Trigonometry

Trigonometry is a mathematical concept which deals with shapes and its respective angles



The diagram below shows triangle ABC in which AC = 18cm, CAB = 30o and ABC = 90o

Trigonometry example 1

Calculate the length of BC

Solution:

AC = 18

BC = ?

Use SOH CAH TOA , which is a short name for three types of trig ratios or functions

Sin = O/H

Cos = A/H

Tan = O/A

Sin30 = O/H

O = BC

H = 18

Sin30 = BC/18

BC = 18×Sin30

BC = 9

How to solve Trigonometry Example

In the diagram ABC and ACD are right angled. < ABC=< CAD= 90°. AC = 20.2cm BC = 9.5cm and AD = 12cm,

How to solve Trigonometry Example

Calculate

(a) AB

(b) < BAC

(c) The area of quadrilateral ABCD

(d) The perimeter of quadrilateral ABCD.

Solutions:

(a)

a2=b2+c2

AB = ? , AC = 20 , BC = 9.5

AC2=AB2+BC2

20.22=AB2+9.52

20.22-9.52=AB2

408.04-90.25=AB2

317.79=AB2

AB2=317.79

AB=317.79

Answer: AB=17.83

(b)

Using SOHCAHTOA we will find angle BAC

hypotenuse = 20.2 , Opposite = 9.5 , adjacent = 17.83

We will use TOA

Tanx = 9.517.83

Tanx = 0.533

Tanx-1=0.533

BAC=28.05°

(c)

To find the area of the diagram above we have to know the shape of the diagram hence using the formula with respect to the diagram

Trapezium

Formula for Trapezium

A=a + b2h

a = AB, b = CD, h = 9.5

We need to find the length CD before we can find the area

a2=b2+c2

hypotenuse = CD, Opposite = 20.2 , adjacent = 12

CD2=20.22+122

CD2=408.04+144

CD2=552.04

CD=552.04

CD=23.49cm

A=AB + CD2BC

AB = 17.83, CD = 23.49, BC = 9.5

A=17.83 + 23.4929.5

A=20.66×9.5

A=20.66×9.5

Answer: Area=196.27cm2

(d)

Perimeter is the total distance around the shape

AB + BC + CD + DA

17.83 + 9.5 + 23.49 + 12

Answer: 62.82

Sine and Cosine Law

When finding the angle and distance of a triangle more than or less than the right angle triangle sine and cosine law are used. Therefore below are ways in which sine and cosine law can be applied

Sine Law

Trigonometry example 3

The diagram above shows how to apply sine law by deriving the following:

Sine Formula

aSinA=bSinB=cSinC

To apply the sine law formula to solve a problem, it requires a complete pair of the length of its opposite angle to be given and also the half pair which can be the length or an angle

Cosine Law

Trigonometry example 4

The diagram above shows a triangle having angles not equal to a right

Cosine Formula

C2=a2+b2-2ab cosC

To apply cosine law, it requires the length of two or all the sides of a triangle to be given with an angle

Area of a triangle using Sine

below is the formula of the area of a triangle, from it derive the formula of the area of triangle using sine

A=12bh

Sine is used to Calculate the area of any angle not just right triangle

Sine formuale of an area

A=12bhsinx

Solving trigonometry using sine law

The diagram below shows the position of a Guava (G) tree, Orange (O) tree and Lemon (L) tree on a farm.

Trigonometry example 5

Given that LG = 10.1m, OG = 14.2m and angle OLG = 40, calculate

(i) angle OGL,

(ii) the area of triangle OGL,

(iii) the shortest distance from L to OG

Solution:

(i)

Find angle LOG first using Sine law

aSinA=bSinB=cSinC

lSinL=oSinO

LOG = 14.2Sin40=10.1SinO

14.2 × SinO=10.1 × Sin40

14.2 SinO=6.49

14.2SinO=6.4914.2

SinO=0.457

SinO-1=0.457

LOG=27.19o

OGL = 180 - GLO + LOG

OGL = 180 - 40 + 27.19

OGL = 180 - 67.19

OGL = 112.81o

(ii)

To find the area use Sine method

A=12bhsinx

A=1210.1(14.2)sin112.81

A=12143.42sin112.81

A=12132.20

Area=66.10m2

(iii)

shortest distance from L to OG is the opposite to angle OGL and adjacent to LG , Therefore using SOHCAHTOA we can find the shortest distance

We will use SOH

Sinx=OH

Sin112.81=shortdistance10.1

Sin112.81=shortdistance10.1

shortdistance=Sin112.81 × 10.1

shortdistance=9.31 m

Solving trigonometry using Cosine law

The diagram below shows the location of the houses for a village Headman (H), his Secretary (S) and a Trustee (T). H is 1.3km from S, T is 1.9km from H and THS = 130o

Trigonometry example 6

Calculate

(i) the area of triangle THS,

(ii) the distance TS,

(iii) the shortest distance from H to TS.

(iv) Find the angle between 0o and 90o which satisfies the equation cosx = 23

Solution

(i)

Area of triangle THS

A=12bhsinx

A=121.9(1.3)sin130

A=122.47sin130

A=121.89

Area=0.95km2

(ii)

Distance TS to be found Cosine must be used

C2=a2+b2-2ab cosC

TS2=TH2+SH2-2(TH)(SH) cosH

TS2=1.92+1.32-2(1.9)(1.3) cos130

TS2=3.61+1.69-2(1.9)(1.3) cos130

TS2=3.61+1.69-(-3.18)

TS2=5.3+3.18

TS2=8.48

TS2=8.48

TS=2.91 km

(iii)

short distance from H to TS

Find angle TSH using sine law then apply SOHCAHTOA to find the shortest distance

aSinA=bSinB=cSinC

hSinH=sSinS

2.91Sin130=1.9SinS

2.91 × SinS=1.9 × Sin130

2.91 SinS=1.46

SinS2.91=1.462.91

SinS=0.50

SinS-1=0.50

TSH=30o

Given angle TSH the opposite is the shortdistance and the hypotenuse is 1.3 from SOHCAHTOA we will use SOH

Sinx=OH

SinTSH=shortdistance1.3

shortdistance=Sin30 × 1.3

shortdistance=Sin30 × 1.3

shortdistance=0.65 km

(iv)

cosx = 23

cosx = 0.667

cosx-1 = 0.667

X = 48.19o