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1. BASIC of TRIGONOMETRY
For X grade Senior High School
By. Alfiramita Hertanti
1111040151_ICP MATH
2011

3. SIMILAR TRIANGLE
A
BC
10
8
6
PQ
R
18
30
Show that triangle ABC and PQR are similar triangles?
Mention the ratio of the corresponding sides on both the
triangles.

4. Measurement of Angle
1
360
round
1
4
round1
2
round1 round
360 𝑜
= 2𝜋 𝑟𝑎𝑑 𝑜𝑟 1 𝑜
=
𝜋
180
𝑟𝑎𝑑 𝑜𝑟 1 𝑟𝑎𝑑 = 57,3 𝑜
DEFIITION 8.2
Degree :“O”
Radian: “Rad”

5. DEFINITON 8.3
• Round to degree
1.
1
4
𝑟𝑜𝑢𝑛𝑑 =
1
4
𝑥 360 𝑜 = 90 𝑜
• Degree to radian
⇔ 90 𝑜 = 90 𝑜 𝑥
𝜋
180
𝑟𝑎𝑑 =
1
2
𝜋 𝑟𝑎𝑑.
2.
1
3
𝑟𝑜𝑢𝑛𝑑 =
1
3
𝑥 360 𝑜
= 120 𝑜 ⇔ 120 𝑜 = 120 𝑥
𝜋
180
𝑟𝑎𝑑 =
2
3
𝜋 𝑟𝑎𝑑
3.
1
2
𝑟𝑜𝑢𝑛𝑑 =
1
2
𝑥 360 𝑜
=180 𝑜 ⇔ 180 𝑜 = 180 𝑥
𝜋
180
𝑟𝑎𝑑 = 𝜋 𝑟𝑎𝑑
4.
3
4
𝑟𝑜𝑢𝑛𝑑 = 4𝑥 360 𝑜 = 270 𝑜 ⇔ 270 𝑜 = 270 𝑥
𝜋
180
𝑟𝑎𝑑 =
3
2
𝜋 𝑟𝑎𝑑

6. Initial side
Initial side

7. 180o
270o
0o,360o
90o
Quadrant II Quadrant I
Quadrant III Quadrant IV
0o - 90o
90o - 180o
180o - 270o 270o-- 360o

8. 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

9. A
B E
G C
F
D
XO
Where :
AB = The height of flagpole (8 m)
BC = The lenght of the pole’s shadow
DE = The height of Mr. Yahya
EC = The length of Mr. Yahya’s Shadow
FG = The height of Dani
GC = The Lenght of Dani’s shadow
6,4
8
1,6
1,2
32 f

10. CE
D
A
B C CG
F
g8
32
1,6
6,4
1,2
1088 43,52
f
𝐹𝐺
𝐷𝐸
=
𝐺𝐶
𝐸𝐶
=
1,2
1,6
=
𝑓
6,4
. f = 4,8
𝐹𝐶 = 𝑔 = 24,48
a. ____ = ____ = ____ = ________ = ________ = ______ = ____________________ =
24,48 43,52 1088
Opposite side the angleFG
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 1088
adjacentGC
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
adjacent
0,25
the tangent of the angle C,
written tan x0 = 0.25

11. 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 𝛼.

12. Where :
AC = The height Of Pine Tree
ED = The height of Undu
DC = The distance between Tree and Undu
1,5 m
8 m
A
B
CD
E 𝜶
9,5 m
SOLUTION
𝑠𝑖𝑛 𝛼? 𝑐𝑜𝑠 𝛼? 𝑡𝑎𝑛 𝛼?
Find EA!
8 2
𝐸𝐴 = 𝐸𝐵2 + 𝐴𝐵2
= 82 + 9,5 − 1,5 2
= 64 + 64
= 128
= 8 2
𝑐𝑜𝑠 𝛼 =
8
8 2
=
1
2
2
𝑡𝑎𝑛 𝛼 =
8
8
= 1
𝑠𝑖𝑛 𝛼 =
8
8 2
=
1
2
2

13. B
P J
Trigonometric ration in Right Triangle

14. 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 =
𝑃𝐵
𝐵𝐽
the cosine of an angle is the length of
the adjacent side divided by the length
of the hypotenuse.
𝑐𝑜𝑠 𝐽 =
𝑃𝐽
𝐵𝐽
the tangent of an angle is the
length of the opposite side
divided by the length of the
adjacent side.
𝑡𝑎𝑛 𝐽 =
𝑃𝐵
𝑃𝐽

15. 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 =
𝐵𝐽
𝑃𝐵
the secant of an angle is the length of
the hypotenuse divided by the length of
the adjacent side.Written:
𝑠𝑒𝑐 𝐽 =
𝐵𝐽
𝑃𝐽
the tangent of an angle is
the length of the adjacent
side divided by the length of
the opposite side. written :
𝑐𝑜𝑡 𝐽 =
𝑃𝐽
𝑃𝐵
cos𝑒𝑐 𝐽 =
1
sin 𝐽
𝑠𝑒𝑐 𝐽 =
1
cos 𝐽
𝑐𝑜𝑡 𝐽 =
1
tan 𝐽

16. S O H C A H T O A
REMEMBER
i
n
p
p
o
s
i
t
e
y
p
o
t
e
n
u
s
e
o
s
d
j
a
c
e
n
t
y
p
o
t
e
n
u
s
e
a
n
p
p
s
o
s
i
t
e
d
j
a
c
e
n
t

17. 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

18. C
BA 3 units
4 un
From the figure below,
𝐴𝐶 = 𝐵𝐶2 + 𝐴𝐵2 = 32 + 42 = 5
𝑆𝑖𝑛 𝐴 =
Cos 𝐴 =
Tan 𝐴 =
𝑡ℎ𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 𝐴
𝑡ℎ𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
=
𝑡ℎ𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑜𝑓 𝑎𝑛𝑔𝑙𝑒 𝐴
𝑡ℎ𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑡ℎ𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 𝐴
𝑡ℎ𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑜𝑓 𝑎𝑛𝑔𝑙𝑒 𝐴
4
5
=
3
5
=
4
3

19. Ratio for Specific Angles
A(x,y)
x
y
r
Y
O
X
Suppose point A (x, y), the length
OA = r and the angle AOX = α.
𝑆𝑖𝑛 α =
Cos 𝛼 =
Tan 𝛼 =
𝑦
𝑟
𝑥
𝑟
𝑦
𝑥
𝛼
A(-x,y)
-x
y
r
Y
O
X
𝑆𝑖𝑛 α =
Cos 𝛼 =
Tan 𝛼 =
𝑦
𝑟
−
𝑥
𝑟
−
𝑦
𝑥
Quadrant II (90o-180o)Quadrant III (180o-270o)
Y
O
X
A(-x,-y) -x
-y
r
𝑆𝑖𝑛 α =
Cos 𝛼 =
Tan 𝛼 =
−
𝑦
𝑟
−
𝑥
𝑟
𝑦
𝑥
O
A(x,-y)
x
-y
r
Y
X
Quadrant IV (270o-360o)
𝑆𝑖𝑛 𝛼 =
Cos 𝛼 =
Tan 𝛼 =
−
𝑦
𝑟
𝑥
𝑟
−
𝑦
𝑥

20. ALL
REMEMBER
SINTACOS

21. EXAMPLE
Suppose given points A(-12,5) and ∠XOA = α.
Determine the value of sin α, cos α and tan α
SOLUTION
x = -12 and y = 5. Quadrant II
A(-12,5)
5
O
Y
X
α
Cos 𝐴 = −
12
13
Tan 𝐴 = −
5
12
𝑆𝑖𝑛 𝐴 =
5
13
12
𝑋𝑂 = 12 2 + 52
= 144 + 25
= 169
= 13
13

22. 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

23. 45o
45o
A
B C
𝐴𝐶 = 𝐴𝐵2 + 𝐵𝐶2
= 1 + 1
= 2
1
1
2
sin 45 𝑜
=
1
2
=
1
2
2
𝑐𝑜𝑠 45 𝑜 =
1
2
=
1
2
2
𝑡𝑎𝑛 45 𝑜 =
1
1
= 1

24. 30o
60o
M
P L
2
1
𝑀𝑃 = 𝑀𝐿2 − 𝑃𝐿2
= 4 − 2
= 3
sin 30 𝑜 =
1
2
𝑐𝑜𝑠 30 𝑜 =
3
2
=
1
2
3
tan 30 𝑜 =
1
3
=
3
3
sin 60 𝑜 =
3
2
=
1
2
3
𝑐𝑜𝑠 60 𝑜 =
1
2
𝑡𝑎𝑛 60 𝑜
=
3
1
= 3
3

25. P(x,y)
1
1NO x
y
X
Y
ᶿ
sin 𝜃 =
𝑦
1
= 𝑦 cos 𝜃 =
𝑥
1
= 𝑥 tan 𝜃 =
𝑦
𝑥
If 𝜃 = 0 𝑜, 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, undefine
If 𝜃 = 90 𝑜
, then P(0,1)

26. Trigonometric ratios of Special Angles
𝛼 0 𝑜 30 𝑜 45 𝑜 60 𝑜 90 𝑜
Sin 𝛼 0
1
2
1
2
2
1
2
3 1
Cos 𝛼 1
1
2
3 1
2
2
1
2
0
Tan 𝛼 0
1
3
3 1 3 ∞

27. Anzar want to determine Angle size from a
trigonometric ratio. Given to her ratio as follows.
sin 𝛼 =
1
2
, He must to determine the value of α
(Angle size)
PROBLEM

28. 1
30o
sin 𝛼 =
1
2
, 𝑡ℎ𝑒𝑛
𝛼 = 𝑎𝑟𝑐 sin
1
2
= 30 𝑜
1
2

29. THANK YOU
FOR ATTENTION