This is a guide to applications and models for trigonometry, explaining everyday problems through math and technology. For each example, we will provide you with a situation that you can solve with trigonometry. These include solving for sides and angles of right triangels and writing equations for harmonic motion. To do an example problem, just watch the video for the problem, look under "Setup" to see how to label your triangle, and look under "Solution" to see step-by-step instructions for how to solve the problem and the answer. Let's get started!

EXAMPLE 1: Pitty O on the Roof



The Setup:
Pitty_O_on_the_Roof_Triangle.jpg



















The Solution:
Let x equal how tall the ladder must be in feet. From the equation sin 70 = 25/x, it follows that

x = 25/sin 70 ≈ 26.6 feet.

So, no, a 26 foot tall ladder would not be tall enough to save Pat. Poor Pat.






EXAMPLE 2: Pat Up a Tree



The Setup:
Pat_Up_a_Tree_Triangle.png.jpg




















The Solution:
Let x equal Pat’s height from the ground and y equal how far Pat is from the top of the tree, in feet. This problem involves two right triangles. From the smaller right triangle, From Lawrence to Pat, use the fact that tan 37° = x/36 to conclude that the Pat’s height from the ground is

x = 36 tan 37°.

Now, for the height of the tree, use the equation

tan 58° = x + y/36

to conclude that y = 36 tan 58° − x. So, the distance from Pat to the top of the tree is

y = 36 tan 58° − x = 36 tan 58° − 36 tan 37° ≈ 30.5 feet.

Pat has a long way to go.





EXAMPLE 3: Lawrence's Slide



The Setup:
Lawrence's_Slide_Triangle.jpg


















The Solution:
Let A equal the angle of depression. Using the tangent function, you see that

tan A = opp/adj = 10/6 ≈ 1.67

So, the angle of depression is A = arctan 2.67 ≈ 59.1°. And Lawrence is going to enjoy all 59.1 of those degrees.





EXAMPLE 4: Ballin' on the Pool Table



The Setup:
Ballin'_on_the_Pool_Table_Triangle_1.jpg

















For this problem, you'll have to draw additional triangles. Here's the setup for these:
Ballin'_on_the_Pool_Table_Triangle_2.jpg
The Solution (refer to triangle above for angle and triangle names):
For triangle BCD, you have B = 90° − 50° = 40°. The two sides of this triangle can be determined to be

b = 55 sin 40° and d = 55 cos 40°.

In triangle ACD, you can find angle A as follows:

tan A = b/d + 16 = 55 sin 40°/55 cos 40° + 16 ≈ 0.6081512

A ≈ arctan 0.6081512 ≈ 31.31°

Finally, from the triangle ACD, you have sin A = b/c, which becomes

c = b/sin A = 55 sin 40°/sin 31.31° = 68.03 inches.

Pat needs a lot of skill to make this shot.





EXAMPLE 5: Swingin'



The Setup:

Picture1.jpg


















The Solution:
Because the swing is at it’s maximum displacement (d) when t = 0, use the equation

d = a cos ωt

Also, because the maximum displacement from zero is 3 and the period is 2.5, you have the following.

Amplitude = │a│ = 3

Period = 2π = 2.5 ω = 2π/2.5

From this, you can derive the equation of harmonic motion:

d = 3 cos 2π/2.5 t

Now Lawrence can swing happily now that he knows his harmonic motion.