Transcript 12.5 Lesson ppt

Yr 2 w-up
For 1-5 solve – you decide what to use, set up ratio, round to the
hundredths place
3.
2.
1.
5 ft xo
16 cm
x
32o
20 cm
17 yd
22 yd
37.690
32.360
10.64 cm
4.
xo
4 ft 42o
17 m
33 cm
xo
5.
15 m
22 m
20o
41 cm
42.890
15.07 cm
x
Problem Solving with
Trigonometry
Lesson 12.5
Do the following when solving trigonometry
word problems…
1. Draw a picture.
2. Decide which trigonometry formula you must use.
3. Solve the word problem.
SOH – CAH – TOA
Right Triangle Ratios
LAW OF SINES
sin A sin B sin C


a
b
c
LAW OF COSINES
SAS TRIANGLE AREA
c  a  b  2ab cos C
1
A  ab sin C
2
2
2
2
A large helium balloon is tethered to the ground by two taut
lines. One line is 100 feet long and makes an 80° angle with the
ground. The second line makes a 40° angle with the ground.
How long is the second line, to the nearest foot? How far apart
are the tethers?
sin 40 sin 80
sin 40 sin 60


100
x
100
y
x sin 40  100 sin 80
y sin 40  100 sin 60
100 sin 60
y
sin 40
x
60°
x = 153 ft
x
y = 135 ft
80°
40°
y
100 sin 80
sin 40
A ship’s sonar locates a treasure chest at a 12° angle of
depression. A diver is lowered 40 meters to the ocean floor.
How far (to the nearest meters) does the diver need to swim
along the ocean floor to get the treasure chest?
40
tan12 
x
x tan12  40
40
x
tan 12
12°
40 m
x = 188 m
12°
x
Farmer Joe needs to fence his triangular plot of land for his cows.
The angle between two sides measures 83°. One side is 122 ft and
the other is 215 ft. How much fencing does farmer Joe need to the
nearest foot? What is the area of his plot of land?
x 2  1222  2152  2122215cos 83
x 2  54,715.73
x  234
234  122  215 
571 ft
83°
1
Area  (122)(215) sin 83
2
Area = 13017 ft2
x
234 ft
HOMEWORK:
12.5 Worksheet and
p. 648 1, 3, 4, 6, 8