Solving Trigonometric Questions | GCSE Success Tips
Introduction
When preparing for your GCSE Maths exams one of the topics that you are guaranteed to meet is trigonometry and there are so many trigonometric questions that you can try.
You need to be aware of Pythagoras’ Theorem in order to be able to determine the longest side given the two shorter sides as well as being able to determine the length of one of the shorter sides, given the other shorter side and the longest side.
In a number of cases you will be asked to find a length or a angle and for this the acronym SOHCAHTOA comes in very useful:
\begin{aligned} & \mathrm{SOH} \rightarrow \sin \alpha=\frac{\text { opposite }}{\text { hypotenuse }} \\ & \mathrm{CAH} \rightarrow \cos \alpha=\frac{\text { adjacent }}{\text { hypoteruse }} \\ & \mathrm{TOA} \rightarrow \tan \alpha=\frac{\text { opposite }}{\text { adjacent }} \end{aligned}Solving Trigonometric Questions - An Example
Take a look at the following question:
In order to answer these questions it is important that you are confident with what sides of the triangle have been given to you. You can see from the diagram that you have been given the adjacent side (side next to the angle) and you are asked to work out the length of BC which is opposite to the angle.
So the calculation would be done by using:
\begin{gathered} \tan \alpha=\frac{\text { opposite }}{\text { adjacent }} \\ \tan 36=\frac{B C}{8.7} \\ B C=\tan 36 \times 8.7=6.32 \mathrm{~cm} \end{gathered}Trigonometric Questions - Example
Example
Take a look at the following question:
Here you want to calculate the missing side correctly to 3 significant figures.
Again it is important to know what sides are given and what side needs to be found. From the diagram the longest side (hypotenuse) is given and the opposite (opposite to the angle) is needed.
This would mean that you need to use:
\begin{gathered} \sin \alpha=\frac{\text { opposite }}{\text { hypotenuse }} \\ \sin 60=\frac{x}{32} \\ x=\sin 60 \times 32=27.7 \mathrm{~cm} \end{gathered}
Example
Take a look at the following trigonometric questions that are below:
With this question the missing side wants to be found correctly to 1 decimal place.
In this question you need to calculate a missing angle. As before it is important to know what sides have been provided. From the diagram you can see that the opposite side and the adjacent side are present.
This would mean using:
\begin{gathered} \tan \alpha=\frac{\text { opposite }}{\text { adjacent }} \\ \tan x=\frac{5}{12.5} \end{gathered}Now in order to find the value of x you need to perform an inverse tan calculation on \frac{5}{12.5}
This is often seen written as x=\tan ^{-1} \frac{5}{12.5} and you need to know how to perform this calculation on your calculator.
In this case x=21.8°
Example
Take a look at the following question:
Looking at the diagram you can see that there are two right angled triangles that have been put together.
The length CD is required. What information would be useful to have? Clearly the length BD.
From the triangle ABD, the length BD can be calculated using Pythagoras’s Theorem.
\begin{aligned} &\therefore 16^2=12^2+(B D)^2\\ &\begin{gathered} (B D)^2=16^2-12^2=112 \\ \therefore B D=\sqrt{112}=10.583 \end{gathered} \end{aligned}Looking at triangle CBD, the opposite is given and it is the hypotenuse that needs to be calculated. And because an angle is also given this would mean using the following:
\begin{gathered} \sin \alpha=\frac{\text { opposite }}{\text { hypotermuse }} \\ \sin 40=\frac{10.583}{C D} \\ C D=\frac{10.583}{\sin 40}=16.5 \mathrm{~cm} \end{gathered}Question Practice
Try the following question on your own before looking at the solution:
Question Practice Solution
The whole length of the base of the triangle is needed and the triangle has been split in two.
To find AC you need to find the length of AD and DC and add these together.
Looking at triangle ABD, the hypotenuse is given and AD the adjacent is needed. Since an angle is also given this means using:
\begin{gathered} \cos \alpha=\frac{\text { adjacent }}{\text { hypotenuse }} \\ \cos 65=\frac{A D}{7} \\ A D=7 \times \cos 65=2.958 \end{gathered}Now looking at triangle BDC only one side is given and the length BD would be useful to have and this can be found from triangle ABD either using Pythagoras (using the length of AD) or because that BD is the opposite side then the following would need to be used:
\begin{gathered} \sin \alpha=\frac{\text { opposite }}{\text { hypoteruse }} \\ \sin 65=\frac{B D}{7} \\ B D=7 \times \sin 65=6.344 \end{gathered}Now DC can be found using Pythagoras:
\begin{gathered} 6.344^2+D C^2=12^2 \\ D C^2=12^2-6.344^2=103.753664 \\ \therefore D C=10.186 \end{gathered}So the length of A C=A D+D C=2.958+10 \cdot 186=13 \cdot 144=13.1 \mathrm{~cm}
If there are any questions that you did not understand then you need to go over them again, trying them, and comparing your solution to what is written in this article.
Some of the more advanced trigonometric questions can be quite challenging and you are often having to use a variety of mathematical techniques.