Finding the Missing Side
41
Finding the Missing Side Year 11
description
Finding the Missing Side. Year 11. Here’s the problem. An anchor cable shelves at 50 º. You are aware that 24m of cable has been let out. What is the depth of water?. 50 º. 1. Draw diagram. x. 24m. 2. Label inside. With. 50 º. H , O & A. x. 24m. 50 º. x. 24m. A. O. H. 3. - PowerPoint PPT Presentation
Transcript of Finding the Missing Side
Finding the Missing Side
Year 11
Here’s the problem
An anchor cable shelves at 50º. You are aware that 24m of cable has been let out. What is the depth of water?
50º
24m x
With
H, O & A50º
24m x
50º
24m x
A
OH
50º
24m x
A
OH
SOHCAHTOA
If given the opposite and the hypotenuse, use: sin = opposite
hypotenuseIf given the adjacent and the hypotenuse, use:
cos = adjacenthypotenuse
If given the opposite and the adjacent, use:tan = opposite
adjacent
50º
24m x
A
OH
As there is nothing on the
Adjacent side, we use
Sin θ =
OH
must
Sine
50º
24m x
A
OH
Sin θ =OH
Sin 50º =x
24
Sin θ =OH50º
24m x
A
OH
50º
24m x
A
OH
Sin θ =OH
Sin 50º =
x24
We need to get x = ……
B = AC
If B = 2 and C = 5, find O
2 = A5
X 5 X 5
10 = A
Multiply both sides by 5
Algebra revision
50º
24m x
A
OH
Sin θ =OH
Sin 50º =x
24
50º
24m x
A
OH
( x 24)(24 x)
Sin θ =OH
Sin 50º =x
24
50º
24m x
A
OH
( x 24)(24 x)
Sin θ =OH
Sin 50º =x
24
A
OH
50º
24m x
Sin θ =
OH
Sin 50º =
x24
x = 24 x Sin 50º
A
H
50º
24m xO
Sin θ =OH
Sin 50º =
x24
x = 24 x Sin 50º
Now use calculator and
round off
50º
24m x
A
OH
x = 18.39m
Sin θ =OH
Sin 50º =
x24
x = 24 x Sin 50º
Therefore…
The depth of water is 18.39m
50º
24m x
The six steps are:-
1. Draw diagram
2. Label inside
3. Select ratio
4. Substitute values
5. Rearrange
6. Use Calculator
An example
Ex 7B 1e
1. Draw diagram
2. Label inside
3. Select ratio
4. Substitute values
5. Rearrange
6. Use Calculator
Your Turn
Page 251 Ex 7B Qu1 a, d, f, g, h, i
A real problem
UHF Communication is a radio system that works on line of sight. Why did the ships helicopter lose communication when only at an altitude of XXm and range of XXm
Colombus could work it out!!
Solution
I
Finding the length of the Hypotenuse
40cm
75º
ySome
problems are tricky
Algebra Revision
B = AC
If B = 3 and A = 12, find C
3 = 12C
X C
3C = 12
Multiply both sides by C
Divide both sides by 3
3 3
C = 4Note how the C and 3 have swapped places
40cm
75º
y
A
OH
Cos θ =
AH
40cm
75º
y
A
OH
Cos θ =
AH
40cm
75º
y
A
OH
Cos 75º =
40y
Cos θ =
AH
40cm
75º
y
A
O HCos 75º
=
40y
Cos θ =
AH
40cm
75º
y
A
O H
( x y )( y x ) Cos 75º =
40y
Cos θ =
AH
40cm
75º
y
A
O H
( x y )( y x ) Cos 75º =
40y
Cos θ =
AH
40cm
75º
y
A
O Hy x Cos 75º = 40
Cos 75º =
40y
Cos θ =
AH
40cm
75º
y
A
O Hy x Cos 75º = 40
Cos 75º =
40y
Cos θ =
AH
40cm
75º
y
A
O H
Cos 75º
Cos 75ºy x Cos 75º = 40
Cos 75º =
40y
Cos θ =
AH
40cm
75º
y
A
O H
Cos 75º
Cos 75ºy x Cos 75º = 40
Cos 75º =
40y
Cos θ =
AH
40cm
75º
y
A
O H
y =40
Cos 75º
Cos 75º =
40y
Cos θ =
AH
40cm
75º
y
A
O H
y = 154.55 cm
y =40
Cos 75º
Cos 75º =
40y
Cos θ =
AH
An example
7B 2e
Your turn
Page 251 Ex 7BQu2 a, c, d, f, i, lQu3 do any five
