Deciding the Trig Rule (Edexcel IGCSE Maths)

Applications of Trigonometry

Spec point — 4.8C

How do I decide which trig rule to use?

• Different rules are required depending on the question

  • You need to be able to decide which is appropriate to use
  • Think about what information you have and what you want to find

• This table summarises the possibilities:

Flowchart to decide the rule

Can I use multiple trig rules in the same question?

• Harder questions will require you to use more than one trig rule

  • For example, you may need the sine rule followed by the cosine rule

• The area formula only works for an angle between two sides

  • If you are not given this setup, you may need to use the sine or cosine rule first

• If it looks like no rule would work, remember that all angles in a triangle sum to 180

  • This often helps to find a missing angle

Exam Hint Look at the number of marks for a question - if it is a lot, you are likely to need more than one trig rule!

Worked Example Find the area of the triangle below.

Answer:

The area of a triangle can be found using the formula $A=\frac{1}{2}ab\mathrm{sin}C$

The three side lengths are known , but we need to find an angle in order to calculate the area Because we know all three sides, any of the angles could be found

Find angle ABC using the cosine rule

Cosine Rule: $a^{2}=b^{2}+c^{2}-2bc\mathrm{cos}A$, where $A$ is the angle opposite side $a$* *

$\begin{gathered} A C^{2} & = A B^{2} + B C^{2} - 2 ( A B ) ( B C ) \cos A B C \\ 4.4^{2} & = 7.4^{2} + 4.8^{2} - 2 ( 7.4 ) ( 4.8 ) \cos A B C \end{gathered}$

Rearrange to make$\mathrm{cos}ABC$ the subject

$\begin{gathered} 4.4^{2} - ( 7.4^{2} + 4.8^{2} ) & = - 2 ( 7.4 ) ( 4.8 ) \cos A B C \\ \cos A B C & = \frac{4.4^{2} - 7.4^{2} - 4.8^{2 } }{- 2 ( 7.4 ) ( 4.8 )} \end{gathered}$

Use your calculator to find the value of $\mathrm{cos}ABC$

$\begin{gathered} \cos A B C & = \frac{487}{592} \end{gathered}$

Use the cos^-1 button on your calculator to find the value of $ABC$

$\begin{gathered} A B C & = \cos ^{- 1} ( \frac{487}{592} ) \\ = 34.65054... \end{gathered}$

Now we can find the area of the triangle using the formula and angle ABC as the known angle

$\begin{gathered} \text{Area} = 1 \frac{1}{2} \times 4.8 \times 7.4 \times \sin 34.65054... \\ = 10.09779... \end{gathered}$

Final answer: $\mathrm{Area}=10.1\mathrm{cm}^{2}$(3.s.f)