+314 How do you know the length of the hypotenuse and side of a right triangle when you are only given one of the sides and an angle? Thanks!
+2440 For the purposes of this problem, I will assume "when you are given one of the sides and an angle" means an angle other than the right angle. Another assumption I am making is that one of the given side lengths is not the hypotenuse as I doubt you would ask for the length of the hypotenuse if you already knew its side length. If these conditions are true, there are still 2 possibilities. I have created a diagram for both scenarios. Here they are:
Let's assume that \(AB=6\) and \(m\angle C=35^{\circ}\). Now, let's "solve" the triangle. "Solving" the triangle means to figure out all the missing side lengths and angle measures. Let's start with the easy one first. You can use the triangle sum theorem to figure out what the measure of the angle located at vertex A:
| \(m\angle A+m\angle B+m\angle C=180\) | This is what the triangle sum theorem says; the sum of the measures of the angles in a triangle will equal 180 degrees. Now, plug in the angle measures that we already know. |
| \(m\angle A+90+35=180\) | Let's simplify the left-hand side of the equation by combining the like terms. |
| \(m\angle A+125=180\) | Finally, isolate the measure of angle a to figure out its exact measure. |
| \(m\angle A=55^{\circ}\) | |
The next part requires the knowledge of right-triangle trigonometry. Let's review that quickly. I understand that
\(\frac{\sin \textcolor{\green}\alpha}{1}=\frac{\textcolor{red}{\text{opposite}}}{\textcolor{blue}{\text{hypotenuse}}}\)
\(\frac{\cos \textcolor{green}{\alpha}}{1}=\frac{\textcolor{red}{\text{adjacent}}}{\textcolor{blue}{\text{hypotenuse}}}\)
\(\frac{\tan \textcolor{green}{\alpha}}{1}=\frac{\textcolor{red}{\text{opposite}}}{\textcolor{blue}{\text{adjacent}}}\)
Since we know that \(AB=6\) and \(m\angle C=35^{\circ}\), we should use the sine function to calculate the length of the hypotenuse and the tangent function for the angle adjacent to the known angle. Let's solve for the hypotenuse!
| \(\frac{\sin \textcolor{\green}\alpha}{1}=\frac{\textcolor{red}{\text{opposite}}}{\textcolor{blue}{\text{hypotenuse}}}\) | We know the measure of the angle, and we know the measure of the opposite side length. |
| \(\frac{\sin \textcolor{green}{35^{\circ}}}{1}=\frac{\textcolor{red}{\text{6}}}{\textcolor{blue}{\text{AC}}}\) | Multiply by the length of AC on both sides. |
| \(AC\sin 35^{\circ}=6\) | Now, divide by \(\sin 35^{\circ}\). |
| \(AC=\frac{6}{\sin 35^{\circ}}\) | \(\frac{6}{\sin 35^{\circ}}\) is the exact measure of the hypotenuse. By isolating its measure, we can now input it into a trusty scientific calculator (like the one provided on this Web site!) to obtain its approximate length. Be sure that the calculator is set to degree mode. |
| \(AC=\frac{6}{\sin 35^{\circ}}\approx10.46\) | Generally, rounding to the hundredths place is a sufficient estimation. |
Now, let's solve for the other side length. This time, we will need the tangent function.
| \(\frac{\tan \textcolor{green}{\alpha}}{1}=\frac{\textcolor{red}{\text{opposite}}}{\textcolor{blue}{\text{adjacent}}}\) | Let's do the exact same process of plugging in and isolating. |
| \(\frac{\tan \textcolor{green}{35^{\circ}}}{1}=\frac{\textcolor{red}{6}}{\textcolor{blue}{BC}}\) | |
| \(BC\tan 35^{\circ}=6\) | |
| \(BC=\frac{6}{\tan 35^{\circ}}\approx8.57\) | |
We have solved the triangle because we have found all the missing side lengths and angle measures. However, there is 1 scenario that we have not covered yet. However, the process is the same. Here is the diagram, with AB=5 and \(m\angle A=25^{\circ}\). You will notice that this is different that the first diagram.
Let's find the easy one again: the measure of the angle of C:
| \(m\angle A+m\angle B+ m\angle C=180\) | |
| \(25+90+m\angle C=180\) | |
| \(115+m\angle B=180\) | |
| \(m\angle B=65^{\circ}\) | |
Oh wait! Now, this looks the exact same as the 1st triangle because we know the measure of the angle across from the known side length. Therefore, to solve this triangle, you would use the exact same process as done above.
