Losing your bearings with Standard 2 trigonometric Functions? This article will elaborate on what the trigonometric functions in the HSC are and how they can be used in a variety of contexts. Trigonometric functions are present in many physical situations in everyday life, from setting up a ladder, to a golf swing, to the slope of a roof.
Trigonometric functions relate the sides of a triangle to its angle, allowing us to infer unknown sides and angles of triangles for different measurements.
In a right triangle, the ratio between different sets of sides form the sine, cosine and tangent ratios which are constant for a constant angle; and by putting right triangles together in different ways we can derive formula which can be used for non-right triangles as well.
Students:
Students should be comfortable with:
Students may have encountered trigonometric ratios in previous years, and this will be beneficial for understanding the content to be presented.
The most basic use of the trigonometric ratios: sine, cosine and tangent, is in finding the length of the unknown side or size of an unknown angle of a right-angled triangle. If we are given an angle and a side, we can find the length of the other 2 sides. Conversely, if we are given the length of 2 sides, we can find the size of a specific angle of the triangle.
Let’s recap what you know, first.
Where:
\begin{align*}
\sinθ&=\frac{opposite}{hypotenuse}\\
\cosθ&=\frac{adjacent}{hypotenuse}\\
\tanθ&=\frac{opposite}{adjacent}\\
\end{align*}
Note: be careful with the definition of the opposite or adjacent side. The hypotenuse is always the longest side of any triangle, whilst the opposite/ adjacent side depends on the angle you are looking at. The opposite side is the side directly opposite the angle, and the adjacent side is the side that sandwiches the given angle with the hypotenuse.
In one type of question, you will be given an angle and the length of one side. You must identify the correct trigonometric ratio to use depending on the given side and the side you are trying to find, then substitute the given information into the trig definition and solve for wanted side – for example:
Example 1: Find the unknown side x
Solution 1:
\begin{align*}
\sin45°&=\frac{o}{h}\\
&=\frac{12}{x} \text{(sub in values)}\\
sin45° \times x&=12\\
x&=\frac{12}{\sin45°}\\
x&=12\sqrt{2}\\
\end{align*}
These questions will always give you the length of 2 sides. You must substitute the given information into the trig definition and solve for angle by using inverse trig.
The inverse trigonometric functions are \(\sin^{-1}x, \cos^{-1}x, \tan^{-1}x\). Normal trigonometric functions take an angle and output a ratio e.g. \(\sin45=\frac{1}{\sqrt{2}}\),whereas inverse trigonometric functions take a ratio and output an angle, e.g. \( tan^{-1}\frac{1}{\sqrt{3}}=30°\). To use inverse functions, you can find an angle \(θ\) by substituting the appropriate trigonometric ratio, for example \(θ=\tan^{-1} \ \frac{opposite}{adjacent}\):
Example 2: Find the unknown angle.
Solution 2:
\begin{align*}
\tanθ &= \frac{opposite}{adjacent}\\
&=\frac{10}{10}\\
θ&=\tan^{-1}\frac{10}{10}\\
&=45°\\
\end{align*}
In addition to right-angled triangles, trigonometry can be applied to triangles more generally to find the area, angle size and side lengths of triangles when they meet certain conditions. There are three general trigonometric rules you need to remember:
Using the area rule, it is possible to find the area of any triangle given the length of two adjacent sides and the internal angle created by them. The area of a triangle is given by the formula:
\(A=\frac{1}{2} ab \ \sinC\)
Where \(a\) and \(b\) are two sides of a triangle and \(C\) is the opposing angle:
Example 1: Find the Area of the following triangle
Solution 1:
\begin{align*}
A&=\frac{1}{2} \times 7 \times 4 \times \sin70\\
&=13.2 \ cm^2\\
\end{align*}
The sine rule helps us to solve questions involving pairs of opposite sides and angles. In the following diagram angle \(A\) is opposite side \(a\), angle \(B\) is opposite side \(b\) and angle \(C\) is opposite side \(c\).
The sine rule states that:
\(\frac{a}{\sinA} =\frac{b}{\sinB} =\frac{c}{\sinC}\)
If we are investigating two of these pairs of opposite angles and sides, we are able to solve for one of the angles or sides if we know the other three.
Example 2: Find \(x\) in the following triangle
Solution 2:
\begin{align*}
\frac{x}{\sin30} &=\frac{8}{\sin70}\\
x&=\frac{8 \sin30}{\sin70}\\
&=4.3 \ cm\\\
\end{align*}
The cosine rule helps us to solve questions in another case. Using the same triangle as for the sine rule, the cosine rule is:
\(c^2=a^2+b^2-2ab \ \cosC\)
Example 3: Find \(x\) in the following diagram, correct to 1 d.p.
Solution 3:
| \begin{align*} c^2&=a^2+b^2-2ab \ \cosC\\ 12^2&=9^2+x^2 -2 \times 9 \times \ \cos110\\ x^2+18 \ \cos110 x-63&=0\\ x&= \frac{-6.16+\sqrt{6.16^2+4\times63 }}{2}\\ &=11.6\\ \end{align*} |
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The basic idea of angles of elevation and depression is measuring the angle between one’s horizontal line of sight to their line of sight to an object above or below the horizontal.
If a person is looking up at an object, the angle between the person’s horizontal line of sight and the object the angle of elevation.
If said person is now looking down at an object, the angle between the person’s horizontal line of sight and the object the angle of depression.
To find the angle of elevation or depression, trigonometry can be applied utilising the regular trigonometric ratios. Questions involving angles of elevation or depression are usually embedded in wordy situation problems.
Example 1:
Jack is standing at the top of a cliff 60 metres above sea level looking down at a sailing boat. The distance between Jack and the boat is 125 metres. At what angle of depression is Jack looking at the boat? (Round to the nearest minute)
Solution 1:
| \begin{align*} <ABJ&= θ (\text{alternate angles in parallel lines})\\ sin(<ABJ)&= \frac{60}{125}\\ <ABJ&= \sin^{-1} \frac{60}{125}=28°41’\\ ∴θ&=28°41’\\ \end{align*} |
We can apply the same kind of logic to questions involving angles of elevation.
Trigonometry can be used for navigation. The angular measurements used for indicating directions are called bearings. There are two ways to measure bearings, and they are known as compass bearings and true bearings.
To determine the compass bearing of an object/location of interest with respect to a reference point in a horizontal plane, we could start by drawing a vertical line and a horizontal line with the reference point as their point of intersection.
On the other hand, to determine the true bearing of an object/location of interest with respect to a reference point in a horizontal plane, we simply measure the angle from North in the clockwise direction. Then we just write down the bearing in the form \(XXX°T\) e.g. \(188°T\).
We also need to be familiar with some common terminology used for indicating directions – “north” means \(0°T\), “north-east” means \(45°T\), “east” means \(90°T\), “south-east” means \(135°T\), “south” means \(180°T\), “south-west” means \(225°T\), “west” means \(270°T\), and “north-west” means \(315°T\).
Example 1: Write down the compass bearing and the true bearing from O to A as shown in the diagram below
Solution 1:
\(A=S55°W\) or \(235°T\)
Navigation problems are an application of trigonometric relations and bearings. They consist of some given locations, bearings, and distances between the locations. Typical exam questions describe an object or person that starts from an initial location and travels a certain distance in a given direction, then changes its direction and travels further until it reaches a second location, and so on. Examples of questions that could be asked include the bearing of the final position from the original one, the distance between the original and final positions, or the total distance travelled.
Example 2: Starting from location A, Jack walks 2 metres south-west to location B, and then 10 metres west to location C. What is the distance between locations A and C?
Solution 2:
From point \(B\), we draw a line \(BD\) that is parallel to \(AN\). Since Jack walks southwest from \(A\) to \(B\), this means that \(∠BAN=45°\). Since the lines \(BD\) and \(AN\) are parallel, \(∠BAN\) and \(∠ABD\) are equal (alternate angles), so \(∠ABD=45°\). Note that \(∠DBC=90°\). This means that
| \( ∠ABC = ∠ABD+∠DBC=45°+90°=135°\) |
To find \(AC\), we can use the cosine rule, which gives :
| \begin{align*} AC^2&=AB^2+BC^2-2 \times AB \times BC \cos∠ABC\\ AC^2&=2^2+10^2-2 \times 2 \times 10 \times \cos135°\\ AC&=11.5\\ \end{align*} |
Therefore, the distance between \(A\) and \(C\) is \(11.5\) metres.
And that’s the Trigonometry part of the Year 12 Standard 2 Syllabus! We’ve also prepared some concept Check Questions for you to try out below!
1. Find \(x\) and \(θ\):
2. If the following triangle has an area of \(12 \ cm^2\), find the length of side \(AC\).
3. A plane \(200\)m in the air is spotted by a child looking up at \(54°\). How far away is the child from the plane at that exact point in time? (Round to the nearest metre)
4. Town \(B \) is at a bearing of \(N20°E\) and \(30\) kilometres from Town \(A\). Town \(C\) is at a bearing of \(S75°E\) and \(50\) kilometres from Town \(B\). Find the distance and bearing of Town \(C\) from Town \(A\).
1. Find \(x\) and \(θ\):
\begin{align*}
\sin30°&=\frac{x}{5}\\
x&=5 \sin30°\\
x&=2.5\\
\cosθ&=\frac{2.5}{7}\\
θ&=cos^{-1} \ \frac{2.5}{7}\\
&=69.08°\\
\end{align*}
2. If the following triangle has an area of \(12 \ cm^2\), find the length of side \(AC\).
Using the area formula:
\begin{align*}
12&=\frac{1}{2} \times 8 \times BC \times \ \sin30° \\
BC&=\frac{12}{4 \sin30°}\\
&=6\\
\end{align*}
Now we use the cosine rule to find the length of side \(AC\):
\begin{align*}
AC^2&=BC^2+AB^2-2AB BC \ \cosB\\
&=36+64-483AC\\
&=4.1cm\\
\end{align*}
3. Let \(C\) be the position of the child and \(P\) be the position of the plane.
\begin{align*}
\sin54°&=\frac{200}{CP}\\
CP&=\frac{200}{\sin54°} = 618 \ m\\
\end{align*}
4. We start with the sketch below (\(D\) and \(F\) are not part of the information given; we have labelled them just to make it easier to refer to angles in the sketch).
Since \(AD\) and \(BF\) are parallel, \(∠ABF=∠BAD=20°\) . Notice that
| \(∠ABC=∠ABF+∠CBF=20°+75°=95°\) |
We can find \(AC\) using the cosine rule:
| \begin{align*} AC^2 &= AB^2 + BC^2 – 2AB BC cos∠ABC \\ AC^2&= 30^2+ 50^2- 2 \times 30 \times 50 cos(95°)\\ AC&=60.51 \ km\\ \end{align*} |
Before finding the bearing of Town \(C\) from Town \(A\), we need to find angle \(BAC\). Applying the cosine rule again gives
| \begin{align*} BC^2&= AB^2+ AC^2- 2AB AC \cos∠BAC \\ 50^2 &= 30^2 + 60.51^2 – 2 \times 30 \times 60.5101 cos∠ BAC\\ ∠BAC&=55.4°\\ \end{align*} |
This means that
| \(∠ DAC =∠ DAB +∠ BAC = 20 ° + 55.40 ° = 75.40 °\) |
so the bearing of Town \(C \) from Town \(A\) is \(N75°E\).
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