SOHCAHTOA - How To Solve These GCSE Maths Questions Quickly
Introduction
What we are going to look at in this article is a branch of trigonometry called SOHCAHTOA that allows us to calculate missing angles or to find missing lengths if we are given an angle.
This is referred to as SOH CAH TOA and these are actually trigonometric ratios. This topic is not the same as Pythagoras which allows us to calculate missing lengths when we are given two other lengths.
In order to be able to do SOH CAH TOA it is important to be able to identify the correct sides of a right angled triangle.
The longest side of a triangle is known as the Hypotenuse. The side which is next to the angle is known as the Adjacent and the side that is opposite the angle is the Opposite.
The trigonometric ratios that you should learn are:
\sin \alpha=\frac{\text { Opposite }}{\text { Hypotenuse }}\cos \alpha=\frac{\text { Adjacent }}{\text { Hypotenuse }}\tan \alpha=\frac{\text { Opposite }}{\text { Adjacent }}When doing SOHCAHTOA questions quite often it is a case of determining what sides are given and to associate it with the appropriate trig ratio.
SOHCAHTOA - Example 1
Looking at the question you can see that the adjacent is given and it is the opposite that you need to find out.
Using SOH CAH TOA means that TOA needs to be used.
Our solution will therefore be:
\begin{aligned} \tan 36 & =\frac{B C}{8.7} \\ B C & =8.7 \times \tan 36 \\ & =6.32091 \ldots \\ \therefore B C & =6.32 \mathrm{~cm} \end{aligned}SOHCAHTOA Example 2 - Finding a missing angle
In this question you are asked to find a missing angle where two lengths have been provided. From the diagram you will notice that the opposite and adjacent lengths have been provided.
From SOH CAH TOA this means that we need to use TOA and so our solution will be as follows:
\begin{aligned} \tan x & =\frac{5}{12.5} \\ x & =\tan ^{-1}\left(\frac{5}{12.5}\right) \\ & =21.8014 \ldots \\ \therefore x & =21.8^{\circ} \end{aligned}
Example 3 – Worded SOHCAHTOA problem
Quite often with trigonometry GCSE maths revision questions you are presented with a worded problem and you need to sketch a diagram to help you solve the question. With enough practice of questions like these you will know that the question is one involving trigonometry.
Our solution to this question will start with a quick clear diagram as follows:
You can see from the diagram that what we are looking for is the opposite side of the triangle. This can be found as follows:
\begin{aligned} \tan 87 & =\frac{x}{4} \\ x & =4 \tan 87 \\ & =76.3 \mathrm{~m} \end{aligned}The word “elevation” simply means the angle that is made when you look up at something from the horizontal.
Example 4 – challenge question
When it comes to questions involving trigonometry, they will not always be straightforward. Look at the question below:
From the diagram you ought to be able to see two triangles, ABD and BCD. Each triangle has the common length BD
From triangle ABD, Pythagoras can be used to find BD and this is calculated as follows:
\begin{aligned} 12^2+B D^2 & =16^2 \\ B D^2 & =16^2-12^2 \\ & =112 \\ B D & =\sqrt{112} \\ & =10.5830 \ldots \end{aligned}From BCD you can now use SOH CAH TOA to find the length CD which is the hypotenuse and BD is the hypotenuse. This is done as follows:
\begin{aligned} \sin 40^{\circ} & =\frac{10.5830}{C D} \\ C D \sin 40^{\circ} & =10.5530 \\ C D & =\frac{10.5830}{\sin 40} \\ C D & =16.46423 . \\ & =16.5 \mathrm{~cm} \end{aligned}Example 5 – Challenge question
Once again we have a question that involves two triangles. Here you are asked to find the length of AC. You need to find the length of AD and the length DC.
From triangle ABD the length of AD can be found using SOH CAH TOA as follows:
\begin{aligned} \cos 65^{\circ} & =\frac{A D}{7} \\ A D & =7 \times \cos 65 \\ & =2.958327 \end{aligned}Now triangle BCD has only one length and a second needs to be found. The length BD is common to both triangles so this can be found as follows using SOH CAH TOA:
\begin{aligned} \sin 65^{\circ} & =\frac{B D}{7} \\ B D & =7 \sin 65^{\circ} \\ & =6.34415^{\circ} . \end{aligned}Now we have two lengths for triangle BCD, the length AD can be found using Pythagoras as follows:
\begin{aligned} & a^2+b^2=c^2 \\ & \text { 6. } 34415 . \ldots^2+D C^2=12^2 \\ & D C^2=12^2-6.34415 \ldots^2 \\ & =103.7517 . . \\ & D C=\sqrt{103.7517} \\ & =10 \cdot 1858 . \\ & \end{aligned}The final length of AC can now be found be adding the lengths AD and DC:
\begin{aligned} & =2.958327 \ldots+10.1858 \ldots \\ & =13.144185 \\ & =13.1 \mathrm{~cm}(3 \mathrm{sf}) \end{aligned}This last question has required us to use SOH CAH TOA twice and then to use Pythagoras. Such questions require you to use the diagram and the information carefully to help you solve the question.
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