The Cosine Rule
The Cosine Rule – Introduction
The use of the Cosine Rule in A Level Maths is significant. It is introduced in the GCSE Higher Paper and remains a crucial skill at the A Level. Common mistakes are made at both levels, but understanding its history enhances comprehension.
At GCSE level, the Cosine Rule is typically introduced as a method to calculate the length of a side or the size of an angle in a non-right-angled triangle. Students learn to use the formula a^2=b^2+c^2-2 b c \cos A where a, b, and c represent the lengths of the sides and C is the angle opposite side c. This formula enables students to solve problems related to triangles that cannot be easily solved using basic trigonometry.
In A Level Maths, the use of the Cosine Rule becomes more sophisticated. It is applied in various topics, such as vectors, calculus, and mechanics. Understanding this rule is crucial for solving complex problems involving non-right-angled triangles and vectors. A Level students need to grasp the concept of the Cosine Rule and apply it accurately in different contexts.
Common mistakes are made by students at both GCSE and A Level when using the Cosine Rule. Some of these mistakes include incorrectly applying the formula, misidentifying the correct values for the sides and angles, and miscalculating the cosine of an angle. These errors can lead to inaccurate solutions and hinder the understanding of the rule’s application.
The history of the Cosine Rule dates back to ancient mathematics. It has been traced back to Indian and Greek mathematicians who developed similar principles independently. The rule was popularised by the Persian mathematician Al-Khwarizmi in the 9th century and later became known as the Law of Cosines in the Western world. Its application and development have evolved over time, becoming an essential tool in modern mathematics.
In conclusion, the Cosine Rule plays a vital role in A Level Maths. Its introduction at the GCSE Higher Paper ensures a foundation for its application in more advanced topics. Understanding its history and common mistakes made at both levels enhances students’ ability to apply the rule accurately and effectively.
You need to know the formula for the cosine rule and you need to be able to rearrange this should you need to find a missing angle.
The cosine rule formula is as follows:
a^2=b^2+c^2-2 b c \cos AAnd rearranging to find an angle is give as follows:
\cos A=\frac{b^2+c^2-a^2}{2 b c}The Cosine Rule - Examples
Example 1
You can always label a triangle as you wish, providing you are calculating the correct side or angle. For this particular triangle the cosine rule gives us:
x^2=y^2+z^2-2 y z \cos XYou can then substitute in the appropriate values to find the missing length as follows:
\begin{aligned} x^2 & =7^2+6^2-2 \times 7 \times 6 \cos 95^{\circ} \\ & =92.32 \ldots \\ >x & =9.608 \ldots \end{aligned}Example 2
Here you want to find an angle given all three sides of the triangle. In this particular case the cosine rule gives us the following:
\begin{aligned} & \cos C=\frac{4^2+5^2-7^2}{2 \times 4 \times 5} \\ & \Rightarrow \cos C=-0.2 \\ & \Rightarrow \quad C=101.53 \ldots \end{aligned}The Cosine Rule - Further Examples
Example 1
If a diagram is not provided then it is important that you take the time to draw a diagram. Doing so will give you the information that you need in a visual form and remember to label the triangle as required.
\begin{aligned} & c=11.729 \ldots \\ & A B=11.7 \mathrm{~cm} \text { (3 s.f.) } \end{aligned}
Example 2
This question is concerned with regards to bearings. As mentioned before if no diagram is provided then it is very important that you take a few moments to draw a clear diagram as shown below:
Make sure that you are performing all the correct calculations as needed and as shown above.
You can now perform the required calculation to find the missing length as follows:
\begin{aligned} a^2 & =b^2+c^2-2 b c \cos A \\ a^2 & =4.8^2+8^2-2 \times 4.8 \times 8 \times \cos 42^{\circ} \\ & =29.966 \ldots \\ a & =5.47 \text { (3 s.f.) } \end{aligned}
The Cosine Rule- Further Examples
Example 3
For this question a diagram is certainly needed to help present all the required information.
\begin{aligned} & b^2=a^2+c^2-2 a c \cos B \\ & 5^2=(x+2)^2+x^2-2 x(x+2) \cos 60^{\circ} \end{aligned}\begin{aligned} 25=2 x^2+4 x & +4-x^2-2 x \\ x^2+2 x-21 & =0 \\ x & =\frac{-2 \pm \sqrt{88}}{2} \\ & =3.69 \text { (3 s.f.) } \end{aligned}From the solution above we have had to initially use the cosine rule and then to arrive at a quadratic equation which has required use of the quadratic formula!
The Cosine Rule – Exam Style Questions
In this section we will look at some exam style questions focused around the cosine rule and discuss their solution in some detail.
Example 1
To be able to answer this question you need to be able to find some missing angles.
This has been done using co interior angles and the fact that angles at a point add to 360 degrees.
\begin{aligned} c^2 & =a^2+b^2-2 a b \cos C \\ c^2 & =120^2+150^2-2 \times 120 \times 150 \cos 90^{\circ} \\ & =14400+22500-0 \\ & =36900 \end{aligned}c=192.0937 \ldotsExample 2
To do part a) of this question it is important that you draw a quick diagram to help visualise the question better.
Which is the required result.
For part b) we are using our answer to part a) in order to solve for x and this is done as follows:
\begin{aligned} \frac{4 x-15}{2 x} & =-\frac{1}{7} \\ 7(4 x-15) & =-2 x \\ 28 x-105 & =-2 x \\ 30 x & =105 \\ x & =\frac{105}{30} \\ & =3 \frac{1}{2} \end{aligned}Example 3)
Part a) is a simple cosine rule substitution question which can be done as follows:
\begin{aligned} & a^2=b^2+c^2-2 b c \cos A \\ & a^2=70^2+50^2-2 \times 70 \times 50 \cos 20^{\circ} \\ & a^2=4900+2500-6577.848 \ldots \\ & a^2=822.15165 \ldots \end{aligned}a=28.673 \ldotsPart b) you need to find the angle at B which can be done using the cosine rule but you are finding an angle. This can be done as follows:
\begin{aligned} & \cos B=\frac{a^2+c^2-b^2}{2 a c} \\ & \cos B=\frac{28.673^2+50^2-70^2}{2(28.673)(50)} \\ & \cos B=\frac{822.15165+2500-4900}{2867.3187} \\ & \cos B=-0.55028 \ldots \\ & \mathrm{B}=\cos ^{-1} 0.55028 \ldots \\ & \quad=123.3867 \ldots \end{aligned}To find the actual bearing you now need to subtract from 180 degrees as follows:
180^{\circ}-123.3867^{\circ}=56.6^{\circ}As you can see you are still required to use some trigonometric skills from GCSE and in another article we will consider the Sine Rule.