This resource is a great reference tool for students calculating missing sides in right-angled triangles using either sine, cosine or tangent. There are three illustrated examples, one for each function. In each case, the numerator of the ratio is to be found.
The idea of SOHCAHTOA is an important concept in Trigonometry. It is specifically used for solving mainly those problems that involve right-angled triangles. If you are a student preparing for your upcoming exams or a teacher who is looking for perfect teaching resources for KS4, this worksheet will help you to master the knowledge- ‘how to use SOHCAHTOA‘ slowly and gradually.
The idea of SOHCAHTOA can be easily defined as a mnemonic that is mostly used in trigonometry. This idea will help your students to remember the formulas for sine (sin), cosine (cos), and tangent (tan) in a right-angled triangle.
This simple formula can be a huge help for your young learners to identify which trigonometric function needs to be used based on the sides and angles given in a problem.
To correctly apply SOHCAHTOA, it is important to understand the different parts of a right-angled triangle:
Only if we can correctly identify these sides can we decide which trigonometric ratio to use.
Step 1: Identify the Given Information
Step 2: Choose the Correct Formula
Step 3: Solve for the Missing Value
Example 1: Finding a Missing Side with Sine
Problem: Find the length of the opposite side in a triangle where the hypotenuse is 7 cm, and the angle is 30°.
Solution:
\(
\sin 30^\circ = \frac{\text{Opposite}}{\text{Hypotenuse}} \\
\sin 30^\circ = \frac{x}{7} \\
x = 7 \times \sin 30^\circ \\
x = 7 \times 0.5 \\
x = 3.5 \text{ cm}
\)
Example 2: Finding an Angle with Cosine
Problem: A triangle has an adjacent side of 4 cm and a hypotenuse of 8 cm. Find the angle.
Solution:
\(
\cos \theta = \frac{4}{8} \\
\cos \theta = 0.5 \\
\theta = \cos^{-1}(0.5) \\
\theta = 60^\circ
\)
Example 3: Finding a Side with Tangent
Problem: A right-angled triangle has an adjacent side of 9 cm, and the angle is 50°. Find the opposite side.
Solution:
\(
\tan 50^\circ = \frac{\text{Opposite}}{9} \\
\text{Opposite} = 9 \times \tan 50^\circ \\
\text{Opposite} = 9 \times 1.1918 \\
\text{Opposite} = 10.7 \text{ cm}
\)
We need to be really careful while using SOHCAHTOA. There can be some common mistakes that might be an issue while using SOHCAHTOA, like-
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