- 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
Use the slider to explore worked examples.
-
Find the unknown angles and side length of the triangle shown.
Using the sine rule
\(\frac{\sin A}{2.1} =\frac{\sin70^\circ }{3.5}=\frac{\sin B}{b}\)From the first equality,
\(\sin A=\frac{2.1\sin 70^\circ }{3.5} = 0.5638\)
\(A=34.32^\circ\)Since angles in triangle add up to \(180^\circ\)
\(B=180^\circ-70^\circ-34.32^\circ=75.68^\circ\)From the sine rule
\(\frac{\sin 70^\circ}{3.5} =\frac{\sin B}{b}\)
\(b = \frac{3.5 \times \sin B}{\sin 70^\circ} =\frac{3.5 \times \sin 75.68^\circ}{\sin 70^\circ} =3.61cm\) -
Find two solutions for the unknown angles and side of the triangle shown.
Using the sine rule:
\(\frac{\sin A}{a} =\frac{\sin B}{b} =\frac{\sin C}{c}\)From the second equality:
\(\sin B=\frac{6\times \sin 42^\circ}{5} = 0.8030\)The graph above shows that between \(0^\circ\) and \(180^\circ\) there are two solutions for \(B\).
These solutions are:
\(B = 53.41^\circ \)
and by symmetry
\(B = 180^\circ – 53.41^\circ = 126.59^\circ\)Solving for \(A\) we have:
\(A = 180^\circ – 42^\circ - B\)
When \(B = 53.41^\circ A = 84.59^\circ\)
When \(B = 126.59^\circ A = 11.41^\circ\)From the sine rule:
\(a = \frac{6 \times \sin A}{\sin B}\)
For \(A = 84.59^\circ B = 53.41^\circ a = 7.44 cm\)
For \(A = 11.41^\circ B = 126.59^\circ a = 1.48 cm\)Looking at the diagram above you can see that only one solution fits
\(a = 7.44 cm\)
-
Find the unknown side and angles of the triangle shown.
To find a, use the cosine rule:
\( a^{2} = b^{2} + c^{2} -2bc \; cos \; A \)
\( a^{2} = 4.9^{2} + 3.7^{2} - 2 \times 4.9 \times 3.7 \times cos 65^{\circ} \)
\( a^{2} = 22.3759 \)
\( a= 4.73 \; (to 2 d.p.) \)To find the angles, use the sine rule:
\( \frac{sin \; C}{3.7} = \frac{sin65^{\circ}}{a} = \frac{sin \; B}{4.9} \)
\( sin \; B = \frac{4.9 \; sin 65^{\circ}}{a} = \frac{4.9 \; sin 65^{\circ}}{4.73} = 0.9389 \)
giving: \( B = 69.86^{\circ} \)
Therefore:
\( A = 65^{\circ} \)
\( B= 69.86^{\circ} \)
\( C= 180^{\circ} - 65^{\circ} - 69.86^{\circ} = 45.14^{\circ} \)
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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
Use the slider to explore worked examples.
-
Find the unknown side and angles of the triangle shown.
Solution
To find \(a\), use the cosine rule:
\(a^{2}= 4.9^{2}+ 3.7^{2}- 2\times 4.9\times 3.7\times \cos65^\circ\)
\(a^{2}= 22.3759\)
\(a= 4.73\) (to 2 d.p.)To find the angles, use the sine rule:
\(\frac{\sin65^\circ}{a}= \frac{\sin B}{4.9}= \frac{\sin C}{3.7}\)
\(\sin B=\frac{4.9\times \sin65^\circ}{a}- \frac{4.9\times \sin65^\circ}{4.73}= 0.9389\)
\(B= 69.86^\circ\)
\(\sin C=\frac{3.7\times \sin65^\circ}{a}- \frac{3.7\times \sin65^\circ}{4.73}= 0.7090\)
\(C= 45.15^\circ\)
(alternatively, use A + B + C = 180\(^\circ\) to Find C)Checking. \(AB++C=65^\circ+ 69.86^\circ+45..15^\circ= 180^\circ\)
The three angles should add to \(180^\circ\), the extra \(.001^\circ\) is due to rounding errors. -
Find the shaded angle in the triangle shown.
Solution
Using the cosine rule.
\(25^{2}= 16^{2}+ 13^{2}- 2\times 13\cos x\)Rearranging.
\(2\times 16\times 13\cos x= 16^{2} + 13^{2}= 25^{2}\)
\(416\cos x= -200\)
\(\cos x= \frac{-25}{52} \)
\(x= \cos^{-1}\left(\frac{-25}{52} \right) ^\circ\)
\(=118.7^\circ\) to \(1\) decimal place using SHIFT and COS buttons on a calculator.Note: As \(\cos x \) is negative, the angle will be obtuse \(\left(90^\circ\lt x\lt 180^\circ\right)\)
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- 1, Chapter 3, Task 2, Worked Example 3,
Worked Example
-
Find the unknown side and angles of the triangle shown.
Solution
To find "a", use the cosine rule:
\(a^2 = b^2 + c^2 - 2bc \cos A\)
\(a^2 = 4.9^2 + 3.7^2 - 2 \times 4.9 \times 3.7 \times \cos 65^ \circ\)
\(a^2 = 22.3759\)
\(a = 4.73 (to 2 d.p.)\)To find the angles, use the sine rule:
\(\frac{\sin C}{3.7} = \frac{\sin 65^ \circ}{a} = \frac{\sin B}{4.9}\)
\(\sin B = \frac{4.9 \sin 65^ \circ}{a} = \frac{4.9 \sin 65^ \circ}{4.73} = 0.9389\)
\(B = 69.86^ \circ\)Therefore
\(A = 65^ \circ\)
\(B = 69.86^ \circ\)
\(C = 180^ \circ - 65^ \circ - 69.86^ \circ = 45.14^ \circ\)
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- 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
Use the slider to explore worked examples.
-
An oil tanker leaves Town X, and travels on a bearing of \( 050^{\circ} \) to Town Z, \( 50 \; km \) away.
The tanker then travels to Town Y, \( 70 \; km \) away, on a bearing of \( 120^{\circ} \).
a) Find the distance of Y from X, giving your answer to 3 sig. figs.
b) b. Calculate the bearing of \(X\) from \(Y\), giving your answer to the nearest degree
Solution
a) In triangle \( XZY \), angle \( XZY = 50^{\circ} + 60^{\circ} = 110^{\circ} \) so, using the cosine rule
\( XY^{2} = 50^{2} + 70^{2} - 2 \times 50 \times 70 \times cos \; 110^{\circ} \)
\( = 2500 + 4900 - 7000 \; cos \; 110^{\circ} \)
\( = 7400 - 7000 \times (-0.3420) = 9794 \)
\( XY = 98.965 = 99.0 \) to 3 sig. figs.b) We need to find the angle \(ZYX\) to determine the bearing of \(X\) from \(Y\).
In triangle \(ZYX\), using the sine rule:
\(\frac{\sin ZYX}{50} = \frac{\sin 110^\circ}{99}\)
\(\sin ZYX \approx 0.4748\)
angle \(ZYX \approx 28.34^\circ = 28^\circ\) to the nearest degree
Hence the bearing of \(X\) from \(Y = 360^\circ - (60^\circ + 28^\circ) = 272^\circ\) -
Find the shaded angle in the triangle shown.
Solution
Using the cosine rule,
\(25^2 = 16^2 + 13^2 - 2 \times 16 \times 13 \times \cos x \)Rearranging
\(2 \times 16 \times 13 \times \cos x = 16^2 + 13^2 - 25^2\)
\(416 \times \cos x = -200\)
\(\cos x = \frac{-25}{52}\)
\(x = \cos -1\left(\frac{-25}{52} \right) = 118.7^\circ\)
1 d.p. using the SHIFT and COS buttons on a calculator
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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,
The area (A) is given by \(A=\frac{1}{2} ab\sin \theta\)
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
Use the slider to explore worked examples.
-
The diagram shows a circle of radius \(64\;cm\). The length of the chord \(AB\) is \(100\;cm\).
- Find the angle \(\theta\) , to \(2 d.p\).
- Find the area of triangle \(OAB\).
Solution
- \(\sin \theta= \frac{50}{64}\) \(\theta=51.38^\circ\)
- \(A=\frac {1}{2}\times\sin \theta = \frac {1}{2}\times 64\times 64\sin 102.76^\circ=1997cm^{2}\)
-
Given that \(\sin \theta=\frac{\sqrt{3} }{2}\), \(0^\circ\leq \theta\leq 90^\circ\)
Find the area of triangle \(CDE\) where\(CD = 30\;units\) and \(DE = 20\;units\).
Solution
\(A=\frac{1}{2} ab\sin \theta=\frac{1}{2}\times 30\times 20\times \frac{\sqrt{3} }{2} = 150\sqrt{3}cm^{2}\)
Find the length of the side \(CE\) using the cosine ruleSolution
\(CE^{2}= 30^{2}+20^{2}- 2\times 30\times 20\times \cos 60^\circ\)
\(CE^{2}= 900+ 400- 1200\times \cos 60^\circ\)
\(CE^{2}= 700\) \(CE=\sqrt{700}\)\(CE=26.46 2.d.p.\)
Solution
\(sin \theta= \frac{\sqrt{3} }{2}\)
\(\sin \theta ^{- 1} \left(\frac{\sqrt{3} }{2} \right)\)
\(\theta= 60^\circ\)
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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.
\(A=\frac{1}{2} ab\sin C\)
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
-
For the triangle shown, find
- the area of the triangle,
- angle \(\theta\).
Solution
- Using Heron's formula
\(Area= \sqrt{s\left(s- a\right)\left(s- b\right)\left(s- c\right) }\)
where: \( s=\frac{a+b+c}{2} \; \) gives \(\; s=\frac{4+5+7}{2} = 8 \)
\(Area= \sqrt{8\left(8- 4\right)\left(8- 7\right)\left(8- 5\right) }\)
\(S=\frac{a+ b+ c}{2}\)
\(S=\frac{4+ 5+ 7}{2}= 8\)
Area of Triangle ABC = \(9.80 \; cm^{2}\) to 2 d.p. - To find angle \(\theta\) we use \(A=\frac{1}{2} ab\sin \theta\)
\(9.80 cm^{2}=\frac{1}{2}\times 4\times 5\times \sin \theta\)
\(\sin \theta= 0.98\)
\(\theta= 78.5^\circ\) tp 1 d.p.
-
- 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
Use the slider to explore worked examples.
-
Find a) \(cos \; 150^{\circ}\)
b) \(sin \; 240^{\circ}\)Solution:
a) \(150^{\circ}\) is in the second quadrant so \(P\) has coordinates \(( -cos \;30^{\circ}, sin \; 30^{\circ} )\)
Hence \(cos \; 150^{\circ} = -cos \;30^{\circ} = \frac{\sqrt{3}}{2}\)
b) \(240^{\circ}\) is in the third quadrant so \(P\) has coordinates \( ( -cos \; 60^{\circ}, -sin \; 60^{\circ} ) \)
Hence \( sin \; 240^{\circ} = -sin \; 60^{\circ} = - \frac{\sqrt{3}}{2} \)
-
If \( cos \; \theta = - \frac{1}{2} \), how many values of the angle \(\theta\) are possible for \( 0 \leq \theta \leq 720^{\circ} \). Find these values of \(\theta\) .
Solution:
The graph of cos𝜃 shows that there are 4 possible values of \(\theta\)
Using the symmetry of the graph, the values of \(\theta\) are:
\( \theta = 120^{\circ}, 240^{\circ}, 480^{\circ}, 600^{\circ} \) -
Use a calculator to solve the equation \(sin \theta = -0.2\). Sketch the sin graph to show this solution. Give the principal value solution
scroll left to see another worked example
Solution:
Using the \( sin^{-1} \) key on a calculator, \( \theta = sin^{-1} (-0.2) = - 11.537^{\circ} \)
The graph of the graph of \(sin \theta\) shows why \(\theta\) is negative
The principal value solution is \( -11.537^{\circ} \)
-
- 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
