Y} = 64°}\)
Therefore, we can determine that \(\boldsymbol{\textcolor{crimson}{\angle \text{X} = 58°}}\), \(\boldsymbol{\textcolor{seagreen}{\angle \text{Y} = 64°}}\), and \(\boldsymbol{\textcolor{royalblue}{\angle \text{Z} = 58°}}\).
ii. In order to find the surface area of the water, we can use the Pythagorean theorem to find the height of the pool. We can begin by splitting the pool in half so that it resembles a right triangle:
\(\textcolor{red}{a}^2 + \textcolor{green}{b}^2 = \textcolor{blue}{c}^2\)
\(\textcolor{red}{a}^2 + (\textcolor{green}{2})^2 = (\textcolor{blue}{3.8})^2\)
\(\textcolor{red}{a}^2 + 4 = 14.44\)
\(\textcolor{red}{a}^2 = 14.44 - 4\)
\(\sqrt{\textcolor{red}{a}^2} = \sqrt{10.44}\)
\(\textcolor{red}{a = 3.231\;[\text{cm}]}\)
Now we can use the formula for surface area of a Triangle to determine the Surface Area of the water. In this instance, \(\textcolor{green}{2b}\) represents the base, \(b\) while \(\textcolor{red}{a}\) represents the height, \(h\):
\(A = \cfrac{\textcolor{green}{b}\textcolor{red}{h}}{2}\)
\(A = \cfrac{(\textcolor{green}{4})(\textcolor{red}{3.231})}{2}\)
\(A = (2)(3.231)\)
\(A = 6.462\;[\text{cm}^2]\)
Therefore, we can determine that the Surface Area of the water is \(\textbf{6.462 [cm²]}\).