- Sections:
- 0, Chapter 1, Introduction,
Introduction
This unit extends the theme of finding angles and lengths in a right angle triangle to finding them in any triangle. After completing this unit you should
- understand how to use the sine and cosine rules in non right angled triangles
- be able to find the area of any triangles, given two sides and the included angle
- be able to use Heron's formula to calculate the area of any triangle, given the lengths of its three sides.
You have five sections to work through.
- Sine Rule
- Cosine Rule
- Sine and Cosine
- Application: Area of Any Triangle
- Heron's Formula
- Angles larger than \(90^{\circ}\)
- 0, Chapter 2, Sine Rule,
Sine Rule
In the triangle \(ABC\) shown below, the side opposite angle \(A\) has length \(a\), the side opposite angle \(B\) has length \(b\) and the side opposite angle \(C\) has length \(c.\)
The sine rule states \(\frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c}\)
The sine rule can be used in any triangle to calculate:
- a side when two angles and an opposite side are known \(\left(AAS\right)\)
- an angle when two sides and an opposite angle are known \(\left(SSA\right)\)
The cosine rule states:
\( a^{2} = b^{2} + c^{2} - 2bc \; cos \; A \)
\( b^{2} = a^{2} + c^{2} - 2ac \; cos \;B \)
\( c^{2} = a^{2} + b^{2} - 2ab \; cos \; C \)
Note that when, for example \( A = 90^{\circ} \) , the formula becomes, as expected, Pythagoras’ Theorem: \( a^{2} = b^{2} + c^{2} \)The cosine rule can be used in any triangle to calculate:
- a side when two sides and the angle in between them are known \( (SAS) \)
- an angle when three sides are known \( (SSS) \)
- 1, Chapter 2, Task 1, Worked Example 1,
Worked Example
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- 0, Chapter 3, Cosine Rule,
Cosine Rule
In the triangle \(ABC\) shown below, the side opposite angle \(A\) has length \(a\), the side opposite angle \(B\) has length \(b\) and the side opposite angle \(C\) has length \(c\).
The cosine rule states:
\(a^2=b^2+c^2-2bc\cos A\)
\(b^2=a^2+c^2-2ac\cos B\)
\(c^2=a^2+b^2-2ab\cos C\)The cosine rule can be used in any triangle to calculate:
- a side when two sides and the angle in between them are known \(\left(SAS\right)\)
- an angle when three sides are known \(\left(SSS\right)\)
- 1, Chapter 3, Task 1, Worked Example 2,
Worked Example
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- 1, Chapter 3, Task 2, Worked Example 3,
Worked Example
- 0, Chapter 4, Sin and Cos,
Sin and Cos
sin\(\theta\) and cos\(\theta\)
The graphs of \(\sin \theta\) and \(\cos \theta\) for any angle are demonstrated iin the diagrams below which show the sine and cosine curves.
The graphs are examples of periodic functions. The graph repeats every \(360^\circ\) (This is called the period , ie the period is \(360^\circ\)). In each period the maximum and minimum value of the functions are \(1\) and \(-1\) respectively.
Below are the values of sine and cosine for angles which often appear in questions.
\( \theta\) \(\sin \theta\) \(\cos \theta\) \(0^\circ\) \(0\) \(1\) \(30^\circ\) \(\frac{1}{2} \) \(\frac{\sqrt{3} }{2} \) \(45^\circ\) \(\frac{1}{\sqrt{2} }\) \(\frac{1}{\sqrt{2} }\) \(60^\circ\) \(\frac{\sqrt{3} }{2} \) \(\frac{1}{2} \) \(90^\circ\) \(1\) \(0\) - 1, Chapter 4.1, Bearing,
Bearing
A bearing is an angle in degrees measured clockwise from North.
Some questions using sine and cosine are about objects (such as cars, ships, or joggers) travelling in certain directions for periods. In these questions you use sine or cosine to determine the eventual position of the object. They usually contain the concept of bearing described above.
Here is an example of such a question.
- 1, Chapter 4, Task 1, Worked Example 4,
Worked Example
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- 1, Chapter 4, Task 2, Exercise 1,
Exercise
Here are some questions to check your progress
Exercise 1
- Find the unknown angle marked \(\theta\).
- Find the unknown side marked \(a\)
\(\frac{\sin\theta}{8.1} = \frac{\sin 85^ \circ}{10.3} \)
\(\sin \theta= \frac{8.1 \sin85^\circ}{10.3}\)
\(\theta = 51.6^\circ 1 d.p.\)
Exercise 2
Find the unknown angles and sides.
(a) \(a^{2}=3^{2}+ 3.5^{2}- 2\times 3\times 3.5\times \cos60^\circ\)
(b) \(\frac{\sin C}{3.5}= \frac{\sin 60^\circ}{3.278}\)
c = 67.6\(^\circ\) (1d.p)
B = 180 – 60 – 67.6 = 52.4\(^\circ\)
Exercise 3
To calculate the height of a church tower, a surveyor measures the angle
of elevation of the top of the tower from two points 50 m apart.- Calculate the distance BC.
- Hence calculate the height CD.
- 0, Chapter 5, Application: Area of Any Triangle,
Application: Area of Any Triangle
An important application of trigonometry is that of finding the area of a triangle with side lengths a and b and included angle \(\theta\)
Area of a triangle when the lengths of two sides and the included angle are known,
We can prove this result by constructing the perpendicular line from point \(B\) to the line \(AC\).
Area \( = \frac{1}{2} \; \times \; base \times \; prependicular height \)
\( = \frac{1}{2} \times b \times p \)
where: \( p= a \; sin \; \theta \)
So Area \( = \frac{1}{2} \times b \times a \; sin \; \theta \)
\( = \frac{1}{2} \; a \; b \; sin \; \theta \) - 1, Chapter 5, Task 1, Worked Example 5,
Worked Example
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- 1, Chapter 5, Task 2, Exercise 2,
Exercise
Here are some questions to check your progress
Exercise 1
Find the area of the shaded region
\(A=\frac{1}{2} ab\sin \theta=\frac{1}{2}\times 9\times 7\times \sin 130^\circ\)
\(A=24.13 cm^{2}\)
Exercise 2
Find the area of the shaded region
\(A=49\pi - \frac{1}{2} ab\sin \theta= \frac{1}{2}\times 7\times 7\times \sin 150^\circ\)
\(A=141.67 cm^{2}\)
Exercise 3
In the diagram ST = 5 cm , TW = 9 cm and \(STW = 52^\circ\)
Calculate
- the length of SW
- the area of \(\triangle\) STW
- 𝑆𝑊 = 7.11 cm
- Area =17.73〖 cm〗^2
- 0, Chapter 6, Heron's Formula,
Heron's Formula
You have already met the formula for the area of a triangle when the lengths
of two sides and the included angle are known,Formula for the area of a triangle when the lengths of all three sides are known.
We will use this result to find the formula for the area of a triangle when the lengths
of all three sides are known.The formula is credited to Heron (or Hero) of Alexandria, and a proof can be found in his book,
Metrica, written in about AD 60.The formula, known as Heron's formula, is given by
\(Area= \sqrt{s\left(s- a\right)\left(s- b\right)\left(s- c\right) }\)
\(S=\frac{a+ b+ c}{2}\)
- 1, Chapter 6, Task 1, Worked Example 6,
Worked Example
- 1, Chapter 6, Task 2, Exercise 3,
Exercise
Here are some questions to check your progress
Exercise 1
Calculate the area of the triangle
\(Area=\sqrt{8 \left(8- 5\right)\left(8- 5\right)\left(8- 6\right) }\)
\(Area= \sqrt{144} \)
\(A = 12 cm^{2}\)
Exercise 2
- the area of the triangle,
- the angle shown by \(\theta\)
- Area = 27.81〖 mm〗^2
- 𝛳 = 44°
- 0, Chapter 7, Angles Larger than \(90^{\circ}\),
Angles Larger than \(90^{\circ}\)
Let us explore how we define the trig rules for angles greater than \(90^{\circ}\)
First consider the \(x-y\) plane is divided into four quadrants by the \(x\) and \(y-axes\) as shown in the picture below
The angle \(\theta\) that a line \(OP\) makes with the positive \(x-axis\) lies between \(0^{\circ}\) and \(360^{\circ}\).Angles between \(0^{\circ}\) and \(90^{\circ}\) are in the first quadrant.
Angles between \(90^{\circ}\) and \(180^{\circ}\) are in the second quadrant.
Angles between \(180^{\circ}\) and \(270^{\circ}\) are in the third quadrant.
Angles between \(270^{\circ}\) and \(360^{\circ}\) are in the fourth quadrant.
Angles larger than \(360^{\circ}\) can be reduced to lie between \(0^{\circ}\) and \(360^{\circ}\) by subtracting multiples of \(360^{\circ}\)
The trigonometric formulae, \(\cos \theta\) and \(\sin \theta\), are defined for all angles between \(0^{\circ}\) and \(360^{\circ}\) as the coordinates of a point, \(P\), where \(OP\) is a line of length \(1\), making an angle \(\theta\) with the positive \(x-axis\).
This is illustrated in the diagram below:
Here are some of the important values of \(\sin \theta\) , \(\cos \theta\) and \(\tan \theta\) which we have meet before; you should already be familiar with these.
- 1, Chapter 7.1, Sin and Cos 1,
Graphs of Sin\(\theta\) and Cos\(\theta\)
The graphs of \(\sin \theta\) and \(\cos \theta\) for any angle are shown below.
The graphs are examples of periodic functions. Each basic pattern repeats itself every \(360^{\circ}\). We say that the period is \(360^{\circ}\).
For any angle, note that \(\sin \left(90^{\circ}- \theta \right) = \cos \theta\)
The graph of \( \tan \theta\) has period \(180^{\circ}\). It is an example of a discontinuous graph.
The trigonometric equations \(sin \theta =a\) , \( \cos \theta =b\) and \( \tan \theta=c\) can have many solutions.
The inverse trig keys on a calculator (that is \(sin^{-1}\), \(\cos^{-1},\tan^{-1}\) give the principal value solution.
For \(\sin \theta=a\) and \(\tan \theta=c\) , the principal value solution is in the range:
\(-90^{\circ} \leq \theta \leq 90^{\circ}\)
For \(\cos \theta = b\), the principal value solution is in the range:
\( 0 \leq \theta \leq 180^{\circ}\)
- 1, Chapter 7.2, Tan,
Graphs of Tan\(\theta\)
The graph of tanθ has period \(180^{\circ}\). It is an example of a discontinuous graph.
- 1, Chapter 7.3, Trigonometric Equations,
Trigonometric Equations
The trigonometric equations \( sin \; \theta = a \) , \( cos \; \theta = b \) and \( tan \; \theta = c \) can have many solutions.
The inverse trig keys on a calculator (that is \( sin^{-1} \), \( cos^{-1} \), \( tan^{-1} \)) give the principal value solution.
For \( sin \; \theta = a \) and \( tan \; \theta = c \) , the principal value solution is in the range:
\( -90^{\circ} \; \leq \; \theta \; \leq \; 90^{\circ} \)For \( cos \; \theta = b \), the principal value solution is in the range:
\( 0^{\circ} \; \leq \; \theta \; \leq \; 180^{\circ} \) - 1, Chapter 7.3, Task 1, Worked Example 7,
Worked Example
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- 0, Chapter 8, Summary,
Summary
the angle of elevation \(\alpha\) is the angle made with the horizontal, as shown in the diagram
(angle \(\beta\) is called the angle of depression). Angle \(\alpha = angle\; \beta\) (alternate angles).\(Angle\; \alpha=\;angle\;\beta\) (alternate angles).
for any triangle ABC, the sine rule states
\(\frac{sin\;A}{a} = \frac{sin\;B}{b} = \frac{sin\;C}{c}\)
and the cosine rule states
\(c^{2}=a^{2}+b^{2} - 2 ab \cos C.\)
and similarly,
\(a^{2} = b^{2} + c^{2} - 2bc \cos A\)
\(b^{2} = c^{2} + a^{2} - 2ca \cos B\)
- 0, Chapter 1, Introduction,
- Interactive Exercises:
- Sine and Cosine Rules Interactive Exercises, https://www.cimt.org.uk/sif/trigonometry/t3/interactive.htm
- Sine and Cosine Rules , https://www.cimt.org.uk/sif/trigonometry/t3/interactive/s1.html
- Area of Any Triangle , https://www.cimt.org.uk/sif/trigonometry/t3/interactive/s2.html
- Heron's Formula, https://www.cimt.org.uk/sif/trigonometry/t3/interactive/s3.html
- YouTube URL: https://youtu.be/i3Q1hvi6I5w
- File Attachments: media/course-resources/Sine-Cosine-Rules_Learing-Objectives.pdfmedia/course-resources/Sine-Cosine-Rules_PowerPoint-Presentation.pptxmedia/course-resources/Sine-Cosine-Rules_Text.pdfmedia/course-resources/Sine-Cosine-Rules_Answers.pdfmedia/course-resources/Sine-Cosine-Rules_Essential-Information.pdf