Trigonometry and Pythagoras

Topic Overview

Pythagoras' Theorem describes the mathematical relationship between three sides of a right-angled triangle. Pythagoras' Theorem states that; in a right-angled triangle the square of the hypotenuse longest side is equal to the sum of the squares of the other two sides. It is written in the formula:

\[{a^2} + {b^2} = {c^2}\]
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As well as Pythagoras' Theorem, there are other formulae which can be used to calculate a unknown side or angle in a triangle; such as trigonometry.

'Trigonometry' is a field of study in mathematics which observes the relationships of the sides and angles of triangles.

The trigonometric functions ‘sine’, 'cosine' and ‘tangent’, or as they are more commonly know ‘sin’, ‘cos’, and ‘tan’, can be used to find missing sides or angles of triangles. These functions are defined as the ratios of the different sides of a triangle.

The symbol "θ" is used to describe an unknown angle. If you know the lengths of two sides of a triangle, you can calculate the value of angle "θ" using trigonometry. The functions of sin, cos and tan can be calculated as follows:

  • Sine Function: sin(θ) = Opposite / Hypotenuse
  • Cosine Function: cos(θ) = Adjacent / Hypotenuse
  • Tangent Function: tan(θ) = Opposite / Adjacent

The term SOH CAH TOA is often used by students to remind themselves of these functions. As long as you remember the correct terms for each trigonometric function, you can use them to calculate the values of unknown angles or sides of various triangles.

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Key Concepts

In the new linear GCSE maths paper, you will be required to solve various mathematical problems involving trigonometry and Pythagoras' theorem. The specific questions you will be expected to answer will vary depending upon which examination board with which you are registered, but as a rule you will be required to:

  • Use Pythagoras' theorem in 2D and 3D shapes
  • Use trigonometric functions, the sine rule and the cosine rule to solve various mathematical problems
  • Calculate the area of a triangle by using 1/2 ab sinC

Listed below are a series of summaries and worked examples to help you solidify your knowledge about trigonometric functions and Pythagoras' theorem.

Worked Examples

1. Using Pythagoras' Theorem
During your GCSE maths exam, you will be required to calculate various mathematical problems using Pythagoras' theorem:

Example
(a) - The following triangle has the values a = 9 and c = 15. Calculate the value of side 'b

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Solution
(a) - Using Pythagoras' theorem

\[{a^2} + {b^2} = {c^2}\]
\[{9^2} + {b^2} = {15^2}\]

Now all you have to do is solve this expression to find the value of ‘b’:

\[{b^2} = 255 - 81 = 144\]
\[b = \sqrt {144} = 12\]

2. Using Pythagoras in 3D shapes - Higher Tier
If you are sitting the Higher Tier GCSE maths exam paper, you may be required to use Pythagoras' theorem to solve problems which involve right-angled triangles within 3D shapes.

Example
(a) - The following triangle has the values a = 9 and c = 15. Calculate the value of side 'b

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Solution
(a) - Although this question may appear complex, you can simplify the process using Pythagoras' theorem.

Firstly, you must use Pythagoras' theorem in triangle ABC in order to find length AC:

\[A{C^2} = A{B^2} + B{C^2}\]
\[A{C^2} = {6^2} + {2^2}\]
\[A{C^2} = 36 + 4=40\]
\[AC = \sqrt {40} \]

Now that you know the value of length AC, you can use Pythagoras' theorem in triangle ACF in order to find the length of AF:

\[A{F^2} = A{C^2} + C{F^2}\]
\[A{F^2} = {(\sqrt {40} )^2} + {3^2}\]
\[A{F^2} = 40 + 9 = 49\]
\[AF = \sqrt {49} = 7\]

Therefore, the length of the diagonal AF = 7cm

3. Trigonometry: using sin, cos and tan
As mentioned in the topic overview, you can use the trigonometric functions sin, cos and tan to find the length of the sides of a triangle; the hypotenuse, opposite and adjacent, as well as unknown angles.

During your GCSE maths exam, you will be required to use these trigonometric functions to find the value of an unknown angle:

Example
(a) - Calculate the angle 'C' and state your answer correct to 1 decimal place to 1dp.

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Solution
(a) - From the diagram above, you know the lengths of the adjacent AC and the hypotenuse BC. Using the phrase SOH CAH TOA, you know that you need to use the cosine function to calculate angle C:

\[\cos (C) = 5/7\]
\[C = {\cos ^{ - 1}}(5/7)\]

In order to calculate the value of angle C, you need to calculate the inverse of cos. To do this, type [SHIFT] cos (5/7) into your calculator to receive the value of,

\[C = {44.4^ \circ }\]

During your GCSE maths exam, you will also be required to use trigonometric functions to calculate the length of one of the sides of a triangle:

Example
(b) - Calculate the length of the side BC.

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Solution
(b) - From the diagram above, you know the values of the hypotenuse AB and the angle CAB . The length you are being asked to calculate (BC) is opposite to angle CAB. Using the sentence SOH CAH TOA, you know you need to use the function 'sin':

\[\sin (CAB) = BC/AB\]
\[\sin (30) = BC/7\]

From here, you can solve the expression to calculate the value of length BC:

\[BC = (\sin (30))/7\]

Using your calculator, type [sin30 ÷ 7] to receive the value of length BC = 3.5cm

4. The sine rule
Sine and cosine each have a rule which can be used to find the size of an angle or the length of a side. You may be asked to use these rules during your GCSE maths exam in order to calculate unknown values:

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The sine rule is:

\[\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}}\]

or it can be written as

\[\frac{{\sin A}}{a} = \frac{{\sin B}}{b} = \frac{{\sin C}}{c}\]

The sine rule can be used if you know either;

2 of the sides of a triangles as well as an angle which is opposite to one of the 2 known sides OR one side and any 2 angles of a triangle.

Example
(a) - Calculate the size of angle R, and give your answer correct to 1 decimal place (1dp).

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Solution
(a) - Using the sine rule, you know that:

\[\frac{{\sin R}}{4} = \frac{{\sin 75}}{9}\]

Now you need to solve this expression to find the value of R:

\[\sin (R) = (\sin (75)/9)*4 = 0.429....\]

Using a calculator, you can work out that R = [SHIFT sin0.429] = 25.4

Therefore angle R = 25.4°to 1dp

5. The cosine rule
Similarly to the sine rule, you may be required to use the cosine rule during your GCSE maths exam in order to find the length of a side or size of an angle:

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The cosine rule is:

\[{a^2} = {b^2} + {c^2} - 2bc\cos A\]

However, depending upon your known values, it can also be written as:

\[{b^2} = {a^2} + {c^2} - 2ac\cos B\]

or as

\[{c^2} = {a^2} + {b^2} - 2ab\cos C\]

These variations of the cosine rule can also be rearranged into the following rules:

\[\cos A = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\]
\[\cos B = \frac{{{a^2} + {c^2} - {b^2}}}{{2ac}}\]
\[\cos C = \frac{{{a^2} + {b^2} - {c^2}}}{{2ac}}\]

The cosine rule can be used if you know either:

2 sides and an angle of a triangle or 3 sides of a triangle but no angles

Example
(a) - Calculate the length of the side BC, and give your answer correct to 2 decimal places to 2dp.

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Solution
(a) - Using the cosine rule

\[{a^2} = {b^2} + {c^2} - 2bc\cos A\]
\[{BC^2} = {7^2} + {3^2} - 2*3*7\cos 35\]

Now you need to solve this expression to find the value of the length BC:

\[{BC^2} = {58} - 42\cos 35\]
\[{BC^2} = 23.5956\]
\[BC = \sqrt {23.5956} \]

Therefore, the length of BC = 4.86cm to 2dp

6. Using the side-angle-side (SAS) formula to calculate the area of a triangle
Trigonometry can also be used to calculate the area of a triangle. Using the Side-Angle-Side (SAS) rule, you can calculate the height of a triangle and use that value to calculate the area of the triangle:

The SAS rule is written in the form:

Area of triangle ABC = 1/2ab sin C

This SAS rule can also be rearranged and written as either:

Area of triangle ABC= 1/2ac sin B OR Area of triangle ABC = 1/2bc sin A

(Note: The SAS rule can only be used if you know two sides of a triangle and the included angle).

Example
(a) - Calculate the area of the triangle ABC and give your answer correct to 1 decimal place 1dp.

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Solution
(a) - Using the SAS rule, you know that the area of the triangle ABC is:

\[1/2*AB*AC*\sin (CAB)\]

So the area of triangle is calculated as,

\[ABC = 1/2*5.2*7.1*\sin (42) = 12.4\]
\[ABC = 12.4c{m^2}\]

Note: When calculating the area of a triangle, remember to present your answer in the correct units of measurement; i.e. cm squared.

Exam tips

  1. Memorise Pythagoras' theorem
  2. Memorise the term SOH CAH TOA
  3. Remember the sine rule
  4. Remember that the cosine rule
  5. Memorise the SAS rule for calculating the area of a triangle
  6. Write down ALL of your working out in order to earn maximum method marks.

Topic Summary

When revising 'Trigonometry and Pythagoras' Theorem', your main focus should be memorising the fundamental principles and formulae. Once you understand and can recall these functions, you can apply a step-by-step process to every trigonometry and Pythagoras question during your actual exam. By doing so, you will substantially increase your chances of scoring higher marks.

You should also write down ALL of your working out. This proof of your method will demonstrate to the examiner who is marking your paper that you have a comprehensive understanding of trigonometric functions and Pythagoras' theorem. Moreover, it will provide you with a golden opportunity to quickly double check your answers and efficiently correct any mistakes you may have otherwise overlooked!

Related topics

  • Polygons
  • Congruent Triangles
  • Construction and Loci Problems
  • Perimeter, Area, Volume
  • Transformations
  • Circle Theorem Problems
  • Proof