angle angular measurement. angle is defined a rotation result from initial side to the terminal side...
TRANSCRIPT
ANGLEANGLE
ANGULAR MEASUREMENT
• Angle is defined a rotation result from initial side to the terminal side
• An angle has “a positive” sign if its rotation is anticlockwise
• An angle has “a negative” sign if its rotation is clockwise
• We only talk about the magnitude of angle~ not observe its signs
Initial side
Terminal side
o
radius
radius
arc=radius
What is a radian?
Angle = 1 rad
Why do mathematicians use radians instead of degrees?
How many times does the radius divide into the circumference?
There are 2 radians in a circle.
1 radian = = 57.3o
4
THE TRIGONOMETRIC RATIOS
180 rad
360 rad 2
rad 180 1
180 rad 1
Convert each angle in radians to degrees.
Convert each angle in degrees to radians.
1. 2c
2. 5c
3. 3 c
4. c
5. c
1. 65o
2. 200o
3. 120o
4. 180o
5. 330o
114.6o
286.5o
540o
90o
240o
1.13c
3.49c
Length of an arc using radians
Area of a sector using radians
Two important formula using radians
Trigonometric Ratios
Objectives/Assignment
• Find the since, the cosine, and the tangent of an acute triangle.
• Use trionometric ratios to solve real-life problems
Finding Trig Ratios
• A trigonometric ratio is a ratio of the lengths of two sides of a right triangle. The word trigonometry is derived from the ancient Greek language and means measurement of triangles. The three basic trigonometric ratios are sin, cos, and tan
Trigonometric Ratios
• Let ∆ABC be a right triangle. The since, the cosine, and the tangent of the acute angle A are defined as follows.
ac
bside adjacent to angle A
Sideoppositeangle A
hypotenuse
A
B
C
sin A =Side opposite A
hypotenuse=
a
c
cos A =Side adjacent to A
hypotenuse=
b
c
tan A =Side opposite A
Side adjacent to A=
a
b
Exercise:
• In the PQR triangle is right angled at R happens cos P = 8/ 17…find the value of tg P and tg Q ?
Exercises
1. PQR triangle is right angled at R & sin P cosQ = . Find out ⅗the value
tgQtgP
c
AB
DE
2 ABC triangle is right angle at A. BC=p,
AD is perpendicular at BC, DE
perpendicular at AC, angle B = Q, prove
that : DE=psin2 Qcos Q
Evaluating Trigonometric Functions
• Acute angle A is drawn in standard position as shown.
Right-Triangle-Based Definitions of Trigonometric Functions
For any acute angle A in standard position,
adjacent side
opposite sidetan
hypotenuse
adjacent sidecos
hypotenuse
opposite sidesin
x
yA
r
xA
r
yA
opposite side
adjacent sidecot
adjacent side
hypotenusesec
opposite side
hypotenusecsc
y
xA
x
rA
y
rA
Finding Trigonometric Function Values of an Acute Angle in a Right Triangle
Example Find the values of sin A, cos A, and tan A
in the right triangle.
Solution– length of side opposite angle A is 7– length of side adjacent angle A is 24– length of hypotenuse is 25
247
tan,2524
cos,257
sin AAA
Trigonometric Function Values of Special Angles
• Angles that deserve special study are 30º, 45º, and 60º.
Using the figures above, we have the exact values of the special angles summarized in the table on the right.
Let’s try to prove it !
O
B
A
Y
X
45O
OA=OBOA2 + OB2 = OC2
OA2 + OA2 = r2
2OA2 = 1OA2 = ½
OA = = OB
How about 300 , 600 and 900
Cofunction Identities
• In a right triangle ABC, with right angle C, the acute angles A and B are complementary.
• Since angles A and B are complementary, and sin A = cos B, the functions sine and cosine are called cofunctions. Similarly for secant and cosecant, and tangent and cotangent.
Bbc
A
Bba
A
Bca
A
cscsec
cottan
cossin
Cofunction Identities
Note These identities actually apply to all angles (not just
acute angles).
If A is an acute angle measured in degrees, then
If A is an acute angle measured in radians, then
)90sin(cos
)90cos(sin
AA
AA
)90csc(sec
)90sec(csc
AA
AA
)90tan(cot
)90cot(tan
AA
AA
AA
AA
2sincos
2cossin
AA
AA
2cscsec
2seccsc
AA
AA
2tancot
2cottan
Reference Angles
• A reference angle for an angle , written , is the positive acute angle made by the terminal side of angle and the x-axis.
Example Find the reference angle for each angle.
(a) 218º (b)
Solution (a) = 218º – 180º = 38º (b)
65
665
Special Angles as Reference Angles
Example Find the values of the trigonometric functions for 210º.
Solution The reference angle for 210º is 210º – 180º = 30º.
Choose point P on the terminal side so that the distance from the origin to P is 2. A 30º - 60º right triangle is formed.
3210cot3
32210sec2210csc
33
210tan23
210cos21
210sin
Finding Trigonometric Function Values Using Reference Angles
Example Find the exact value of each expression.(a) cos(–240º) (b) tan 675º
Solution(a) –240º is coterminal with 120º.
The reference angle is 180º – 120º = 60º. Since –240º lies in quadrant II, the cos(–240º) is negative.
(a) Similarly, tan 675º = tan 315º = –tan 45º = –1.
21
60cos)240cos(
Finding Angle MeasureExample Find all values of , if is in the
interval [0º, 360º) and Solution Since cosine is negative, must lie in either quadrant II or III. Since So the reference angle = 45º.
The quadrant II angle = 180º – 45º = 135º, and thequadrant III angle = 180º + 45º = 225º.
.cos 22
.45cos,cos 221
22
Ex. 1: Finding Trig Ratios• Compare the sine, the
cosine, and the tangent ratios for A in each triangle beside.
• By the Similarity Theorem, the triangles are similar. Their corresponding sides are in proportion which implies that the trigonometric ratios for A in each triangle are the same.
15
817
A
B
C
7.5
48.5
A
B
C
Ex. 1: Finding Trig RatiosLarge Small
15
817
A
B
C
7.5
48.5
A
B
C
sin A = opposite
hypotenuse
cosA = adjacent
hypotenuse
tanA = opposite
adjacent
8
17≈ 0.4706
15
17≈ 0.8824
8
15≈ 0.5333
4
8.5≈ 0.4706
7.5
8.5≈ 0.8824
4
7.5≈ 0.5333
Trig ratios are often expressed as decimal approximations.
Ex. 2: Finding Trig Ratios
S
sin S = opposite
hypotenuse
cosS = adjacent
hypotenuse
tanS = opposite
adjacent
5
13≈ 0.3846
12
13≈ 0.9231
5
12≈ 0.4167
adjacent
opposite
12
13 hypotenuse5
R
T S
Ex. 2: Finding Trig Ratios—Find the sine, the cosine, and the tangent of the indicated angle.
R
sin S = opposite
hypotenuse
cosS = adjacent
hypotenuse
tanS = opposite
adjacent
12
13≈ 0.9231
5
13≈ 0.3846
12
5≈ 2.4
adjacent
opposite12
13 hypotenuse5
R
T S
Notes:
• If you look back, you will notice that the sine or the cosine of an acute triangles is always less than 1. The reason is that these trigonometric ratios involve the ratio of a leg of a right triangle to the hypotenuse. The length of a leg or a right triangle is always less than the length of its hypotenuse, so the ratio of these lengths is always less than one.
Using Trigonometric Ratios in Real-life
• Suppose you stand and look up at a point in the distance. Maybe you are looking up at the top of a tree as in Example 6. The angle that your line of sight makes with a line drawn horizontally is called angle of elevation.
Ex. 6: Indirect Measurement• You are measuring the height of
a Sitka spruce tree in Alaska. You stand 45 feet from the base of the tree. You measure the angle of elevation from a point on the ground to the top of the top of the tree to be 59°. To estimate the height of the tree, you can write a trigonometric ratio that involves the height h and the known length of 45 feet.
The math
tan 59° =opposite
adjacent
tan 59° =h
45
45 tan 59° = h
45 (1.6643) ≈ h
75.9 ≈ h
Write the ratio
Substitute values
Multiply each side by 45
Use a calculator or table to find tan 59°
Simplify
The tree is about 76 feet tall.
Ex. 7: Estimating Distance
• Escalators. The escalator at the Wilshire/Vermont Metro Rail Station in Los Angeles rises 76 feet at a 30° angle. To find the distance d a person travels on the escalator stairs, you can write a trigonometric ratio that involves the hypotenuse and the known leg of 76 feet.
d76 ft
30°
Now the math d76 ft
30°sin 30° =
opposite
hypotenuse
sin 30° =76
d
d sin 30° = 76
sin 30°
76d =
0.5
76d =
d = 152
Write the ratio for sine of 30°
Substitute values.
Multiply each side by d.
Divide each side by sin 30°
Substitute 0.5 for sin 30°
Simplify
A person travels 152 feet on the escalator stairs.