Task 2 : Types of triangles
Exploration 1
State and draw types of triangle which you have learnt. You are encouraged to use ICT.
Answer:
1) Equilateral Triangle:
All 3 sides have equal lengths.
All 3 angles are equal which is 60°.
2) Isosceles Triangle:
2 Sides have equal lengths.
2 of its angles are also equal.
3) Acute Triangle:
All angles are less than 90°.
4) Right Triangle:
It has one right angle, 90°.
5) Obtuse Triangle:
It has one angle more than 90°.
c
b a
BA
C
h
Exploration 2
1) By using the triangle diagram below, prove the “Sine Rule”.
D
Answer:
Proof of Sine Rule: asin A =
bsin B
In triangle CAD: sin A = h b
This rearranges to give h = b sin A
In triangle CBD: sin B = h a
This rearranges to give h = a sin B
When h = b sin A and h = a sin B, then b sin A = a sin B
This rearranges to give asin A =
bsin B
2) State when the “sine rule” should be used and include diagram in your explanation.
Answer:
The sine rule
Sine rule can be used to find the angle or the length of sides of a triangle.
The sine rule are:
asin A =
bsin B = csin C or
sin Aa =
sin Bb = sin C
c
When to use it?
a) Two sides and an angle opposite to one of the two sides are given.
b) One side and any two angles are given.
3) Hence state the relation between included angle and non-included angle in a triangle.
Illustrate.
Answer:
DEFINATION OF INCLUDED ANGLE
i. An included angle is the angle
between two sides of a triangle.
ii. It can be any angle of the triangle,
depending on its purpose.
iii. The included angle is used in proofs
of geometric theorems dealing
with congruent triangles.
iv. Congruent triangles are two triangles whose sides and angles are equal to each
other.
v. You can also use the included angle to determine the area of any triangle as long
as you know the lengths of the sides surrounding the angle.
DEFINATION OF NON - INCLUDED ANGLE
i. A non – included angle is the angle which is not between two sides of a triangle.
ii. Sine rule can be used to solve the problems regarding this triangle.
non - included angle
a
b
6 cm
P
Q
R
30°
110°
Task 3: Sine Rule
Case 1
Based on the diagram above, explain how a sine rule can be applied to calculate the length of
RQ. Hence, find the length of RQ.
Answer:
PQsin R = RQ
sin P
6 sin 30° =
RQsin 110°
6 sin 110° = RQ sin 30°
RQ = 6 sin 110°sin 30°
RQ = 11.28 cm
Case 2
Tools needed
Ruler, Protractor and Compass.
By using the geometry tools above , construct a triangle with whose two sides are 8 cm and 10
cm with a non-included angle of 520 on the 8 cm side in a A4 paper. Hence, find the opposite
angle of the side of 8 cm.
Answer:
Exploration 3
Using the diagram of a triangle, explain the sine rule which involves the ambiguous case.
Answer:
The Ambiguous case
When using the rule of sines to find a side of a triangle, an ambiguous case occurs when
two separate triangles can be constructed from the data provided (i.e., there are two different
possible solutions to the triangle). In the case shown below they are triangles ABC and AB'C'.
Given a general triangle the following conditions would need to be fulfilled for the case
to be ambiguous:
The only information known about the triangle is the angle A and the sides a and c.
The angle A is acute (i.e., A < 90°).
The side a is shorter than the side c (i.e., a < c).
The side a is longer than the altitude h from angle B, where h = c sin A (i.e., a > h).
If all the above conditions are true, then both angles C or C' produce a valid triangle;
meaning both of the following are true:
From there we can find the corresponding B and b or B' and b' if required, where b is the
side bounded by angles A and C and b' bounded by A and C' . Without further information it is
impossible to decide which is the triangle being asked for.
Example 1
In , a = 20, c = 16, and m<A = 30º. How many distinct triangles can be drawn given these
measurements?
Use the sine rule:
C = sin-1 (0.4) = 24º (to the nearest degree) - in Quadrant I.
Sine is also positive in Quadrant II. If we use the reference angle 24º in Quadrant II, the angle C
is 156º.
But, with m<A = 30º and m<C = 156º the sum of the angles would exceed 180º.
Not possible!!!!
Therefore, m<C = 24º, m<A = 30º, and m<B = 126º and only ONE triangle is possible.
Example 2
In , a = 7, c = 16, and m<A = 30º. How many distinct triangles can be drawn given these
measurements?
Use the sine rule:
Since sin C must be < 1, no angle exists for angle C.
NO triangle exists for these measurements.
Example 3
In , a = 10, b = 16, and m<A = 30º. How many distinct triangles can be drawn given these
measurements?
Use the sine rule:
B = sin-1(.8) = 53.13010 = 53º.
Angles could be 30º, 53º, and 97º : sum 180º
The angle from Quadrant II could create angles 30º, 127º, and 23º : sum 180º
TWO triangles possible.
Two triangles are possible
This example is the Ambiguous Case. The information given is the postulate SSA (or
ASS, the Donkey Theorem), but the two triangles that were created are clearly not congruent.
We have two triangles with two sides and the non-included angle congruent, but the triangles are
not congruent to each other.
Case 1
GivenPQ = 6 cm, QR = 4 cm, ∠QPR = 30° . Solve PQR.
Answer:
Side/Angle cm / ° Side/Angle cm / °
PQ 6 cm PQ 6 cm
QR 4 cm QR’ 4 cm
PR 7.842 cm PR’ 2.550 cm
∠P 30° ∠ P 30°
∠Q 101.41° ∠Q 18.59°
∠R 48.59° ∠R’ 131.41°
c
b a
BA
C
h
Task 4: Cosine Rule
Exploration 4
1) By using the triangle diagram below, prove the “Cosine Rule”.
Answer:
Proof of Cosine Rule: a2 = b2 + c2 – 2bc cos A
In triangle CAD: sin A = h b
This rearranges to give h = b sin A
In triangle CAD: cos A = x b
This rearranges to give x = b cos A
Using the Pythagorean Theorem in triangle CBD, we have: a2 = h2 + (c – x)2
When h = b sin A and x = b cos A, then a2 = (b sin A)2 + (c – b cos A)2
a2 = b2 sin2 A + c2 – 2bc cos A + b2 cos2 A
a2 = b2 (sin2 A + cos2 A) + c2 – 2bc cos A
Basic identity shows that sin2 A + cos2 A = 1, so a2 = b2 (1) + c2 – 2bc cos A
c - xx D
This rearranges to give a2 = b2 + c2 – 2bc cos A
This same process could be used to produce other lettered statement of this rule.
b2 = a2 + c2 – 2ac cos B
c2 = a2 + b2 – 2ab cos C
2) State when the cosine rule should be applied. Include diagram(s) in your explanation.
Answer:
The cosine rule
Cosine rule can be used to find the angle or the length of sides of a triangle.
The cosine rule are:
a2 = b2 + c2 – 2bc cos A
b2 = a2 + c2 – 2ac cos B
c2 = a2 + b2 – 2ab cos C
When to use it?
a) Two sides and an angle in a triangle are given
13 cm
10 cmC
B
A 50°
b) Three sides without any angle in a triangle are given.
Case 1
The diagram shows triangle ABC. Calculate the length of BC.
Answer:
a2 = b2 + c2 – 2bc cos A
BC2 = AB2 + AC2 – 2(AB)(AC) cos A
BC2 = 132 + 102 – 2(13)(10) cos 50°
BC = √269−260 cos50 °
BC = 10.09 cm
B
A
C
7 cm
9 cm
6 cm
Case 2
The diagram above shows triangle ABC. Find∠BAC .
Answer:
a2 = b2 + c2 – 2bc cos A
BC2 = AB2 + AC2 – 2(AB)(AC) cos A
62 = 92 + 72 – 2(9)(7) cos A
36 = 130 – 126 cos A
cos A = 94126
∠A = cos-1 94126
∠A = 41.75°
∠BAC = 41.75°
Case 3
The diagram above shows triangle PQR . Find the length of QR.
Answer:
PRsin Q =
PQsin R
18sin Q = 10
sin 22°
sin Q = 18 sin 22°10
∠Q = 42.40°
When ∠Q = 42.40°, then ∠P = 180° - (42.40° + 22°)
∠P = 115.60°
QRsin P = PQ
sin R
QRsin 115.60° =
10sin 22°
QR = 24.07 cm
P
Q
R
10cm
18cm
22°
Task 5: Area of Triangle
Exploration 5
With the help of diagram(s) state and give examples of the use of the formula area of a triangle
with the characteristics involved.
Answer:
The formula area of triangle
1) The basic formula:
12
× b × h
This formula can be used when we know the base and height of the triangle.
Example:
What is the area of this triangle?
(Note: 12 is the height, not the length of the left-hand side)
Height = h = 12
Base = b = 20
Area = ½ bh = ½ × 20 × 12 = 120 cm2
2) The trigonometry based formula:
Area = 12 ab sin C
This formula can be used when we know two sides and the included angle of the
triangle.
Depending on which sides and angles we know, the formula can be written in three
ways.
12 ab sin C or
12 bc sin A or
12 ac sin B
Example:
Find the area of this triangle:
We know angle C = 25º, and sides a = 7 and b = 10.
Start with : Area = (½)ab sin C
Put in the values we know : Area = ½ × 7 × 10 × sin(25º)
Do some calculator work : Area = 35 × 0.4226
Area = 14.79 cm2
How does it work:
Well, we know that we can find an area when we know a base and height:
Area = ½ × base × height
In this triangle:
the base is: c
the height is: b × sin A
Putting that together gets us:
Area = ½ × (c) × (b × sin A)
Which is (more simply):
Area =1
bc sin A2
7 cm
5cm
45°
B
A
C
B
A
C
5.4cm9.2cm
Case 1
Find the area of triangle ABC above.
Answer:
Area = 12 ac sin B
= 12 (5)(7) sin 45°
= 12.37 cm 2
Case2
Find the area of triangle ABC above
12.8cm
Answer:
a2 = b2 + c2 – 2bc cos A
BC2 = AB2 + AC2 – 2(AB)(AC) cos A
(12.8)2 = (5.4)2 + (9.2)2 – 2(9.2)(5.4) cos A
cos A = [(5.4)2 + (9.2)2 – (12.8)2] ÷ [2(9.2)(5.4)]
A = 120.24°
Area = 12 ac sin B
= 12 (AC)(AB) sin A
= 12 (5.4)(9.2) sin 120.24°
= 21.46cm 2
9 km
8km
6 km
Task 6: HOTS challenge
Advanced Exploration
Kampung Murni
Uncle Ghani who lives in Kampung Murni has a piece of land measuring just like the picture
above. The local government wants to make the acquisition of a land to build a road connecting
Kampung Murni to the city. Uncle Ghani’s land which adjacented triangle-shape is beside a
rubber factory is 34km2 and acute angled at the three- way intersection.
(i) Find the length of Uncle Ghani’s land which the government wants to acquire.
(ii) If the local government wants to acquire the land of width 20 m, calculate the
remaining of Uncle Ghani’s land.
(iii) If he wants to build a new fence around his land with a radius of 7 km, find the length
of the fence that he has to buy.
(iv) If the area is to be planted with corn, estimated the number of seeds that is needed if
300 seeds is for300m2.
Land Acquisition Plan
Rubber Factory
7km
City
Appreciation
After tiring days and a lot of efforts, I finished this Additional Mathematics Project Work
2015. This project work is interesting and it is related to the concept of Mathematics in our daily
life. First and foremost, I would like to thank my parents for providing everything such as money
to buy anything that are related to this project work. They had also advised me and provided me
with facilities which are very important for this project work such as internet, computer and
books. They also supported and encouraged me to complete this task so that I will not
procrastinate in doing it.
Secondly, I would like to thank my Additional Mathematics teacher, Puan Ruhil Hayati
Binti Abd. Ghani for giving me guidance and support throughout this project. Even I had some
difficulties in completing this task, but she patiently taught me until I managed to understand
about this project work. She is also very kind – hearted until she helped me to do this project
work until late in the evening at school.
Lastly, I would like to thank my friends especially my classmates who are always
supporting me. Even this project is an individually project but we cooperated between each other
to complete this project. Sometimes, we stayed back after the school hours to fulfill this task. We
had also organized a few discussions at library to share ideas among us about this project work. I
would like to thank to god for giving me friends like them and hopefully our friendship will last
long forever. I am also would like to thank everyone who has involved directly or indirectly in
completing this project work.
Objective
We students taking Additional Mathematics are required to carry out a project work
while we are in Form 5. Upon completing this Additional Mathematics Project Work, we are
able to gain valuable experiences and able to :-
Apply Mathematics to everyday situation.
Develop positive attitudes towards Mathematics.
Improve high order think skills.
Promote effective mathematical communication.
Appreciate the importance and beauty of Mathematics in everyday life.
Provide learning environment that stimulates and enhances effective learning.
Develop mathematical knowledge through problem solving in a way that increases
students’ interests and confidence.
Prepare ourselves for the demand of our future undertakings and in workplace.
Train ourselves not only to independent learners but also to collaborate and to
cooperate and to share knowledge in an engaging and healthy environment.
Use technology especially ICT appropriately and effectively.
Improve computer skills and handling the geometrical apparatus skills.
Improve the skills of drawing various types of triangles.
Conclusion
After doing research, discussion with friends, answering question, drawing triangles and
some problem solving, I saw that the usage of solution of triangle is important in daily life. It is
not only used by students in primary schools, secondary schools and higher level of studies, but
also widely used in construction field. Engineers used the concept of solution of triangle to build
buildings and bridges in the shape of triangle. For example the middle part of the Penang bridge
was built in a shape of triangle. Solution of triangle concept is used to determine the length and
angle of the triangle to build a strong and stable bridge. Without it, sure a strong bridge could not
be built. So we should be thankful to the people who contribute the idea of solution involving
triangles.
Reflection
After spending countless hours, days and nights to finish this project and also sacrificing
my time for video games and magazines in this holiday, there are a few things that I would like
to say …
Oh Additional Mathematics…
From the day I born,
From the day I was able to hold pencil,
From the day I start learning,
And…
From the day I heard your name,
I always thought you will be my great obstacle and rival in excelling…
Sometimes I regret of learning you,
But now… I was totally wrong,
I realized something really important in you,
I know everything just because of you…
I really love you, and
You are my soul mate…
Thank you Additional Mathematics.
I ADDITIONAL MATHEMATICS
Task 1: Exploration of triangle
Mathematics is always perceived as a difficult subject. Hearing the word Mathematics would
make some of us to feel uncomfortable with the terms which are considered hard and difficult to
understand.
In reality, Mathematics is very close and is connected to daily your lives. We can witness the
different shapes of buildings, vehicles, roads and many more which are connected with
Mathematics. They are all implemented with the basic fundamental of Mathematics. So
Mathematics is already involved unintentionally in our daily lives.
Triangle are found everywhere. You are asked to draw a mural on the school walls by using the
geometry basic themes which must consists of triangle shapes. Hence, you are to draw a draft on
a A4 paper with the measurement of 10 cm X 20 cm of the mural that you would like on the
school walls.
Introduction
A triangle is a closed three-sided, three-angled figure, and is the simplest example of
what mathematicians call polygons (figures having many sides). Triangles are among the most
important objects studied in mathematics owing to the rich mathematical theory built up around
them in Euclidean geometry and trigonometry , and also to their applicability in such areas as
astronomy, architecture, engineering, physics, navigation, and surveying. In Euclidean geometry,
much attention is given to the properties of triangles. Many theorems are stated and proved about
the conditions necessary for two triangles to be similar and/or congruent. Similar triangles have
the same shape but not necessarily the same size, whereas congruent triangles have both the
same shape and size.
One of the most famous and useful theorems in mathematics, the Pythagorean Theorem,
is about triangles. Specifically, the Pythagorean Theorem is about right triangles, which are
triangles having a 90° or "right" angle. The theorem states that if the length of the sides forming
the right angle are given by a and b, and the side opposite the right angle, called the hypotenuse,
is given by c, then c 2 = a 2 + b 2. It is almost impossible to over-state the importance of this
theorem to mathematics and, specifically, to trigonometry.
Triangles feature very frequently in the discipline of mathematics. For example, almost
all two dimensional shapes (apart from a circle) can be made up of a series of triangles arranged
in a certain way. Furthermore, in the field of mathematics, triangle-related formulae such as
Pythagoras' theorem (a^2+b^2=c^2) form the basis of a great deal higher levels of mathematical
education, in branches such as trigonometry.
As mentioned above, Pythagoras' theorem is an incredibly important theory. However, its
importance goes beyond the field of pure mathematics and spans other fields including
engineering and architecture. Disciplines such as these, which focus a great deal upon the safe
distribution of weight or force, for example, rely heavily on Pythagoras' theorem, which is
entirely concerned with triangles.
Furthermore, many bridges and other similar structures are
often designed to include triangle shapes, as these shapes are able
to withstand a great amount of pressure (in a similar way to
arches). Because of the way that triangles disperse pressure
throughout their shape, they are able to withstand more pressure
than a differently-shaped object (for example, a square) of the
same size.
Principles of trigonometry, or the study of triangles, are used widely in fields such as
astronomy, space travel and communication in ways that I, as a non-astronomer, cannot even
begin to understand. However, my research suggests that trigonometry plays a role in aspects of
astronomy such as deciding how far about the earth a satellite dish should be placed.
In conclusion, we can see that actually triangle has many uses in our daily life. So we
should explore about the types of triangles, characteristics of triangles and the solution of
problems involving triangles.
