# 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.

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.

Part 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 = (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.

# Right Triangle Trigonometry and SOHCAHTOA

Sohcahtoa isn’t actually an Egyptian god, but if it helps to remember him that way, you’ll have an easier time recalling right angle trig relationships.

Right triangles are extremely common in science homework. Even though they are common, they can be confusing to new students. That is why we have the Egyptian god SOHCAHTOA.

SOHCAHTOA is a handy mnemonic trigonometry students learn to remember which sides of a triangle are used for the three main trig functions: sine, cosine, and tangent.

These functions are defined by the ratios of various lengths of the sides of a right triangle. Let’s look at this right triangle.

This triangle is made up of three sides of lengths a, b and c. Note the angle marked θ. This angle is formed by the intersection of b and c. The hypotenuse is always the longest of the three sides and opposite of the right angle. The side b is ‘adjacent’ to the angle, so this side is known as the adjacent side. It follows the side ‘opposite’ of the angle is known as the opposite side. Now that we have all our sides labelled, we can use SOHCAHTOA.

### SOHCAHTOA

S – Sine
O – Opposite
H – Hypotenuse

C – Cosine
H – Hypotenuse

T – Tangent
O – Opposite

SOH = sin θ = opposite over hypotenuse = ac
CAH = cos θ = adjacent over hypotenuse = bc
TOA = tan θ = opposite over adjacent = ab

Easy to remember. Now let’s see how easy it is to apply.

### Example Problem

Consider this triangle.

The hypotenuse has a length of 10 and one angle of the triangle is 40º. Find the lengths of the other two sides.

Let’s start with the side with length a. This side is opposite the angle and we know the length of the hypotenuse. The part of SOHCAHTOA with both hypotenuse and opposite is SOH or sine.

sin 40º = opposite / hypotenuse
sin 40º = a / 10

solve for a by multiplying both sides by 10.

10 sin 40º = a

Punch 40 into your calculator and hit the sin key to find the sine of 40º.

sin 40º = 0.643

a = 10 sin 40º
a = 10 (0.643)
a = 6.43

Now let’s do side b. This side is adjacent to the angle, so we should use CAH or cosine.

cos 40º = adjacent / hypotenuse
cos 40º = b / 10

solve for b

b = 10 cos 40º

Enter 40 and hit the cos button on your calculator to find:

cos 40º = 0.766

b = 10 cos 40º
b = 10 (0.766)
b = 7.66

The sides of our triangle are 6.43 and 7.66. We can use the Pythagorean equation to check our answer.

a2 + b2 = c2
(6.43)2 + (7.66)2 = c2
41.35 + 58.68 = c2
100.03 = c2
10.00 = c

10 is the length of the triangle’s hypotenuse and matches our calculation above.

As you can see, our friend SOHCAHTOA can help us calculate the angles and lengths of the sides of right triangles with very little information. Make him your friend too.

# Trig Identities Study Sheet

This study sheet has ten groups of trig identities for the basic trigonometry functions. These identities include the reciprocal and co-function relationships between trig functions. There are also half-angle and double angle identities, along with sum and product relationships. This table is an extremely useful thing to keep on hand when working with trig functions. Print a copy and keep it with your textbook today.

This table contains values for sine, cosine and tangent for angles between 0 and 90º. All values are rounded to four decimal places.

The downloadable trig table PDF is optimized to fit on a single 8½ x 11″ sheet of paper.

# Special Right Triangles

Right triangles are central to trigonometry. Two special right triangles appear over and over in standardized exams and homework problems. These triangles are “special” because they have simple ratios between the lengths of each side.

The first is the 30-60-90 triangle.

The 30-60-90 is named after the interior angles of the triangle. If the shortest side is length x, the longest side or hypotenuse is twice as long. The remaining leg of the triangle is √3 times the length of the short leg.

The trigonometric ratios for these angles are easy to figure out. Let’s look at the 30º angle:

The second special triangle is the 45-45-90 triangle. This triangle is formed by cutting a square along its diagonal. If the sides of the square are length x, the hypotenuse will be x√2.

Both angles are 45º, so the trigonometric ratios are the same for both interior angles.

Memorizing these values would be time well spent. These triangles will show up again and again in exams and homework.

For more general right triangles, check out Right Triangle Trigonometry.

# Right Triangles – Trigonometry Basics

The right triangle is a special type of triangle that forms the basis for nearly all trigonometry functions.

As you know, a triangle is a closed shape consisting of three sides. Every two of these sides form an angle between them for a total of three interior angles. These three angles together add up to 180º. If one of these angles is 90º, also known as a right angle, the triangle is known as a right triangle.

This is a right triangle. The three angles A, B and C are marked and each side’s length is denoted by a, b, and c. The side length letters are the sides opposite of the angle with the corresponding capital letter. The side opposite the right angle C is known as the hypotenuse of the right triangle.

The length of the hypotenuse can be calculated using the Pythagorean Theorem

a2 + b2 = c2

Each triangle has six parts, 3 sides and 3 angles. If you know two of these values, you can calculate any of the other four using the six trigonometric ratios. These ratios are sine, cosine, tangent, cosecant, secant, and cotangent. Using the variables in the right triangle, the formulas for these ratios for angle A are:

The common method to remember these formulas is the mnemonic SOH CAH TOA.

S – Sine
O – Opposite
H – Hypotenuse

C – Cosine