This module builds on: M2 on Pythagoras' theorem and M3 on Trigonometry.
Trigonometry in 3 dimensions
It may be necessary to use Pythagoras' theorem and trigonometry to solve a problem.
The trigonometric ratios can be used to solve 3-dimensional problems which involve calculating a length or an angle in a right-angled triangle.
Example
The shape ABCDEFGH is a cuboid.
Length AB is 6 cm, length BG is 3 cm and length FG is 2 cm.
The length of the diagonal AF is 7 cm.
Calculate the angle between AF and the plane ABCD. Give the answer to 3 significant figures.
The plane ABCD is the base of the cuboid. The line FC and the PlaneA flat, two-dimensional surface. ABCD form a right angle.
Draw the right-angled triangle AFC and label the sides. The angle between AF and the plane is \(x\).
Use: \(\sin{x} = \frac{o}{h}\)
\(\sin{x} = \frac{3}{7}\)
\(\sin{x} = 0.428571 \dotsc\) – do not round this answer yet.
To calculate the angle use the inverse sin button on the calculator \(\text{(}\sin^{-1}\text{)}\)
\(x = 25.4^\circ\)
Question
The shape ABCDV is a square-based pyramidA base with triangular faces meeting at a a point called an apex.. O is the midpoint of the square base ABCD.
Lengths AD, DC, BC and AB are all 4 cm.
The perpendicularIf the angle between two lines is a right angle, the lines are said to be perpendicular.height of the pyramid (OV) is 3 cm.
Calculate the angle between VC and the plane ABCD. Give the answer to 3 significant figures.
Example
The plane ABCD is the base of the pyramid. The line VO and the plane ABCD form a right angle.
Draw the right-angled triangle OVC and label the sides. The angle between VC and the plane is (\(y\))
It is not possible to use trigonometry to calculate the angle because the length of another side is required.
Pythagoras can be used to calculate the length OC.
Draw the right-angled triangle ACD and label the sides.
To find hypotenuse (Pythagoras) \( a^2 + b^2 = c^2\)
\( \text{CD}^2 + \text{AD}^2 = \text{AC}^2\)
\( 4^2 + 4^2 = c^2\)
\( 32 = c^2\)
\( c = \sqrt{32}\)
\( \sqrt{32}\) – do not round this answer yet.
The length AC is \( \sqrt{32}\)
The point O is in the centre of the length AC so OC is half of the length AC.
The length OD is \( \frac{\sqrt{32}}{2}\) cm.
Use \(\tan{x} = \frac{o}{a}\)
\(\tan{x} = \frac{3}{\frac{\sqrt{32}}{2}}\)
\(\tan{x} = 1.06066 \dotsc\) – do not round this answer yet.
To calculate the angle use the inverse tan button on the calculator \(\text{(}\tan^{-1}\text{)}\)
\(x = 46.7^\circ\)
The sine rule
The angles are labelled with capital letters. The opposite sides are labelled with lower case letters. Notice that an angle and its opposite side are the same letter.
The sine rule is: \(\frac{a}{\sin{A}} = \frac{b}{\sin{B}} = \frac{c}{\sin{C}}\)
This version is used to calculate lengths.
It can be rearranged to: \( \frac{\sin{A}}{a} = \frac{\sin{B}}{b} = \frac{\sin{C}}{c}\)
This version is used to calculate angles.
Example
Calculate the angle PRQ. Give the answer to three significant figures.
Use the form \(\frac{\sin{A}}{a} = \frac{\sin{B}}{b} = \frac{\sin{C}}{c}\) to calculate the angle.
\(\frac{\sin{x}}{4} = \frac{\sin{75}}{9}\).
\(\sin{x} = 0.429300 \dotsc\) – do not round this answer yet.
To calculate the angle use the inverse sin button on the calculator
\(x = 25.4^\circ\).
Question
Calculate the length QR. Give the answer to three significant figures.
Use the form \(\frac{a}{\sin{A}} = \frac{b}{\sin{B}} = \frac{c}{\sin{C}}\) to calculate the length.
\(\frac{\text{QR}}{\sin{53}} = \frac{7.1}{\sin{105}}\)
\(\text{QR} = \frac{7.1 \times \sin{53}}{\sin{105}}\)
QR = 5.87 cm
The cosine rule
The cosine rule is: \(a^2 = b^2 + c^2 - 2bc \cos{A}\).
This version is used to calculate lengths. \(\cos{A} = \frac{b^2 + c^2 - a^2}{2bc}\)
It can be rearranged to: \(\cos{A} = \frac{b^2 + c^2 - a^2}{2bc}\)
This version is used to calculate angles.
Example
Calculate the length BC. Give the answer to three significant figures.
Use the form: \( a^2 = b^2 + c^2 - 2bc \cos{A}\) to calculate the length.
\( \text{BC}^2 = 3^2 + 7^2 - 2 \times 3 \times 7 \cos{35}\)
\( \text{BC}^2 = 23.59561414 \dotsc\) – do not round this answer yet.
BC = 4.86 cm
Question
Calculate the angle QPR. Give the answer to three significant figures.
Use the form: \(\cos{A} = \frac{b^2 + c^2 - a^2}{2bc}\) to calculate the angle.
\(\cos{y} = \frac{4^2 + 6.9^2 - 4.2^2}{2 \times 4 \times 6.9}\)
\(\cos{y} = 0.8327898 \dotsc\) Do not round this answer yet.
To calculate the angle use the inverse cos button on the calculator \(\text{(}\cos^{-1}\text{)}\).
(\(y\)) = 33.6°
Solving problems using the sine and cosine rule
To solve problems involving non-right-angled triangles, the correct rule must first be chosen.
The cosine rule can be used in any triangle to calculate:
- a side when two sides and the angle in between them are known
- an angle when three sides are known
The sine rule can be used in any triangle to calculate:
- a side when two angles and an opposite side are known
- an angle when two sides and an opposite angle are known
Example
Ship A leaves port P and travels on a bearing A direction measured in degrees clockwise from North. Due North is 000, due East is 090, due South is 180 and due West is 270. of 200°. A second ship B leaves the same port P and travels on a bearing of 165°. After half an hour ship A has travelled 5.2 km and ship B has travelled 5.8 km. How far apart are the ships after half an hour? Give the answer to three significant figures.
Remember a bearing is an angle measured clockwise from north. The angle APB = \(200 - 165 = 35^\circ\).
Two sides and the angle in between are known. Calculate the length AB.
Use the cosine rule.
\(a^2 = b^2 + c^2 - 2bc \cos{A}\).
\( text{AB}^2 = 5.2^2 + 5.8^2-2 \times 5.2 \times 5.8 \cos{35}\).
\(\text{AB}^2 = 11.26874869\).
AB = 3.36 km
The area of a triangle
The area of any triangle can be calculated using the formula:
\(\text{Area of a triangle} = \frac{1}{2} ab \sin{C}\).
To calculate the area of any triangle the lengths of two sides and the angle in between are required.
Example
Calculate the area of the triangle. Give the answer to 3 significant figures.
Use the formula:
\(\text{area of a triangle} = \frac{1}{2} bc \sin{A}\)
\(\text{area} = \frac{1}{2} \times 7.1 \times 5.2 \sin{42}\)
area = \(12.4 \text{cm}^{2}\)
It may be necessary to rearrange the formula if the area of the triangle is given and a length or an angle is to be calculated.
Question
The area of the triangle is \(5.45 \text{cm}^{2}\). Calculate the size of the angle YXZ. Give the answer to the nearest degree.
Use the formula:
\(\text{area of a triangle} = \frac{1}{2} ab \sin{C}\)
\(5.45 = \frac{1}{2} \times 5.3 \times 3.2 \sin{C}\)
\(5.45 = 8.48 \sin{C}\)
Rearrange the equation to make the subject.
Divide both sides by 8.48.
\(\sin{C} = 0.642688 \dotsc\) –do not round this answer yet.
To calculate the angle use the inverse sin button on the calculator \(\text{(}\sin^{-1}\text{)}\).
C = 40°
The angle YXZ is 40°.
Test yourself
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