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Course: Trigonometry>Unit 1

Lesson 6: Modeling with right triangles

Right triangle trigonometry review

Review right triangle trigonometry and how to use it to solve problems.

What are the basic trigonometric ratios?

$\mathrm{sin}\left(\mathrm{\angle }A\right)=$$\frac{\text{opposite}}{\text{hypotenuse}}$
$\mathrm{cos}\left(\mathrm{\angle }A\right)=$$\frac{\text{adjacent}}{\text{hypotenuse}}$
$\mathrm{tan}\left(\mathrm{\angle }A\right)=$$\frac{\text{opposite}}{\text{adjacent}}$

Practice set 1: Solving for a side

Trigonometry can be used to find a missing side length in a right triangle. Let's find, for example, the measure of $AC$ in this triangle:
We are given the measure of angle $\mathrm{\angle }B$ and the length of the $\text{hypotenuse}$, and we are asked to find the side $\text{opposite}$ to $\mathrm{\angle }B$. The trigonometric ratio that contains both of those sides is the sine:
$\begin{array}{rl}\mathrm{sin}\left(\mathrm{\angle }B\right)& =\frac{AC}{AB}\\ \\ \mathrm{sin}\left({40}^{\circ }\right)& =\frac{AC}{7}\phantom{\rule{1em}{0ex}}\mathrm{\angle }B={40}^{\circ },AB=7\\ \\ 7\cdot \mathrm{sin}\left({40}^{\circ }\right)& =AC\end{array}$
Now we evaluate using the calculator and round:
$AC=7\cdot \mathrm{sin}\left({40}^{\circ }\right)\approx 4.5$
Problem 1.1
$BC=$

Want to try more problems like this? Check out this exercise.

Practice set 2: Solving for an angle

Trigonometry can also be used to find missing angle measures. Let's find, for example, the measure of $\mathrm{\angle }A$ in this triangle:
We are given the length of the side $\text{adjacent}$ to the missing angle, and the length of the $\text{hypotenuse}$. The trigonometric ratio that contains both of those sides is the cosine:
$\begin{array}{rl}\mathrm{cos}\left(\mathrm{\angle }A\right)& =\frac{AC}{AB}\\ \\ \mathrm{cos}\left(\mathrm{\angle }A\right)& =\frac{6}{8}\phantom{\rule{1em}{0ex}}AC=6,AB=8\\ \\ \mathrm{\angle }A& ={\mathrm{cos}}^{-1}\left(\frac{6}{8}\right)\end{array}$
Now we evaluate using the calculator and round:
$\mathrm{\angle }A={\mathrm{cos}}^{-1}\left(\frac{6}{8}\right)\approx {41.41}^{\circ }$
Problem 2.1
$\mathrm{\angle }A=$
${}^{\circ }$

Want to try more problems like this? Check out this exercise.

Practice set 3: Right triangle word problems

Problem 3.1
Howard is designing a chair swing ride. The swing ropes are $5$ meters long, and in full swing they tilt in an angle of ${29}^{\circ }$. Howard wants the chairs to be $2.75$ meters above the ground in full swing.
How tall should the pole of the swing ride be?
meters

Want to try more problems like this? Check out this exercise.

Want to join the conversation?

• This is not correct. The path of the swing is an arc so at the point where it is parallel to the support pole it would closer to the ground than at the point of full swing which is 2.75 meters. To make this example correct the 2,75 meters needs to be applied to the point where the swing is parallel to the supporting pole.
• You are correct that it is an arc. However, the key to the question is the phrase "in full swing". The swing will be closer than 2.75 meters at the bottom of the arc. That is an interesting point that I hadn't considered, but not what the question is asking.
• This is really fun to do
• I agree with you :)
• Shouldn't we take in account the height at which the MIB shoots its laser. I'm guessing it would be somewhere from his shoulder. Maybe the answer wouldn't differ that much but it might make it a little more challenging to figure out. You would even be able to calculate the height the agent is holding his gun at with stretched arms when you know the angle he's keeping his arms at, his arm length and the length from his shoulders to the ground.
• Good point, let's estimate :D.

Men In Black are generally rather tall so it is fair to estimate the man is about two meters tall. The average arm length of an adult human is ~25 inches which equates to about 0.6 meters. If we assume that the man holds his arms directly above his head (not technically realistic but a fair assumption) then we can estimate the height of the LASER to be about 2.5 meters. If we subtract 2.5 from 324 we get 321.5. Arctan(321.5/54) = 80.465.

That deviates from the ground angle by only 0.09%, so it probably wouldn't affect how he aimed the laser.

If we assume he shot from shoulder height with his arms straight out, then that would be arctan(322/53.5) = ~80.567 which deviates from the ground angle by only 0.04%.
• What is the value of sine, cosine, and tangent?
• The Sine, Cosine, and Tangent are three different functions. They do not have a value outright, it would be like trying to ask what the value of f(x) = x + 1 is. The trig functions give outputs in terms of the ratios of two sides of a triangle when we feed them the input of an angle measure.
Sine outputs the ratio of opposite to hypotenuse
Cosine outputs the ratio of adjacent to hypotenuse
Tangent outputs the ratio of opposite to adjacent
• I am so confused...I try my best but I still don't get it . I need someone to Break it down further for me? I never not understand math but this one really has me stuck.Thank you.
• You might not be taking trig anymore, but there is SOHCAHTOA - Sine = Opposite * Hypotenuse. sine is when you are given the opposite side from the angle and the hypotenuse of the triangle. Cosine = Adjacent * Hypotenuse. cosine is when you are given the adjacent side of the angle and the hypotenuse of the triangle. Finally, Tangent = Opposite * Adjacent. This is when you are given the opposite and adjacent sides of the angle. A side could also have x as its value and you solve for x.
• My problem is that I do not know which one is adjacent and opposite you the one closest to the angle is adjacent but if it doesn't show the angle then how am I supposed to know which one.
• Either the problem will tell you which angle is the reference angle or it will give two sides and you can choose which of the two acute angles you can use as the reference angle.
• when solving for an angle why does cos have a -1 on top?
• Trig functions like cos^-1(x) are called inverse trig functions. THey are the inverse functions of the normal trig functions. Where cos(x) would take in an angle and output a ratio of side lengths, cos^-1(x) takes in the ratio of adjacent/hypotenuse and gives you an angle, which is why we use it when solving for unknown angles.
Note that cos^-1(x), (cos(x))^-1, and cos(x^(-1)) give three completely separate results.