Topics 5 Revision Notes
Trigonometry & Geometry
IB Math Studies
Topic 5.1 Midpoint and Distance
Two useful formulae that you are given with regards to coordinate geometry are:
Midpoint
The midpoint M of the line segment
joining A(x1, y1) and B(x2, y2) is
 x1  x2 y1  y2 
,


2 
 2
Distance
The distance, d, between points
A(x1, y1) and B(x2, y2) is
d
 y2  y1    x2  x1 
2
2
Topic 5.2 , Linear Functions
Slope-intercept form of a line
Standard form of a line
Formula for slope
y  mx  b
ax + by + d = 0 where a, b and d
are integers, and a > 0.
y y
m 2 1
x2  x1
Parallel lines have slopes that are equal. For example, y 
2
2
x  5 and y  x  4 are parallel.
3
3
Perpendicular lines have slopes that are opposite reciprocals. For example, y 
perpendicular. Also y  x  5 and y   x  1 are perpendicular.
2
3
x  5 and y   x  1 are
3
2
Finding the intersection between two lines
To find the point of intersection between two lines there are 3 popular methods:
1. Put each equation in slope-intercept form
Graphing
2. Graph each equation
3. Determine where they intersect.
1. Solve one of the equations for one of the variables.
2. Substitute into the other equation
Substitution
3. Solve
4. Substitute back to solve for the remaining variable.
1. Choose a variable to eliminate
2. Make coefficients opposite numbers by multiplying
Elimination
3. Add the equations
4. Substitute back to solve for the remaining variable.
Writing the equation of a line
To write the equation of a line you need one of the following:
The slope and a point
Two points
1. Find the slope m using the slope formula
1. Start with y  mx  b
2. Starting with y  mx  b
2. Substitute the point for x and y
3. Substitute the slope for m
3. Substitute the point for x and y
4. Solve to find b
4. Substitute the slope for m
5. Solve to find b
5. Write the answer in the form y  mx  b
6. Write the answer in the form y  mx  b
1
Topics 5 Revision Notes
Trigonometry & Geometry
IB Math Studies
Converting to Standard Form
Often on the IB exam they will ask you to leave the answer in standard form. They typically don’t use the
phrase “standard form” and instead will say, “write your equation in the form ax + by + d = 0 where a, b and
d are integers, and a > 0.” To do this you should
1. Write the equation in slope-intercept form y  mx  b
2. If necessary multiply the entire equation by the common denominator to eliminate any fractions.
3. Put all terms on one side of the equals side so that it is equal to zero.
4. If necessary, multiply the entire equation by -1 in order that the coefficient in front of x is positive.
Topic 5.3 Right-angled triangles
Pythagorean Theorem
You need to be able to calculate missing sides in right angled triangles, given two sides by using Pythagoras’
theorem.
c2 = a2 + b2
b
*For all questions make sure you
are in Degree MODE.
c
a
opposite
SOHCAHTOA
If you need to find missing angles and missing sides in right triangles use SOHCAHTOA (Sin Opposite
Hypotenuse, Cosine Adjacent Hypotenuse, Tangent Opposite Adjacent)
opposite
hypotenuse
adjacent
cos 
hypotenuse
opposite
tan  
adjacent
sin  
hypotenuse
θ
adjacent
Topic 5.4 Non-right angled triangles**
Formulae for non-right angled triangles are based on the diagram and notation below.
B
a
c
A
b
C
For the exam you should be able to apply the sine rule, cosine rule, and the area of a triangle. All of these
formulae are in the IB formula booklet.
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Topics 5 Revision Notes
Trigonometry & Geometry
IB Math Studies
The sine rule (non-right angled triangle)
The sine rule is used when:
• You are given two sides and an angle opposite one of the sides. (SSA → this is the ambiguous case)
• You are given two angles and a side and you are missing a side. (AAS, ASA)
a
b
c


sin A sin B sin C
Guided example
Find the missing side in the triangle below.
7
x
46°
25°
ANSWER:
7
x

sin 25 sin 46
7 sin 46
x
sin 25
x = 11.9
The cosine rule (non-right angled triangle)
The cosine rule is used when:
• You are given three sides and you are missing an angle. (SSS)
• You are given two sides and the included angle. (SAS)
Missing side:
a2 = b2 + c2 – 2bc cosA
Missing angle:
b2  c2  a2
cos A 
2bc
Guided example
Find the missing angle in the triangle below
7
θ
8
10
ANSWER
cos  
8 2  7 2  10 2
13
 13 
  cos  
   cos 1 
  θ = 83°
112
2(8)(7)
 112 
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Topics 5 Revision Notes
Trigonometry & Geometry
IB Math Studies
Area of a triangle (non-right angled triangle)
The area of a triangle can be found if two sides are known and the angle between them is given. The formula is
given below.
Area of a triangle =
1
ab sin C
2
Guided example
Find the area of the triangle below.
9
40°
15
ANSWER
1
(9)(15) sin 40 = 43.4 units2
2
Bearings
Bearings are always 3 figure angles. The basic rules for finding a bearing are:
 Locate where you ‘are’. This will be the location that follows from. E.g. If you are asked to find the
bearing of Bangkok from Columbo, you will be at Columbo.
 Draw a line between the two locations.
 Draw a line going north from where you are.
 Draw an arc starting at the line going north and stopping when you reach the destination line.
 The angle of the arc is the bearing you need. Always give 3 figures in your final answer, so if the angle
measure 35º, the bearing will be 035º.
Guided example
Two ships set sail from the port of Palermo. One sails to Cagliari a distance of 390 km on a bearing of 285º.
The other ship sails to Naples on a bearing of 010º and a distance of 320 km.
(a) Draw a diagram to show the information given above.
(b) Use you diagram to find the area in between the towns of Palermo, Cagliari, and Naples.
(c) Find the bearing and distance from Cagliari to Naples.
Answer (a)
Answer (b)
As the triangle is not a right-angled triangle we must use the formula: area 
area 
1
ab sin C
2
1
(390)(320) sin(85)  62200 km2
2
Answer (c)
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Topics 5 Revision Notes
Trigonometry & Geometry
IB Math Studies
The direct distance between C and N can be calculated using the cosine rule:
CN 2  3202  3902  2(320)(390) cos85
CN  482 km
The bearing is slightly harder to find. Look at the diagram below and the bearing is marked.
The angle 75º is calculated using the fact that the two lines north (from P and from C) are
parallel and therefore alternate interior angles are equal. The bearing will then be equal to 180°
– (75° + θ).
The angle θ can be calculated using the sine rule:
320
482

sin  sin 85
320  sin 85
sin  
482
sin   0.6613...
  sin 1 (0.6613...
  41.4
The bearing is therefore 180° – (75° +41.4°)  63.6º
Topic 5.5 3-Dimenstional Geometry
3-d trigonometry basically involves breaking down the 3-d shape into 2-dimension right triangles and using
either the Pythagorean Theorem or SOHCAHTOA.
Examples of this can be seen in the worked example below.
Guided example
In the diagram below AB = 4 cm, BC = 5 cm, and CG = 7 cm.
(a) Calculate the length of AC.
(b) Calculate the length AG.
(c) Find the angle made between the base of the cuboid ABCD and the line AG.
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Topics 5 Revision Notes
Trigonometry & Geometry
IB Math Studies
Answer (a)
By drawing the right triangle on the floor, triangle ABC, we can see that a simple use of the
Pythagorean Theorem will result in AC  42  52  41  6.40. I suggest leaving the value as
41 for the remainder of the calculations.
Answer (b)
By drawing the right triangle ACG we can see that again using the Pythagorean Theorem:
2
AG2  41  72
AG  9.49 .
Answer (c)
To find the angle made between the base of the cuboid ABCD and the line AG. (Really they are
asking for angle GAC). Use the same right triangle ACG and calculate the value of θ by using
tangent function from SOHCAHTOA.
7
41
 7 
  tan 1 

 41 
  47.5
tan  
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Topics 5 Revision Notes
Trigonometry & Geometry
IB Math Studies
Guided Example
In the diagram below, PQRS is the square base of a solid right pyramid with vertex V. The sides of
the square are 8 cm, and the height VG is 12 cm. M is the midpoint of [QR].
Diagram not to scale
V
VG = 12 cm
P
Q
G
(a)
(b)
(i)
(ii)
Find
(i)
(ii)
Answer (a) (i)
(ii)
8 cm
M
S
R
8 cm
Write down the length of [GM].
Calculate the length of [VM].
the total surface area of the pyramid;
the angle between the face VQR and the base of the pyramid.
GM = 4 cm
VM2 = 42 + 122
= 16 + 144
= 160
VM = 160 = 12.6 cm
Answer (b) (i)
Surface Area = area of square base + 4 (area of triangular face)
1
= 8 × 8 + 4 × × 8 × 160
2
= 64 + 202.4
= 266 cm2
Answer (b) (ii)
V
160
12
x
G
4
M
12
4
x = 71.6°
tan x =
7