Download - Chapter 2 Solution
![Page 1: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/1.jpg)
Step 3:
SOLVING THEPROBLEMS
![Page 2: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/2.jpg)
The first thing we must do to solve thisproblem is find how long Wilfred and John willbe on the ride before Francis and Kieran getoff.
![Page 3: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/3.jpg)
The first thing we must do to solve thisproblem is find how long Wilfred and John willbe on the ride before Francis and Kieran getoff.
We can do this by subtracting the time ofWilfred and John’s boarding of the ride fromthe time when Francis and Kieran get off.
![Page 4: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/4.jpg)
The first thing we must do to solve thisproblem is find how long Wilfred and John willbe on the ride before Francis and Kieran getoff.
We can do this by subtracting the time ofWilfred and John’s boarding of the ride fromthe time when Francis and Kieran get off.
35Time that Francis and Kieran will get off
![Page 5: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/5.jpg)
The first thing we must do to solve thisproblem is find how long Wilfred and John willbe on the ride before Francis and Kieran getoff.
We can do this by subtracting the time ofWilfred and John’s boarding of the ride fromthe time when Francis and Kieran get off.
35 - 21Time that Francis and Kieran will get off
How long ‘F’ and ‘K’ have been on the ride when ‘W’ and ‘J’ get on
![Page 6: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/6.jpg)
The first thing we must do to solve thisproblem is find how long Wilfred and John willbe on the ride before Francis and Kieran getoff.
We can do this by subtracting the time ofWilfred and John’s boarding of the ride fromthe time when Francis and Kieran get off.
35 - 21 = 14Time that Francis and Kieran will get off
How long ‘F’ and ‘K’ have been on the ride when ‘W’ and ‘J’ get on
![Page 7: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/7.jpg)
This tells us that Francis and Kieran will be getting off the ferris wheel in 14 minutes.
![Page 8: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/8.jpg)
This tells us that Francis and Kieran will be getting off the ferris wheel in 14 minutes.
Therefore we can conclude that Wilfred and John will have been on the ferris wheel for 14 minutes.
![Page 9: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/9.jpg)
Now that we know how long they have been onthe ride, we must find what height Wilfred and John are at when Francis and Kieran get off.
![Page 10: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/10.jpg)
Now that we know how long they have been onthe ride, we must find what height Wilfred and John are at when Francis and Kieran get off.
We can do this by solving one of the equationsthat we derived a few minutes ago with t=14substituted into it.
![Page 11: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/11.jpg)
Now that we know how long they have been onthe ride, we must find what height Wilfred and John are at when Francis and Kieran get off.
We can do this by solving one of the equationsthat we derived a few minutes ago with t=14substituted into it.
***I will be using both the sine and cosine equation for this part.***
![Page 12: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/12.jpg)
Sine Equation
![Page 13: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/13.jpg)
Sine Equation
H(t)=75sin((2π/35)(t-10.75))+76
The Original Equation:
![Page 14: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/14.jpg)
Sine Equation
H(14)=75sin((2π/35)(14-10.75))+76
Substitute t=14 into the equation.
![Page 15: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/15.jpg)
Sine Equation
H(14)=75sin((2π/35)3.25)+76
Simplify the brackets.
![Page 16: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/16.jpg)
Sine Equation
H(14)=75sin(6.5π/35)+76
Simplify the brackets.
![Page 17: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/17.jpg)
Sine Equation
H(14)=75(0.5509)+76
Calculate the sine of the value inside the brackets.[Accurate to 4 decimal places.]
![Page 18: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/18.jpg)
Sine Equation
H(14)=41.3173+76
Simplify the equation.
![Page 19: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/19.jpg)
Sine Equation
H(14)=117.3173
Simplify the equation.
![Page 20: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/20.jpg)
Sine Equation
H=117.3173m
![Page 21: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/21.jpg)
Cosine Equation
![Page 22: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/22.jpg)
Cosine Equation
H(t)=-75cos((2π/35)(t-2.0))+76
The Original Equation:
![Page 23: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/23.jpg)
Cosine Equation
H(14)=-75cos((2π/35)(14-2.0))+76
Substitute t=14 into the equation.
![Page 24: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/24.jpg)
Cosine Equation
H(14)=-75cos((2π/35)12.0)+76
Simplify the brackets.
![Page 25: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/25.jpg)
Cosine Equation
H(14)=-75cos(24.0π/35)+76
Simplify the brackets.
![Page 26: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/26.jpg)
Cosine Equation
H(14)=-75(-0.5509)+76
Calculate the cosine of the value inside the brackets.[Accurate to 4 decimal places.]
![Page 27: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/27.jpg)
Cosine Equation
H(14)=41.3173+76
Simplify the equation.
![Page 28: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/28.jpg)
Cosine Equation
H(14)=117.3173
Simplify the equation.
![Page 29: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/29.jpg)
Cosine Equation
H=117.3173m
![Page 30: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/30.jpg)
This means that Wilfred and John were approximately 117 m off the ground when Francis and Kieran got off.
![Page 31: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/31.jpg)
We have just answered the first part of the question.
![Page 32: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/32.jpg)
We have just answered the first part of the question.
Now we must use this to answer the secondpart of the question:
![Page 33: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/33.jpg)
We have just answered the first part of the question.
Now we must use this to answer the secondpart of the question:
How much longer will it be until they reach the same height again?
![Page 34: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/34.jpg)
Some people might be confused by thisquestion, so lets take a look at it.
![Page 35: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/35.jpg)
Some people might be confused by thisquestion, so lets take a look at it.
This is the graph that we constructed earlier.
![Page 36: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/36.jpg)
Some people might be confused by thisquestion, so lets take a look at it.
The point on the graph is the point that we just found. (t=14)
![Page 37: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/37.jpg)
Some people might be confused by thisquestion, so lets take a look at it.
The height of this point is 117m.
![Page 38: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/38.jpg)
Some people might be confused by thisquestion, so lets take a look at it.
However, there is another point with a heightof 117m.
![Page 39: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/39.jpg)
Some people might be confused by thisquestion, so lets take a look at it.
Therefore there is another time that theywill reach a height of 117m.
![Page 40: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/40.jpg)
So, we are looking to find that second time which has a height of 117.3173m.
We can do this by making the equation equal to 117.3173m and solving it.
![Page 41: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/41.jpg)
So, we are looking to find that second time which has a height of 117.3173m.
We can do this by making the equation equal to 117.3173m and solving it.
***This time I will only use the Sine equation and explain the difference between the two methods afterwards.***
![Page 42: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/42.jpg)
Sine Equation
![Page 43: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/43.jpg)
Sine Equation
H(t)=75sin((2π/35)(t-10.75))+76
The Original Equation:
![Page 44: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/44.jpg)
Sine Equation
117.3173=75sin((2π/35)(t-10.75))+76
Substitute H(t)=117.3173 into the equation.
![Page 45: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/45.jpg)
Sine Equation
117.3173=75sinØ+76
Let ((2π/35)(t-10.7))=Ø to make it a little easier to work with.
![Page 46: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/46.jpg)
Sine Equation
41.3173=75sinØ
Simplify the equation.
![Page 47: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/47.jpg)
Sine Equation
0.5509=sinØ
Simplify the equation.
![Page 48: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/48.jpg)
Sine Equation
0.5834=Ø
Calculate the arc sine of 0.5509 [to four decimal places].
![Page 49: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/49.jpg)
Sine Equation
Now we have an angle.
The measure of this angle is 0.5834 radians
Another thing we know about this angle is that it’s sine is positive.
WHY IS THIS HELPFUL???
![Page 50: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/50.jpg)
Sine Equation
Well, when the sine of an angle is positive, it must either be in Quadrant 1 or 2.
This particular angle is in Quadrant 1.
HOW DO WE KNOW THIS???
Well, when the sine of an angle is positive, it must either be in Quadrant 1 or 2.
This particular angle is in Quadrant 1.
HOW DO WE KNOW THIS???
![Page 51: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/51.jpg)
Sine Equation
QIQII
QIII QIV
!/2
0!
3!/2
2!
Here we have the Unit Circle.
![Page 52: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/52.jpg)
Sine Equation
QIQII
QIII QIV
!/2
0!
3!/2
2!
The angle that separates Quadrants I and II is π/2.
[or 1.5707...]
![Page 53: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/53.jpg)
Sine Equation
QIQII
QIII QIV
!/2
0!
3!/2
2!
0.5834 is much less than 1.5708 so it lies in Quadrant I.
![Page 54: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/54.jpg)
Sine EquationThis means that to find the other point with the exact same sine value, it’s angle would have to be in Quadrant II.
![Page 55: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/55.jpg)
Sine EquationThis means that to find the other point with the exact same sine value, it’s angle would have to be in Quadrant II.
To find this angle we simply subtract our angle from π (End of Quadrant II).
![Page 56: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/56.jpg)
Sine EquationThis means that to find the other point with the exact same sine value, it’s angle would have to be in Quadrant II.
To find this angle we simply subtract our angle from π (End of Quadrant II).
π - Ø = Ø’
![Page 57: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/57.jpg)
Sine Equation
2.5582=Ø’
Here we have our Quadrant II angle.
![Page 58: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/58.jpg)
Sine Equation
sinØ=sinØ’
Now the two angles we have are related angles. This means their sine values are equivalent.
![Page 59: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/59.jpg)
Sine Equation
Ø=Ø’
And if we take the arc sine of both sides and we see that:
![Page 60: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/60.jpg)
Sine Equation
Ø=2.5582
Now we can continue where we left off.
Since...
![Page 61: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/61.jpg)
Sine Equation
Ø= ((2π/35)(t-10.7))
...AND...
![Page 62: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/62.jpg)
Sine Equation
2.5582= ((2π/35)(t-10.7))
...Then.
![Page 63: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/63.jpg)
Sine Equation
14.25= (t-10.7)
Now just simplify.
![Page 64: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/64.jpg)
Sine Equation
25= t
Now just simplify.
![Page 65: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/65.jpg)
Sine Equation
t=25Cosine Equation
![Page 66: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/66.jpg)
Cosine Equation
The Cosine equation works quite similarly to that of Sine.
![Page 67: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/67.jpg)
Cosine Equation
The Cosine equation works quite similarly to that of Sine.
The equation is a little bit different as we saw earlier but that’s about all that applies to the first part of the procedure.
![Page 68: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/68.jpg)
Cosine Equation
One of the major differences is when you simplify down to cosØ.
![Page 69: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/69.jpg)
Cosine Equation
One of the major differences is when you simplify down to cosØ.
First of all, this value is negative which means the angle is either in Quadrant II or III.
![Page 70: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/70.jpg)
Cosine Equation
One of the major differences is when you simplify down to cosØ.
First of all, this value is negative which means the angle is either in Quadrant II or III.
This one is in Quadrant II because it is 2.1542 which is less than π (3.1415), the end of Quadrant II.
![Page 71: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/71.jpg)
Cosine Equation
This means Ø’ is in Quadrant III and we must subtract Ø (angle that passes Q I and into II) from the full circle (2π) so it passes through Q IV and into III.
![Page 72: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/72.jpg)
Cosine Equation
This means Ø’ is in Quadrant III and we must subtract Ø (angle that passes Q I and into II) from the full circle (2π) so it passes through Q IV and into III.
After that, continue as shown in the Sine procedure until you get t=25 as your answer.
![Page 73: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/73.jpg)
From here, we now can finish the problem off.
t=25 minutes
![Page 74: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/74.jpg)
Since it takes 14 minutes to get from thestart to 117.3173 m the first time.
14
![Page 75: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/75.jpg)
Since it takes 14 minutes to get from thestart to 117.3173 m the first time...
...And it takes 25 minutes to get from the startto 117.3173 m the second time...
14 - 25
![Page 76: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/76.jpg)
...Then the time it takes to get from 117.3173 the first time to 117.3173 m the second timemust equal...
14 - 25 = ?
![Page 77: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/77.jpg)
11 minutes
![Page 78: Chapter 2 Solution](https://reader034.vdocuments.site/reader034/viewer/2022050703/5559c0e3d8b42a236c8b51d8/html5/thumbnails/78.jpg)
When Francis and Kieran get off “The Wheel”, Wilfred and John will be 117.3173 m from the ground.
It will take them 11 more minutes to get to that exact same height again.