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Solving equationsHere you will learn how to use the ratio of the sides of a right triangle, SOHCAHTOA, and right triangle trigonometry to find unknown sides and angles in right angled triangles. You will learn how to label the sides of right-angled triangles and understand what each of the side ratios, sin, cos and tan are, and what their inverses are (\sin^{-1}, \cos^{-1}, \tan^{-1}) in order to problem solve.

Students first learn about right triangle trigonometry in geometry and expand their knowledge as they learn about the unit circle in algebra 2 and precalculus.

**SOHCAHTOA** is a mnemonic device that gives you an easy way to remember the three main **trigonometric ratios **of the sides of a **right triangle.** The trig ratios are used to find the missing sides of the triangle (right triangle) and the missing **acute angles.**

The names of the three ratios are:

**Sine**(Sin)**Cosine**(Cos)**Tangent**(Tan)

When writing the trigonometric ratios, you must write them with respect to one of the acute angles of the right triangle.

Let’s use angle at A as the acute angle by which to write the 3 main trigonometric ratios. Label this angle \theta. Since all the ratios will be written with respect to angle \theta, the side opposite \theta is called the **‘opposite’** side

The side that is part of angle \theta or next to angle \theta is called the **‘adjacent’** side.

The side across from the right angle is always the **‘hypotenuse’.**

What happens if you use the angle at C instead of the angle at A to write the trigonometric ratios?

Notice how the adjacent side and the opposite side switch since the angle being used is the angle at C, labelled \theta.

Use this quiz to check your grade 9 to 12 students’ understanding of Trigonometry. 15+ questions with answers covering a range of 9th to 12th grade trigonometry topics to identify areas of strength and support!

DOWNLOAD FREEUse this quiz to check your grade 9 to 12 students’ understanding of Trigonometry. 15+ questions with answers covering a range of 9th to 12th grade trigonometry topics to identify areas of strength and support!

DOWNLOAD FREENow that you understand how to label the sides of the right triangle, let’s write the ratios.

- The sine or sin ratio with respect to angle \theta is the opposite side length divided by the hypotenuse side length.

\sin(\theta)=\cfrac{\text{opposite}}{\text{hypotenuse}}~(\mathrm{SOH})

- The cosine or cos ratio with respect to angle \theta is the adjacent side length divided by the hypotenuse side length.

\cos(\theta)=\cfrac{\text{adjacent}}{\text{hypotenuse}}~(\mathrm{CAH})

- The tangent or tan ratio with respect to angle \theta is the opposite side length divided by the adjacent side length.

\tan(\theta)=\cfrac{\text{opposite}}{\text{adiacent}}~(\mathrm{TOA})

You can also remember the abbreviation for SOHCAHTOA using triangles.

Sine, Cosine and Tangent are therefore **trigonometric functions** as the sine, cosine, or tangent of an angle \theta is equal to the division of the two respective sides O\div{H}, A\div{H} or O\div{A}.

Each SOHCAHTOA triangle can be used to determine the missing side, or missing angle of a right-angled triangle. Cover up the letter you want to find and calculate the product or division of the other two letters using one of the three formulae highlighted in the table.

How does this relate to high school math?

**High School Geometry – Similarity, Right Triangles, and Trigonometry HSG-SRT.C.6**Understand that by similarity, side ratios in right triangles are properties of the angles in the triangle, leading to definitions of trigonometric ratios for acute angles.

**High School Geometry – Similarity, Right Triangles, and Trigonometry HSG-SRT.C.8**

Use trigonometric ratios and the Pythagorean Theorem to solve right triangles in applied problems.

In order to use SOHCAHTOA to find the missing side length of a right triangle:

**Label the sides of the right triangle with respect to one of the acute angles.****Determine the trigonometric ratio to use and write the formula with the correct subject.****Substitute known values into the trig ratio to find the missing side.****Calculate the unknown side.**

Calculate the side of the right triangle labeled x.

**Label the sides of the right triangle with respect to one of the acute angles.**

Using the 40^{\circ} angle, the side marked x is the opposite side and the side marked 8\mathrm{~cm} is the length of the hypotenuse because it is across from the right angle.

2**Determine the trigonometric ratio to use and write the formula with the correct subject.**

The hypotenuse (H) is known, and the opposite side (O) is what you want to find.

The adjacent side (A) is unknown and so not useful for this problem.

Looking at SOHCAHTOA, the sine function (S) is the ratio of the opposite side (O) and the hypotenuse (H).

You therefore use the sine function:

\sin(\theta)= \cfrac{O}{H}Using the formula triangle, cover up the opposite side (O) as you want to calculate this length. This gives the formula:

O=H\times\sin(\theta)3**Substitute known values into the formula to find the missing side.**

As the opposite side O=x\mathrm{~cm}, the hypotenuse H=8\mathrm{~cm}, and \theta=40^{\circ},

x=8\times\sin(40)4**Calculate the unknown side.**

Use your calculator to find the solution

x=8\times\sin(40)=5.14\mathrm{~cm}\text{ (2dp)}Always check that the answer is reasonable:

As 5.14\mathrm{~cm} is shorter than the length of the hypotenuse (8\mathrm{~cm}), the answer is reasonable.

Calculate the side of the right triangle labeled x.

**Label the sides of the right triangle with respect to one of the acute angles.**

Using the angle labeled 36^{\circ}, the side marked x is the adjacent side (A) because it is next to the angle and the side marked as 15\mathrm{~cm} is the hypotenuse (H) because it is across from the right angle.

**Determine the trigonometric ratio to use and write the formula with the correct subject.**

The hypotenuse (H) is known, and the adjacent side (A) is what you want to find.

The opposite side (O) is unknown and so not useful for this problem.

Looking at SOHCAHTOA, the cosine function (C) is the ratio of the adjacent side (A) and the hypotenuse (H).

You therefore use the cosine function:

\cos(\theta)= \cfrac{A}{H}

Using the formula triangle, cover up the adjacent side (A) as you want to calculate this length. This gives the formula:

A=H\times\cos(\theta)

**Substitute known values into the trig ratio to find the missing side.**

As the adjacent side A=x\mathrm{~cm}, the hypotenuse H=15\mathrm{~cm}, and \theta=36^{\circ},

x=15\times\cos(36)

**Calculate the unknown side.**

Use your calculator to find the solution

x=15\times\cos(36)=12.13\mathrm{~cm}\text{ (2dp)}

Always check that the answer is reasonable:

The value of the side being 12.13\mathrm{~cm} makes sense with the sketched out triangle

Calculate the side labeled x .

**Label the sides of the right triangle with respect to one of the acute angles.**

Using the angle marked 38^{\circ} , the side marked as x is the adjacent side because it is next to the angle, and the side marked as 10\mathrm{~cm} is the opposite side because it is opposite the angle.

**Determine the trigonometric ratio to use and write the formula with the correct subject.**

The opposite side (O) is known, and the adjacent side (A) is what you want to find.

The hypotenuse (H) is unknown and so not useful for this problem.

Looking at SOHCAHTOA, the tangent function (T) is the ratio of the opposite side (O) and the adjacent side (A).

You therefore use the tangent function:

\tan(\theta)=\cfrac{O}{A}

Using the formula triangle, cover up the adjacent side (A) as you want to calculate this length. This gives the formula:

A=\cfrac{O}{\tan(\theta)}

**Substitute known values into the trig ratio to find the missing side.**

As the opposite side O=10\mathrm{~cm}, the adjacent side A=x\mathrm{~cm}, and \theta=38^{\circ},

x=\cfrac{10}{\tan(38)}

**Calculate the unknown side.**

Use your calculator to find the solution

x=\cfrac{10}{\tan(38)}=12.80\mathrm{~cm}\text{ (2dp)}

Always check that the answer is reasonable:

The longer side is opposite the larger angle. As the sum of angles in a triangle is 180^{\circ}, the other angle in the triangle is 52^{\circ}. The longer side is opposite the larger angle so this solution is reasonable.

In order to find missing angles using SOHCAHTOA:

**Label the sides of the right triangle with respect to one of the acute angles.****Determine the trigonometric ratio to use and write the inverse trig formula with \bf{\theta} as the subject.****Substitute the two known side lengths into the formula.****Calculate the unknown angle.**

Calculate the angle labeled \theta.

**Label the sides of the right triangle with respect to one of the acute angles.**

The side marked as 5\mathrm{~cm} is opposite the angle, so this is the opposite side.

The side marked as 9\mathrm{~cm} is the hypotenuse because it is across from the right angle.

**Determine the trigonometric ratio to use and write the inverse trig formula with \bf{\theta} as the subject.**

Looking at SOHCAHTOA, the sine function (S) is the ratio of the opposite side (O) and the hypotenuse (H).

As you are finding the angle, use the inverse sine function:

\theta=\sin^{-1}\left(\cfrac{O}{H}\right)

Note, you can use the formula for the sine function (\sin(\theta)=\cfrac{O}{H}), but you will need to rearrange it by finding the inverse sine of each side to obtain \theta=\sin^{-1}\left(\cfrac{O}{H}\right).

**Substitute the two known side lengths into the formula.**

As the opposite side O=5\mathrm{~cm} and the hypotenuse H=9\mathrm{~cm},

\theta=\sin^{-1}\left(\cfrac{5}{9}\right)

**Calculate the unknown angle.**

\theta=\sin^{-1}\left(\cfrac{5}{9}\right)=33.75^{\circ}\text{ (2dp)}

Remember we can get \sin^{-1}( on the calculator by pressing *SHIFT* or 2 nd and then* sin*. Don’t forget to close the bracket as well.

Always check that the answer is sensible:

You can estimate from the sketch that \theta is an acute angle.

As 33.7^{\circ} is less than 90^{\circ}, the answer is reasonable.

Calculate the angle labeled \theta .

**Label the sides of the right triangle with respect to one of the acute angles.**

The side marked as 9\mathrm{~cm} is next to the angle, so this is the adjacent side.

The side marked as 20\mathrm{~cm} is the hypotenuse because it is across from the right angle.

Looking at SOHCAHTOA, the cosine function (C) is the ratio of the adjacent side (A) and the hypotenuse (H).

As you are finding the angle, use the inverse cosine function:

\theta=\cos^{-1}\left(\cfrac{A}{H}\right)

Note, you can use the formula for the cosine function (\cos(\theta)=\cfrac{A}{H}), but you will need to rearrange it by finding the inverse cosine of each side to obtain \theta=\cos^{-1}\left(\cfrac{A}{H}\right).

**Substitute the two known side lengths into the formula.**

As the adjacent side A=9\mathrm{~cm} and the hypotenuse H=20\mathrm{~cm},

\theta=\cos^{-1}\left(\cfrac{9}{20}\right)

**Calculate the unknown angle.**

\theta=\cos^{-1}\left(\cfrac{9}{20}\right)=63.26^{\circ}\text{ (2dp)}

Remember we can get \cos^{-1}( on the calculator by pressing *SHIFT* or 2 nd and then *cos*. Don’t forget to close the bracket as well.

Always check that the answer is sensible:

You can estimate from the sketch that \theta is an acute angle.

As 63.3^{\circ} is less than 90^{\circ}, the answer is reasonable.

Calculate the angle labeled \theta .

**Label the sides of the right triangle with respect to one of the acute angles.**

The side marked as 15\mathrm{~cm} is opposite the angle \theta , so this is the opposite side.

The side marked as 7\mathrm{~cm} is the adjacent side because it is next to the angle.

Looking at SOHCAHTOA, the tangent function (T) is the ratio of the opposite side (O) and the adjacent side (A).

As you are finding the angle, use the inverse tangent function:

\theta=\tan^{-1}\left(\cfrac{O}{A}\right)

Note, you can use the formula for the tangent function (\tan(\theta)=\cfrac{O}{A}), but you will need to rearrange it by finding the inverse tangent of each side to obtain \theta=\tan^{-1}\left(\cfrac{O}{A}\right).

**Substitute the two known side lengths into the formula.**

As the opposite side O=15\mathrm{~cm} and the adjacent side A=7\mathrm{~cm},

\theta=\tan^{-1}\left(\cfrac{15}{7}\right)

**Calculate the unknown angle.**

\theta=\tan^{-1}\left(\cfrac{15}{7}\right)=64.98^{\circ}\text{ (2dp)}

Remember we can get \tan^{-1}( on the calculator by pressing *SHIFT* or 2 nd and then *tan*. Don’t forget to close the bracket as well.

Always check that the answer is sensible:

Draw an accurate sketch of the triangle and measure the angle using a protractor.

- Spend time looking at different right-angle triangles and labeling sides in relation to the angle so that students practice working out which trigonometric function to use sine, cosine or tangent.

- Use revision cards and practice questions including real life word problems.

- Solve real life problems that require them to apply the trig ratios out of the classroom: find the height of a tree / building, estimate the length of a football pitch, work out the angle of elevation from a horizontal distance to a window / sign etc.

**Using the incorrect mode on the calculator**

When calculating the side lengths or angle measurements using one of the trig ratios, be sure the calculator is in ‘degree’ mode, not ‘radian’ mode. Radians are a different unit of angle measurement which will be explored in a precalculus class.

**Thinking the hypotenuse is the opposite side**

When working with a right triangle, the hypotenuse is always the side across from the right angle. The opposite side is across from the acute angle being used to write the ratio.

**Using the right angle when writing a trig ratio**

To calculate side lengths or angle measurements using the trig ratios, the angle being used to write the ratio must be one of the acute angles.

For example, if trying to calculate the adjacent side:

\tan(90)=\cfrac{24}{\text{adjacent}} \rightarrow This is an incorrect ratio.

You cannot use the 90^{\circ} angle when writing the ratio and 24 is NOT the opposite side.

The hypotenuse is always the side across from the right angle and the longest side.

The correct ratio is: \cos(57)=\cfrac{\text{adjacent}}{24} OR \sin(57)=\cfrac{\text{opposite}}{24}

- Trigonometry
- Trig Functions
- How to find the exact value of a trig function
- Sin cos tan
- Law of sines
- Trig formula for area of a triangle
- Trig formulas
- Trig identities
- Law of cosines
- Trig tables

1. Calculate the length of the side labeled x. Round your answer to the nearest tenth.

7.8\mathrm{~cm}

18.4\mathrm{~cm}

60.4\mathrm{~cm}

16.9\mathrm{~cm}

Label the triangle with respect to the angle marked as 23^{\circ}

Looking at SOHCAHTOA, the sine function (S) is the ratio of the opposite side (O) and the hypotenuse (H).

Use the sine function:

\sin(\theta)=\cfrac{O}{H}

Using the formula triangle, cover up the opposite side (O) as you want to calculate this length. This gives the formula:

O=H\times\sin(\theta)

Substituting in known values,

O=H\times\sin(\theta)=20\times\sin(23)=7.8\mathrm{~cm}\text{ (1dp)}

2. Calculate the length of the side labeled x. Round your answer to the nearest hundredth.

22.42\mathrm{~cm}

37.50\mathrm{~cm}

20.18\mathrm{~cm}

11.15\mathrm{~cm}

Label the triangle:

The adjacent side (A) is known, and the hypotenuse (H) is what you want to find.

The opposite side (O) is unknown and so not useful for this problem.

Looking at SOHCAHTOA, the cosine function (C) is the ratio of the adjacent side (A) and the hypotenuse (H).

You therefore use the cosine function:

\cos(\theta)= \cfrac{A}{H}

Using the formula triangle, cover up the hypotenuse (H) as you want to calculate this length. This gives the formula:

H=\cfrac{A}{\cos(\theta)}

Substituting in known values,

H=\cfrac{A}{\cos(\theta)}=\cfrac{15}{\cos(42)}=20.18\mathrm{~cm}\text{ (2dp)}

3. Calculate the length of the side labeled x. Round your answer to the nearest tenth.

6.0\mathrm{~cm}

4.8\mathrm{~cm}

11.0\mathrm{~cm}

2.8\mathrm{~cm}

Label the triangle:

The opposite side (O) is known, and the adjacent side (A) is what you want to find.

The hypotenuse (H) is unknown and so not useful for this problem.

Looking at SOHCAHTOA, the tangent function (T) is the ratio of the opposite side (O) and the adjacent side (A).

You therefore use the tangent function:

\tan(\theta)=\cfrac{O}{A}

Using the formula triangle, cover up the adjacent side (A) as you want to calculate this length. This gives the formula:

A=\cfrac{O}{\tan(\theta)}

Substituting in known values,

A=\cfrac{O}{\tan(\theta)}=\cfrac{5.3}{\tan(62)}=2.8\mathrm{~cm}\text{ (1dp)}

4. Calculate the size of the angle labeled \theta. Round your answer to the nearest tenth.

43.3^{\circ}

46.7^{\circ}

36.0^{\circ}

0.0126^{\circ}

Label the triangle:

As you are finding the angle, use the inverse sine function:

\theta=\sin^{-1}\left(\cfrac{O}{H}\right)

Substitute in the two known side lengths

\theta=\sin^{-1}\left(\cfrac{8}{11}\right)=46.7^{\circ}\text{ (1dp)}

5. Calculate the size of the angle labeled \theta. Round your answer to the nearest whole degree.

64^{\circ}

1^{\circ}

26^{\circ}

23^{\circ}

Label the triangle:

Looking at SOHCAHTOA, the cosine function (C) is the ratio of the adjacent side (A) and the hypotenuse (H).

As you are finding the angle, use the inverse cosine function:

\theta=\cos^{-1}\left(\cfrac{A}{H}\right)

Substitute in the two known side lengths.

\theta=\cos^{-1}\left(\cfrac{5.4}{12.3}\right)=64^{\circ}\text{ (0dp)}

6. Calculate the size of the angle labeled \theta. Round your answer to the nearest tenth.

60.0^{\circ}

0.5^{\circ}

30.0^{\circ}

26.6^{\circ}

Label the triangle:

Looking at SOHCAHTOA, the tangent function (T) is the ratio of the opposite side (O) and the adjacent side (A).

As you are finding the angle, use the inverse tangent function:

\theta=\tan^{-1}\left(\cfrac{O}{A}\right)

Substitute in the two known side lengths

\theta=\tan^{-1}\left(\cfrac{12.6}{25.2}\right)=26.6^{\circ}\text{ (1dp)}

There are six trig ratios in total. The sine (sin), cosine (cos), tangent (tan), cosecant (csc), secant (sec), and cotangent (cot). The cosecant is the reciprocal of the sine ratio, the secant is the reciprocal of the cosine ratio, and the cotangent is the ratio of the tangent ratio.

Yes, you will learn how to graph the trig functions in algebra 2 and precalculus. They are periodic cures.

The trigonometric identities are based on the six trigonometric ratios. There are various distinct trigonometric identities that are used to solve equations and verify trigonometric proofs.

- Circle math
- Angles of a circle
- Circle theorems

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