Usage of Trigonometry in Daily Life

PROJECT WORK FOR ADDITIONAL MATHEMATICS 1/2012

SMK KAPIT NO. 2

Name : Christie Balan Ak Reba

IC No. : 951128-13-5611

Class : 5 Science 1

Teacher’s Name : Mr Jason Lee

1

-Content-No. Contents Page

1 Introduction 32 Acknowledgement 43 History of Trigonometry 54 Task Specification 75 Problem Solving ( a ) 96 ( b ) 117 ( c ) 128 ( d ) 149 ( e ) 159 ( f ) 17

11 ( g ) 1812 Further Exploration 1913 Conclusion 2014 Reflection 21

2

IntroductionWe students taking Additional Mathematics are required to carry out a project work while

we are in Form 5. We are given two tasks and asked to choose and complete only ONE task based on our area of interest. This project can be done in groups or individually. Upon completion of the Additional Mathematics Project Work, we are to gain valuable experiences and able to :

Apply and adapt a variety of problem solving strategies to solve routine and non-routine problems

Experience classroom enviroments which are challenging, interesting and meaningful and hence improve their thinking skills

Experience classroom enviroments where knowledge and skills are applied in meaningful ways in solving real-life problems

Experience classroom enviroments where expressing ones mathematical thinking, reasoning and communication are highly encouraged and expected

Experience classroom enviroments that stimulates and enhances effective learning Acpuire effective mathematical communication through oral and writing, and to use

the mathematics to express mathematical ideas correctly and precisely Enhance acquisition of mathematical knowledge and skills through problem-solving

in ways that increase interest and confidence Prepare ourselves for the demand of our future undertakings and in workplace Realise that mathematics is an important and powerful tool in solving real-life

problems and hence develop positive attitude towards mathematics Train ourselves not only to be independent learners but also to collaborate, to

cooperate, and to share knowledgein an engaging and healthy environment Use technology especially the ICT appropriately and effectively Train ourselves to appreciate the intrinsic values of mathematics and to become

more creative and innovative Realize the importance and the beauty of mathematics

3

AknowledgementFirst and foremost, I would like to thank God that finally, I have succeeded in finishing this

project work. I would like to thank my beloved Additional Mathematics teacher, Mr Jason Lee for all the guidance he had provided us during the process in finishing this project work.

I also appreciate his patience in guiding us completing this project work. i would like to give a thousand thanks to my parent for giving me their full support in this project work, financially and mentally. They gave me moral support when I needed it. I would also like to give my thanks to my fellow friends who had helped me in finding the information that I clueless of, and the time I spent together in study groups on finishing this project work.

Last but not least, I would like to express my highest gratitude to all those who gave me the possibility to complete this coursework. I really appreciate all the help I got.

Again, thank you very much.

4

History of TrigonometryTrigonometry (from Greek trigōnon "triangle" + metron "measure" is a branch of mathematics that studies triangles and the relationships between their sides and the angles between these sides. Trigonometry defines the trigonometric functions, which describe those relationships and have applicability to cyclical phenomena, such as waves. The field evolved during the third century BC as a branch of geometry used extensively for astronomical studies. It is also the foundation of the practical art of surveying.

Trigonometry basics are often taught in school either as a separate course or as part of a precalculus course. The trigonometric functions are pervasive in parts of pure mathematics and applied mathematics such as Fourier analysis and the wave equation, which are in turn essential to many branches of science and technology. Spherical trigonometry studies triangles on spheres, surfaces of constant positive curvature, in elliptic geometry. It is fundamental to astronomy and navigation. Trigonometry on surfaces of negative curvature is part of Hyperbolic geometry.

Sumerian astronomers introduced angle measure, using a division of circles into 360 degrees.They and their successors the Babylonians studied the ratios of the sides of similar triangles and discovered some properties of these ratios, but did not turn that into a systematic method for finding sides and angles of triangles. The ancient Nubians used a similar methodology.The ancient Greeks transformed trigonometry into an ordered science.

Classical Greek mathematicians (such as Euclid and Archimedes) studied the properties of chords and inscribed angles in circles, and proved theorems that are equivalent to modern trigonometric formulae, although they presented them geometrically rather than algebraically. Claudius Ptolemy expanded upon Hipparchus' Chords in a Circle in his Almagest. The modern sine function was first defined in the Surya Siddhanta, and its properties were further documented by the 5th century Indian mathematician and astronomer Aryabhata. These Greek and Indian works were translated and expanded by medieval Islamic mathematicians. By the 10th century, Islamic mathematicians were using all six trigonometric functions, had tabulated their values, and were applying them to problems in spherical geometry.[citation needed] At about the same time, Chinese mathematicians developed trigonometry independently, although it was not a major field of study for them. Knowledge of trigonometric functions and methods reached Europe via Latin translations of the works of Persian and Arabic astronomers such as Al Battani and Nasir al-Din al-Tusi. One of the earliest works on trigonometry by a European mathematician is De Triangulis by the 15th century German mathematician Regiomontanus. Trigonometry was still so little known in 16th century Europe that Nicolaus Copernicus devoted two chapters of De revolutionibus orbium coelestium to explaining its basic concepts.

Driven by the demands of navigation and the growing need for accurate maps of large areas, trigonometry grew to be a major branch of mathematics. Bartholomaeus Pitiscus was the first to use the word, publishing his Trigonometria in 1595.Gemma Frisius described for the first time the method of triangulation still used today in surveying. It was Leonhard Euler who fully incorporated complex numbers into trigonometry. The works of James Gregory in the 17th century and Colin Maclaurin in the 18th century were influential in the development of trigonometric series. Also in the 18th century, Brook Taylor defined the general Taylor series.

5

Sine function (sin), defined as the ratio of the side opposite the angle to the hypotenuse.

Sin A = oppositehypotenuse

= ac

Cosine function (cos), defined as the ratio of the adjacent leg to the hypotenuse.

Cos A = adjacenthypotenuse

= bc

Tangent function (tan), defined as the ratio of the opposite leg to the adjacent leg.

Tan A = oppositeadjacent

= ab

= sin Acos A

Area of Trigonometry

Area of triangle, A = 12xb x h

sin c=hb

b sin c=h

Since that, A = 12

ab sin C ( included angle )

Phytagoras Theorem a2+b2=c2

6

Task SpecificationADDITIONAL MATHEMATICS PROJECT WORK 1/2012

Pak Samy has a piece of unused land besides his house. This piece of land is surrounded by the river and the mountain. After his retirement, he decided to clear up that piece of land to plant vegetables. The outlook of the land is as shown in the Diagram 1.

Pak Samy thinks that it will be good if he can fence up that land. He measured the diagonal distance from the river to the foot of the mountain (A to B) is 500 m and the distance along the mountain side till it almost meets the streams of the river (B to C) is 800 m. Pak Samy also built a block made from sand bags along the river for flood prevention during heavy rain. The angle subtended between the diagonal distance of AB and the sand block is 30° as shown in diagram 2.

(a) Pak Samy planned to dig a well with cross section of the shape of a sector with centre point B, to make watering job easier for him. He needs to build the top part of the well with radius 1 m and height 1 m. You are required to help him to calculate the angle of in order that he could build the well. Provide at least two methods in your solutions. ABC

(b) Pak Samy wants to build the thickest wall for the well. You are supposed to help him to calculate the number of bricks to buy in order to build the well. As shown in diagram 3. For the curve surface, calculation should base on the internal surface area. Given 400 = 4 bricks inclusive of cement in between bricks. If each brick is 40 cents each, help Pak Samy to calculate how much he needs to spend on bricks. 2 cm

(c) Pak Samy is poor in calculation, he wanted to fence up that piece of land in a triangular shape. You need to help him to calculate the total length of the fencing materials needed. Use at least 2 methods of solution.

(d) After clearing up and fencing the land, he needs to buy seeds of vegetables and plants. However, before he could buy those seeds, he needs to know how big the land is so that he could buy sufficient of various seeds. As an additional mathematics student, you need to help him to find the area of the land by using at least two different methods.

(e) After sometimes, the fence is rundown, Pak Samy wishes to build the new fence with minimum cost. He wishes to minimize the fencing materials but the area of the piece of land must remain the same. The point B and C are fixed because they are by the mountain side. Only point A is movable. Make a conjecture on the position of point A. (Tabulate a few sets of values of AB and AC, and find the minimum length of the fence.)

7

(f) A year later, seeing the price of oil palm is so attractive, Pak Samy is planning to divide the land in to two parts, he wants to plant oil palm which is the golden plant at the farther part of the land. He had another piece of fencing material that is same length with the length of AB. He wants to build the fence from point B outside the well in a straight line until it reaches the line AC. You are required to help him to construct the location of the fence in graphical form. He decided to differentiate the color of the two fences for the two types of plants beside the river; you are asked to help Pak Samy to find the length of the fences of vegetable and oil palm.

(g) Pak Samy hires some workers to clear the piece of land to plant oil palm. The workers ask for RM 60 per 400 as wages. At the end of the clearing process, Pak Samy pays a total sum of RM 6126.40 to the workers as wages. Find the area of the land that Pak Samy uses to plant oil palm. 2 m

Further Exploration:-

After 5 years the oil palm trees are mature and ready for harvest, Pak Samy finds that the return is so attractive. Now he wishes to convert the whole land to plant oil palm. He wishes to build more solid fence for the whole land. He bought 2000 m of fencing materials and the side by the mountain side is fixed (BC), find the dimension of triangle ABC such that the area enclosed is maximum so that he could plant the most oil palm trees. Hence, find the maximum area of the plantation.

8

Problem Solvinga) METHOD 1

B 800m C

500m

30°

A

sinC500

= sin 30 °800

sinC=500sin30 °800

∠C=sin−1 500sin30 °800

= 18° 13´

∠ABC = 180° - 30° - 18° 13’

= 131° 47’ / 131.79°

9

METHOD 2

B 800m C

500m

30° D

A

Draw a perpendicular line from B to AC to intersect the line AC at D.

∠ABD = 60°

In triangle ABD

Sin 30° = BD500

BD = 500 sin 30°

= 250m

In triangle BDC

Cos ∠DBC = 250800

∠DBC =cos−1 250800

= 71° 47’

∠ABC = 60° + 71° 47’

= 131° 47’ / 131.79°

10

b)

The length of internal arc

= (π180

x131 ° 47' ¿ x1m

= 2.3m

= 230cm

Total surface area of the internal arc = 230 x 100

= 23000cm²

Area of the 2 squared walls = 100 x 100 x 2

= 20000cm²

Total area = 23000+ 20000

= 43000cm²

The thickest wall by the dimension of one brick,

Given 400cm² = 4 bricks

Total bricks required = (43000/400) x 4

= 430 bricks.

Given cost of one bricks is 40 cents

Total cost required = 430 x 0.4

= RM 172.00

11

c)

METHOD 1

B 800m C

500m

30° D

A

Sin 30° = BD500

BD = 250

In triangle ABD, AD² = 500² - 250²

= 187500

AD = √187500

= 433.01m

In triangle BDC, CD² = 800² - 250²

= 577500

CD = √577500

= 759.93m

Total length of the fence = 500 + 800 + 433.01 + 759.93

= 2492.94m

12

METHOD 2

In triangle ABC

Using Cosine Rule,

AC² = 500² + 800² - 2 x 500 x 800 x cos 131° 47’

= 1 423 052.47

AC = √1423052.47

= 1 192.92m

Total length of fences = 500 + 800 + 1192.92

= 2492.92m

13

d)

METHOD 1

Area of the land = 12

x AC x BD

= 12

x 1192.92 x 250

= 149115m²

METHOD 2

Area of the land = 12x500 x 800x sin 131.79 °

= 149118m²

14

e)

Conjecture: The length of the fence is minimum when triangle ABC is an isosceles triangle,

where AB = AC and BC = 800m.

Area of triangle ABC = 149134m²

149134 = 12

x 800 x h

= 372.84m

B x D 800-x C

372.84m

A

Let AD be the height of triangle ABC with BC as the corresponding base.

AD = 372.84m

Let BD = x m

Therefore, DC = 800 – x m

15

BD DC AB AC Perimeter100 700 386.01 793.10 1979.11200 600 423.10 706.41 1929.51300 500 478.55 623.71 1902.26400 400 546.82 546.82 1893.64500 300 623.71 478.55 1902.26600 200 706.41 423.10 1929.51700 100 793.10 386.01 1979.11

397 403 544.6271 549.0161 1893.6432398 402 545.3565 548.2825 1893.6390399 401 546.0867 547.5497 1893.6364400 400 546.8178 546.8178 1893.6356401 399 547.5497 546.0867 1893.6364402 398 548.2825 545.3565 1893.6390403 397 549.0161 544.6271 1893.6432

Perimeter minimum

From the tables above, the perimeter is minimum when AB = AC = 546.8178m

16

399.7 400.3 546.5983 547.0373 1893.635602399.8 400.2 546.6715 546.9641 1893.635559399.9 400.1 546.7446 546.8909 1893.635534400 400 546.8178 546.8178 1893.635525400.1 399.9 546.8909 546.7446 1893.635534400.2 399.8 546.9641 546.6715 1893.635559400.3 399.7 547.0373 546.5983 1893.635602

f)

B 800m C

500m

30° AA’ A’ A'C

A

Length of vegetable fence ( AA’ )

A A'

sin 120° =

500sin °30

AA’ = 866.0254m.

Length of Oil Palm fence ( A’C )

A 'Csin 11.79°

= 800sin 150

A’C = 326.92m

Or AC – AA = 1192.9176 – 866.0254

= 326.8922m

17

g)

Wages RM 60 for 400m²

Total wages paid is RM 6126.40

Area of the land = ( 6126.40 / 60 ) x 400

= 40 842.7m²

18

Further Exploration :-

B 800m C

θ

A

Let BC = 800m and ∠ABC = θ °

The perimeter, AB + BC + AC = 2000m

AB AC θ Area of ∆ ABC (m2)300 900 99.59° 118322400 800 75.52° 154919500 700 60° 173205600 600 48.19° 178885650 550 43.05° 177482750 450 33.56° 165831850 350 24.25° 139642

Area maximum

19

597 603 48.51078 178880.407598 602 48.40357 178883.2021599 601 48.29654 178884.8792600 600 48.18969 178885.4382601 599 48.08301 178884.8792602 598 47.97651 178883.2021603 579 47.87019 178880.407

559.7 600.3 48.22172 178885.3879559.8 600.2 48.21104 178885.4158599.9 600.1 48.20036 178885.4326600 600 48.18969 178885.4382600.1 599.9 48.17901 178885.4326600.2 599.8 48.16834 178885.4158600.3 599.7 48.15766 178885.3879

From the table of values, it is noted that for the area is maximum when triangle is an isosceles triangle with BC = 800m and AB = AC = 600m

The maximum area of triangle ABC is 178 885m²

ConclusionAfter doing research, answering questions, drawing table and some problem solving, I saw

that the usage of triangle is important in our daily life. It is not just widely used in architecture but also in interpreting the area of a specific location. Especially in measuring a location of area which to be built something or else.

Without it, all this measurement activities can’t be conducted accurately. So, I should thankful of the people who contribute in the idea of making this way of measurement.

20

Reflection

Additional Mathematics..

I struggle so hard to finish you

You taught me the meaning of calculating

You taught me the way to think

Everything is about you

Additional Mathematics..

Your question was so hard to answer

But I survive this storm and rain just to answer you

I swim all this river to get to you

You are everything..

21