SOHCAHTOA Example Problem – Trigonometry Help

SOHCAHTOA is the mnemonic used to remember which sides of a right triangle are used to find the ratios needed to determine the sine, cosine or tangent of an angle. Here are a pair of SOHCAHTOA example problems to help show how to use these relationships. If you have no idea what SOHCAHTOA means, check out this introduction to SOHCAHTOA.

SOHCAHTOA Example Problem 1

A mathematically inclined squirrel sits atop a 14-foot tall tree. He spies a nut on the ground some distance away. After careful measurements, he determines the nut is 74º from the base of his tree.
How far away is the nut from:
A) The base of the tree?
B) The math squirrel?

Here is a layout of the problem.

Squirrel and Nut SOHCAHTOA Example Problem

Part A: How far is the nut from the base of the tree?

Looking at our triangle, we see we know the angle and the length of the adjacent side. We want to know the length of the opposite side of the triangle. The part of SOHCAHTOA with these three parts is TOA, or tan θ = opposite / adjacent.

tan θ = opposite / adjacent
tan ( 74º ) = opposite / 14 ft

solve for ‘opposite’

opposite = 14 ft ⋅ tan ( 74º )
opposite = 14 ft ⋅ 3.487
opposite = 48.824 ft

The nut is 48.824 ft from the base of the tree.

Part B: How far is the nut from the math squirrel?

This time, the needed distance is the hypotenuse of the triangle. The part of SOHCAHTOA with  both adjacent and hypotenuse is CAH, or cos θ = adjacent / hypotenuse.

cos θ = adjacent / hypotenuse
cos ( 74º ) = 14 ft / hypotenuse
hypotenuse = 14 ft / cos ( 74º )
hypotenuse = 14 ft / 0.276
hypotenuse = 50.791 ft

We can use the Pythagorean Theorem to check our work.

a2 + b2 = c2
(14 ft)2 + (48.824 ft)2 = c2
196 ft2+ 2383.783 ft2= c2
2579.783 ft2 = c2
50.792 ft = c

The two numbers are close enough to account for rounding errors.

SOHCAHTOA Example Problem 2

A prince has arrived to rescue a princess standing in a tower balcony 15 meters up across a moat 12 meters wide. He knew the princess was high up and there was a moat, so he brought the longest ladder in the kingdom, a 20-meter monster.

A) Is the ladder long enough?
B) If the top of the 20 m ladder touches the balcony, how far away from the tower wall is the bottom of the ladder?
C) What is the angle between the ground and the ladder?

Here is an illustration of our situation.

Princess rescue trigonometry examplePart A: Is the ladder long enough?  To know if the ladder is long enough, we need to know how long it needs to be. At the very least, the prince’s ladder must reach the 15-meter balcony from the edge of the 12-meter moat. Use the Pythagorean theorem to find out how far up the tower wall a 20-meter ladder will reach.

a2 + b2 = c2
height2 + moat2 = ladder2
height2 = ladder2 – moat2
height2 = (20 m)2 – (12 m)2
height2 = 400 m2 – 144 m2
height2 = 256 m2
height = 16 m

A 20-meter ladder can reach 16 meters up the side of the tower. This is a meter longer than the 15 meters needed to reach the princess balcony.

Part B: When the ladder touches the edge of the balcony, how far away from the tower wall is the bottom of the ladder?

The first part measured the height the 20-meter ladder reached when the ladder was placed at the edge of the moat. We found we had more ladder than needed. If the ladder touches the balcony, we know the height it reaches is 15 meters. The ladder is still 20 meters long. We just need to find out how far from the tower to stick the bottom of the ladder. Use the Pythagorean theorem again.

a2 + b2 = c2
(balcony height)2 + (ground distance)2 = ladder2
(ground distance)2 = ladder2 – (balcony height)2

(ground distance)2 = (20 m)2 – (15 m)2
(ground distance)2 = 400 m2 – 225 m2
(ground distance)2 = 175 m2
ground distance = 13.23 m

The ladder’s base is 13.23 m away from the tower.

Part C: What is the angle between the ground and the ladder?

We were given the height of the balcony, which in this case is the ‘opposite’ side of the triangle from the needed angle. We also know the length of the ladder that forms the hypotenuse of the triangle. The part of SOHCAHTOA that has both of these parts is SOH, or sin θ = opposite/hypotenuse. Use this to solve for the angle.

sin θ = opposite/hypotenuse
sin θ = 15 m / 20 m
sin θ = 0.75
θ = sin-1(0.75)
θ = 48.59°

You can check your work in Part B if you use the answer we got as the ‘adjacent’ side of the triangle.

If you use the ladder as the hypotenuse, use CAH, or cos θ = adjacent / hypotenuse.

cos θ = adjacent / hypotenuse
cos θ = 13.23 m / 20 m
cos θ = 0.662
θ = cos-1(0.662)
θ = 48.59°

The angle matches the value above.

If you choose the balcony height as the opposite side, use TOA, or tan θ = opposite / adjacent

tan θ = opposite / adjacent
tan θ = 15 m / 13.23 m
tan θ = 1.134
θ = tan-1(1.134)
θ = 48.59°

Pretty neat how it doesn’t really matter which sides of the right triangle you use, as long as you use the correct part of SOHCAHTOA.

 

SOHCAHTOA Example Problem – Trigonometry Help
Last modified: September 8th, 2016 by Todd Helmenstine