BASIC of TRIGONOMETRYFor X grade Senior High School

By. Alfiramita Hertanti1111040151_ICP MATH 2011

SIMILAR TRIANGLE

A

BC

108

6

PQ

R

1830

Show that triangle ABC and PQR are similar triangles?Mention the ratio of the corresponding sides on both the triangles.

Measurement of Angle

round round round1 round

360𝑜=2𝜋𝑟𝑎𝑑𝑜𝑟 1𝑜=𝜋180

𝑟𝑎𝑑𝑜𝑟 1𝑟𝑎𝑑=57,3𝑜DEFIITION 8.2

Degree :“O”Radian: “Rad”

DEFINITON 8.3

• Round to degree

1.

• Degree to radian

⇔ 2. ⇔

3. 12𝑥360𝑜=¿180𝑜 ⇔ 180 𝑥

𝜋180

𝑟𝑎𝑑=¿𝜋𝑟𝑎𝑑

4. 4 𝑥360𝑜=¿270𝑜 ⇔ 270 𝑥𝜋180

𝑟𝑎𝑑=¿32𝜋 𝑟𝑎𝑑

Initial side

terminal side

terminal side

Initial side

Positive Angle

Negative Angle

180o

270o

0o,360o

90o

Quadrant II Quadrant I

Quadrant III Quadrant IV

0o - 90o90o - 180o

180o - 270o 270o-- 360o

BASIC CONSEPT OF ANGLE

Mr. Yahya was a guard of the school. The Height of Mr. Yahya is 1,6 m. He has a son, his name is Dani. Dani still class II elementary school. His body height is 1, 2 m. Dani is a good boy and likes to ask. He once asked his father about the height of the flagpole on the field. His father replied with a smile, 8 m. One afternoon, when he accompanied his father cleared the weeds in the field, Dani see shadows any objects on the ground. He takes the gauge and measure the length of his father shadow and the length of flagpole’s shadow are 6,4 m and 32 m. But he couldn’t measure the length of his own because his shadow follow ing his progression.

PROBLEM

A

B E G C

F

D

XO

Where :AB = The height of flagpole (8 m)BC = The lenght of the pole’s shadowDE = The height of Mr. YahyaEC = The length of Mr. Yahya’s ShadowFG = The height of DaniGC = The Lenght of Dani’s shadow

6,4

8

1,6

1,2

32

flagpole

Mr. Yahya Dani f

CE

D

XO

A

B C

XOCG

F

XO

g8

32

1,6

6,4

1,2

√1088 √ 43,52

f

𝐹𝐺𝐷𝐸

=𝐺𝐶𝐸𝐶

=1,21,6

=𝑓6,4. f = 4,8

𝐹𝐶=𝑔=√24,48

a.

√24,48√ 43,52√1088Opposite side the angle

FG

GC

DE EC EC

AB 1,2 1,6 8 Hytenuse of triangles

0,24

the sine of the angle C, written sin x0 = 0.24

b.

√24,48√ 43,52√1088adjacentGC

FC

EC DC AC

BC 4,8 6,4 32 Hypotenuse of triangle

0,97

the cosine of the angle C, written cos x0 = 0.97

c. 4 ,8 6,4 32

Opposite side the angleFG

GC

DE EC BC

AB 1,2 1,6 8

adjacent0,25

the tangent of the angle C,written tan x0 = 0.25

PROBLEM

1,5 m

8 m

9,5m

𝛼

Undu standing 8 m in front of the pine tree with height of 9.5 m. If the height of Undu is 1,5 m. Determine the trigonometric ratio of Angle .

Where :AC = The height Of Pine TreeED = The height of UnduDC = The distance between Tree and Undu

1,5 m

8 m

A

B

CD

E 𝜶

Undu Tree

9,5 m

SOLUTION

? ? ?

Find EA!

8 √2𝐸𝐴=√𝐸𝐵2+𝐴 𝐵2

¿√82+(9,5−1,5 )2

¿√64+64¿√128¿8 √2

𝑐𝑜𝑠𝛼=¿88√2

=12√2

𝑡𝑎𝑛𝛼=¿88=1

B

P J

Trigonometric ration in Right Triangle

the sine of an angle is the length of the opposite side divided by the length of the hypotenuse.

DEFINITION

B

P J

sin 𝐽=𝑃𝐵𝐵𝐽

the cosine of an angle is the length of the adjacent side divided by the length of the hypotenuse. 𝑐𝑜𝑠 𝐽=

𝑃𝐽𝐵𝐽

the tangent of an angle is the length of the opposite side divided by the length of the adjacent side.𝑡𝑎𝑛 𝐽=

𝑃𝐵𝑃𝐽

the cosecant of an angle is the length of the hypotenuse divided by the length of the opposite side. Written :

DEFINITION

B

P J

cos𝑒𝑐 𝐽=𝐵𝐽𝑃𝐵

the secant of an angle is the length of the hypotenuse divided by the length of the adjacent side.Written:𝑠𝑒𝑐 𝐽=

𝐵𝐽𝑃𝐽

the tangent of an angle is the length of the adjacent side divided by the length of the opposite side. written :

𝑐𝑜𝑡 𝐽=𝑃𝐽𝑃𝐵

cos𝑒𝑐 𝐽=1sin 𝐽

𝑠𝑒𝑐 𝐽=1

cos 𝐽

𝑐𝑜𝑡 𝐽=1

tan 𝐽

S O H C A H T O A

REMEMBER

in pposite

ypotenuse

os djacent

ypotenuse

an ppsosite

djacent

EXAMPLE

Given right triangle ABC, right-angled at ∠ ABC. If the length of the side AB = 3 units, BC = 4 units. Determine sin A, cos A, and tan A.C

BA 3 units

4 units

C

BA 3 units4 units

From the figure below,

5 units

𝐴𝐶=√𝐵𝐶2+𝐴𝐵2=√32+42=5

𝑆𝑖𝑛 𝐴=¿

cos 𝐴=¿

tan 𝐴=¿

h𝑡 𝑒 h𝑙𝑒𝑛𝑔𝑡 𝑜𝑓 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒𝑠𝑖𝑑𝑒 h𝑡 𝑒𝑎𝑛𝑔𝑙𝑒 𝐴h𝑡 𝑒 h𝑙𝑒𝑛𝑔𝑡 𝑜𝑓 h𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒

=¿

h𝑡 𝑒 h𝑙𝑒𝑛𝑔𝑡 𝑜𝑓 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑜𝑓 𝑎𝑛𝑔𝑙𝑒 𝐴h𝑡 𝑒 h𝑙𝑒𝑛𝑔𝑡 𝑜𝑓 h𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒

h𝑡 𝑒 h𝑙𝑒𝑛𝑔𝑡 𝑜𝑓 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒𝑠𝑖𝑑𝑒 h𝑡 𝑒𝑎𝑛𝑔𝑙𝑒 𝐴h𝑡 𝑒 h𝑙𝑒𝑛𝑔𝑡 𝑜𝑓 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑜𝑓 𝑎𝑛𝑔𝑙𝑒 𝐴

45

¿35

¿43

Ratio for Specific Angles

A(x,y)

xyr

Y

O X

Suppose point A (x, y), the length OA = r and the angle AOX = α.𝑆𝑖𝑛 α=¿

cos𝛼=¿

tan𝛼=¿

𝑦𝑟

𝑥𝑟

𝑦𝑥

𝛼

A(-x,y)

-xy r

Y

O X

𝑆𝑖𝑛 α=¿

cos𝛼=¿

tan𝛼=¿

𝑦𝑟

−𝑥𝑟

−𝑦𝑥

Quadrant II (90o-180o)Quadrant III (180o-270o)

Y

OX

A(-x,-y) -x-yr

𝑆𝑖𝑛 α=¿

cos𝛼=¿

tan𝛼=¿

−𝑦𝑟

−𝑥𝑟

𝑦𝑥

O

A(x,-y)

x-yr

Y

X

Quadrant IV (270o-360o)

𝑆𝑖𝑛𝛼=¿

cos𝛼=¿

tan𝛼=¿

−𝑦𝑟

𝑥𝑟

−𝑦𝑥

ALL

REMEMBER

SINTACOSQuadrant I

Quadran

t IIQua

dran

t III

Quadr

ant I

V

EXAMPLE Suppose given points A(-12,5) and XOA = ∠ α. Determine the value of sin α, cos α and tan αSOLUTION

x = -12 and y = 5. Quadrant II

A(-12,5)5

O

Y

X α

cos 𝐴=−1213

tan 𝐴=−512

𝑆𝑖𝑛 𝐴=513

12𝑋𝑂=√ (12 )2+52

¿√144+25¿√169¿13

13

Trigonometric Ration For Special Angles

0o, 30°, 45°,60° and 90o

45o

45o

30o

60o 60o

M

K LP

A

B C

22

1 1

45o

45o

A

B C

𝐴𝐶=√ 𝐴𝐵2+𝐵𝐶2

¿√1+1¿√2

1

1√2

sin 45𝑜=1

√2=12

√2

𝑐𝑜𝑠45𝑜=1

√2=12

√2

𝑡𝑎𝑛45𝑜=11=1

30o

60o

M

P L

2

1

𝑀𝑃=√𝑀𝐿2− 𝑃𝐿2

¿√ 4−2

¿√3sin 30𝑜=

12

𝑐𝑜𝑠30𝑜=√32

=12√3

tan 30𝑜= 1√3

=√33

sin 60𝑜=√32

=12

√3

𝑐𝑜𝑠60𝑜=12

𝑡𝑎𝑛60𝑜=√31

=√3√3

P(x,y)

1

1NO x

y

X

Y

ᶿ

sin 𝜃=𝑦1

=𝑦 cos𝜃=𝑥1=𝑥 tan𝜃=

𝑦𝑥

If , then P(1,0)• sin 0° = y = 0• cos 0° = x = 1• tan 0° = y/x = 0/1=0

• sin 90° = y = 1• cos 90° = x = 0• tan 90° =y/x =1/0, undefineIf , then P(0,1)

Trigonometric ratios of Special Angles

Sin 0 1

Cos 1 0

Tan 0 1

Anzar want to determine Angle size from a trigonometric ratio. Given to her ratio as follows., He must to determine the value of α (Angle size)

PROBLEM

SOLUTION

A

B

C

D

130o

sin𝛼=12, h𝑡 𝑒𝑛

𝛼=𝑎𝑟𝑐 sin ( 12 )¿30𝑜

12

THANK YOUFOR ATTENTION