Chapter 1 Practice Test

1. For each of the following problems

i. Determine the unknown quantities and represent these with appropriate variables.

ii. Determine the equations of the linear system that models the problem. Do not solve.

(a) Bart’s cell phone package costs $15 per month and an additional $0.10 per minute. Zoe pays $10 per monthand $0.12 per minute. Determine a linear system that could be used to find out when their monthly billswould be the same.

(b) Lloyd invested $1200 in an RRSP. He invested part in a technology fund that pays 15% per year and theremainder in bonds that pay 5% per year. Determine a linear system that could be used to find the amount heinvested in each part if he earned $160 in interest for the year.

(c) A jet flew from Toronto to Thunder Bay, a distance of 1290 km, in 2 h and 15 min. The plane flew into aheadwind on the trip to Thunder Bay, and with a tail wind on the trip back to Toronto. If the return trip took 2h, determine the speed of the plane and the speed of the wind.

(d) Monique is starting her own on-line business selling jewelry that she makes. It cost her $3000 to buy a newcomputer, and her monthly expenses for Internet service and materials average about $175 per month. If herrevenue for each month averages about $420, how long will it be before she starts to turn a profit?

2. Solve the system of equations graphically.

y = –2x + 8

3x – 4y = 12

3. For the following system of equations

i. determine, without solving, the number of solutions this system has. Show your work.

ii. verify that your answer in (i) is correct by solving the system using substitution.

2x + y = 12

15x + 3y = 18

4. Solve the system of equations using elimination.

6x – 8y = –58

4x + 5y = 13

5. Choose any two problems from question 1and solve them. Show all your work and provide a complete solution.

6. The following table shows the revenue generated from sales of skis and snowboard equipment in Canada from1994 to 1998 in thousands of dollars.

Year Ski EquipmentSales ($)

SnowboardEquipment Sales ($)

1994 60 000 29 7001995 45 910 21 6401996 41 530 39 9701997 32 100 31 5901998 41 100 49 000

(Source: Statistics Canada)

(a) Create a single scatter plot for both data sets.

(b) Draw the lines of best fit and determine the equation of each line.

(c) Which set of data is closer to being linear? Explain.

(d) Which set of data is changing at a faster rate? Explain how you know.

(e) Use the model to determine when snowboard sales surpassed ski sales in Canada.

(f) If you were an owner of a sports store, how would you use this information?

Chapter 1 Test: Solutions1. (a) Let C represent the monthly cost and t the number of minutes the cell phone was used for in the

month.

C = 0.10t + 15 represents Bart's costs

C = 0.12t + 10 represents Zoe's costs

(b) Let t represent the amount Lloyd invested in the technology fund and b the amount he invested inbonds.

t + b = 1200 represents the total money invested

0.15t + 0.05b = 160 represents the interest earned

(c) Let j represent the speed of the jet and w the speed of the wind.

Distance(km)

Time (h) Speed (km/h)

Toronto to Thunder Bay 1290 2.25 j – wThunder Bay to Toronto 1290 2 j + w

distance = speed × time

1290 = (j – w)2.25 represents the trip from Toronto to Thunder Bay

1290 = (j + w)2 represents the trip from Thunder Bay to Toronto

(d) Let m represent the money Monique either puts out or takes in from sales and t the number ofmonths.

m = 175t + 3000 represents her expensesm = 420t represents her revenue

2. y = –2x + 8 is most easily graphed with table of values or the slope–y-intercept method: slope = –2, y-intercept is 8.

3x – 4y = 12 is most easily graphed with x- and y-intercepts: x-intercept = 4, y-intercept = –3.

3. i. 2x + y = 12 15x + 3y = 18

y = –2x + 12 3y = –15x + 18

y = –2x + 12 y = –5x + 6

Both equations have different slopes, so there is only one point of intersection. This linear system hasone solution.

ii. y = –2x + 12 and y = –5x + 6 When x = –2, y = –2(–2) + 12

–2x + 12 = –5x + 6 = 4 + 12

–2x + 5x = 6 – 12 = 16

3x = –6

x = –2 The solution is (–2, 16).

4. 6x – 8y = –58 (×5) → 30x – 40y = –290

4x + 5y = 13 (×8) → 32 x + 40 y = 104 Add. 62x = –186

x = – 3When x = 3, 4(–3) + 5y = 13 –12 + 5y = 13

5y = 13 + 12 5y = 25 y = 5

The solution is (–3, 5).

5. (a) Using substitution

0.10t + 15 = 0.12t + 10 0.10t – 0.12t = 10 – 15

–0.02t = –5 t = 250

When t = 250, C = 0.10(250) + 15 = 25 + 15 = 40

The cell phone plans cost the same when 250 min of calls are placed. This results in a monthly bill of$40.00 under each plan.

(b) Using elimination

t + b = 1200 (×5) → 5t + 5b = 6000

0.15t + 0.05b = 160 (×100) → 15 t + 5 b = 16 000 Subtract. –10t = –10 000 t = 1000

When t = 1000, 1000 + b = 1200 b = 1200 – 1000 b = 200

$1000 was invested in the technology fund and $200 was invested in bonds.

(c) 1290 = 2.25j – 2.25w (×2) → 2580 = 4.5j – 4.5w

1290 = 2j + 2w (×2.25) → 2902.5 = 4.5 j + 4.5 w Add. 5482.5 = 9j

609.2 jWhen j 609.2,

1290 = (609.2 + w)2 1290 = 1218.4 + 2w

1290 – 1218.4 = 2w

71.6 = 2w

35.8 wThe speed of the jet is about 609 km/h and the speed of the wind is about 36 km/h.

(d) Using substitution 420t = 175t + 3000420t – 175t = 3000

245t = 3000 t 12.24

When t 12.24, m = 420(12.24) = 5140.80

She will make a profit shortly after 12 months. This will cover her expenses.

6. Answers will vary depending on whether the graph is drawn by hand or using technology. The solutionpresented here was done using the TI-83 Plus calculator.

(a)

(b)

Ski Sales Snowboard Sales

y = –5161x + 10345484 y = 4855x – 9656200

(c) The ski sales data are closer to being linear. The correlation coefficient is closer to 1/–1.

(d) The ski sales data are changing at a faster rate. The line of best fit has a slope of –5161, whereas thesnowboard line has a slope of 4855. This means that ski sales are decreasing at a faster rate thansnowboard sales are increasing.

(e) The model indicates that snowboard sales surpassed ski sales late in 1996.

(f) This information indicates that more people are spending their money on snowboard equipmentcompared to ski equipment. A sports store should reduce the amount of ski equipment it carries andincrease the amount of snowboard equipment. The model indicates that this trend will continue.

Chapter 2 Test

1. A straight path is to be built between points A(–35, 32) and B(29, –16), where the units represent metres.(a) How long is the path?(b) A bench is to be placed halfway along the path. Find the coordinates of the point at which the bench will

be placed.

2. A fish catching a small insect on the surface of a still pond causes a circular ripple. The radius of thecircle increases at a constant rate of 4 cm/s.

(a) Write an equation that describes the ripple exactly 5 s after the fish catches the insect.(b) How long does it take for the ripple to reach the edge of a rock that is 1 m east and 0.75 m north of the

point where the fish caught the insect?

3. Triangle ABC has vertices at A(1, 1), B(4, 5), and C(9, –5). Find the lengths and slopes of its sides anddetermine what type of triangle it is.

4. A circle has its centre at the origin and points P(–1, 13) and Q(11, a) are on the circle. Point Q is in thefirst quadrant.

(a) Find the value of a.(b) Find the equation of the perpendicular bisector of chord PQ and show that it passes through the centre of

the circle.

5. Quadrilateral PQRS has vertices P(2, 3), Q(8, 5), R(11, –4), and S(5, –6).(a) Show that the diagonals bisect each other.(b) Determine what type of quadrilateral PQRS is.

6. A house is to be connected to a new water main that runs along the line y = x – 1. The connection point atthe house has coordinates (2, 9), where the units represent metres. What length of plastic pipe is neededto connect to the water main at the closest point?

7. Jean-Luc is designing a triangular component for a snowmobile. On his design, the triangle has verticesS(–5, –3), T(–1, 9), and U(21, 3). Find the coordinates of the centroid of this triangle. Why might Jean-Luc need to locate this point?

8. Triangle JKL has vertices at J(–9, 2), K(3, 10), and L(11, –6). Prove that each midsegment is parallel tothe opposite side and is half the length of the parallel side.

Chapter 2 Test: Solutions1. (a) The length of the path is = 80 m.(b) The midpoint of AB is the point (, ). The bench should be placed at point (–3, 8).

2. (a) Five seconds after the fish catches the insect, the radius of the ripple is 5 × 4 or 20 cm. If the origin is

the point at which the fish catches the insect, the equation of the ripple is x2 + y2 = 202 or x2 + y2 =400.

(b) Keeping the origin at the same place, the rock is at point (100, 75) and its distance from the origin is =125 cm. At 4 cm/s, it will take or 31.25 s for the ripple to reach the rock.

3. Find the lengths and slopes of the sides.

For AB, length = = 5, slope = = .For BC, length = 11.2, slope = = –2.For AC, length = = 10, slope = = –.No sides have the same lengths, but the slopes of sides AB and AC are negative reciprocals. AB isperpendicular to AC. The triangle is a right triangle.

4. (a) The equation of a circle with its centre at the origin may be written x2 + y2 = r2. If P(–1, 13) is on the circle,

(–1)2 + 132 = r2

170 = r2

The equation of the circle is x2 + y2 = 170.If Q(11, a) is on this circle,

112 + a2 = 170

a2 = 49Since Q is in the first quadrant, a = 7 (we ignore the negative root of 49).

(b) To find the equation of the perpendicular bisector, find its slope and a point on this line.The slope of PQ is = –.The slope of the perpendicular bisector of PQ is the negative reciprocal of –, which is 2. The

perpendicular bisector passes through the midpoint of PQ.The midpoint of PQ is (, ) or (5, 10).The equation of the perpendicular bisector is y = 2x + b, and since it passes through (5, 10),

10 = 2(5) + b0 = b

The equation of the perpendicular bisector is y = 2x. This is a line with slope 2 and y-intercept 0. Thisline passes through the origin. Since the centre of the circle is at the origin, the perpendicular bisectorof PQ passes through the centre.

5.

(a) The midpoint of PR = (, ) or (6.5, –0.5).The midpoint of QS is (, ) or (6.5, –0.5).Since the midpoint of each diagonal is the same point, the diagonals must cross at this point. The

diagonals bisect each other.

(b) Since the diagonals of PQRS bisect each other, we know that it is a parallelogram. It might be arhombus, a rectangle or a square. To determine what type of quadrilateral PQRS is, find the lengthsand slopes of the sides.

Length PQ = 6.3, slope of PQ = = Length QR = 9.5, slope QR = = –3Length RS = 6.3, slope RS = = Length SP = 9.5, slope SP = = –3From the slope and length information we see that opposite sides of PQRS are equal and parallel, so

PQRS is at least a parallelogram. Also, the slope information shows that the slopes of adjacent sidesare negative reciprocals. So adjacent sides are perpendicular, which means PQRS is a rectangle or asquare. However, the adjacent sides are not equal in length, so it is not a square. PQRS is a rectangle.

6. Let W be the connection point with the water main and let H be the connection point with the house. Weneed to find the coordinates of W. For this we need to find the equation of the line containing HW.

The slope of line segment HW is –, since HW is perpendicular to the given line and its slope must be thenegative reciprocal of . The equation of the line containing HW is y = –x + b. Since H(2, 9) is on the line, then

9 = –+ b12 = b

The equation is y = –x + 12.A graphing calculator could be used to find the point of intersection

Or, using algebra, line HW will intersect the line of the water main wherex – 1 = –x + 12

x = 13x = 6y = – 1y = 3

The connection should be at W(6, 3). The length of plastic pipe required is 7.2 m.

7.

Method AThe coordinates of the centroid may be found by taking the mean of the coordinates of the vertices.The median is at (, ) or (5, 3).

Method B The coordinates of M are (, ) or (8, 0).Slope of TM = = 1The equation of line containing TM is y = –x + b. Since (–1, 9) is on this line,

9 = –1(–1) + b8 = b

The equation is y = –x + 8.The coordinates of N are (, ) or (10, 6). Slope of SN = = 0.6The equation of line containing SN is y = 0.6x + b. Since (–5, –3) is on this line,

–3 = 0.6(–5) + b0 = b

The equation is y = 0.6x.At the point where these two medians intersect,

–x + 8 = 0.6x

8 = 1.6x

5 = xy = –5 + 8y = 3

The median is at point (5, 3).A graphing calculator can be used to find the intersection point.

When vehicles, equipment, appliances or household items are designed, an important feature to beconsidered is the centre of mass. To locate its position, the centres of mass of all the component parts areused. The centroid is the position of the centre of mass for a thin triangular component. Jean-Luc wouldneed its coordinates to calculate the centre of mass of the snowmobile.

8. Label the midpoints of sides JK, KL, and LJ as P, N, and M as shown.

The coordinates of these points are P(, ) or P(–3, 6), N(, ) or N(7, 2), and M(, ) or M(1, –2).Compare slopes.Slope PN = = – and slope JL = = –These slopes are the same. Midsegment PN is parallel to the opposite side JL.Slope PM = = –2 and slope KL = = –2These slopes are the same. Midsegment PM is parallel to the opposite side KL.Slope MN = = and slope JK = = These slopes are the same. Midsegment MN is parallel to the opposite side JK. Each midsegment isparallel to the opposite side.Compare lengths.Length PN = 10.77Length JL = 21.54The length of midsegment PN is one half the length of the parallel side.Length PM = 8.94Length KL = 17.89The length of midsegment PM is one half the length of the parallel side.Length MN = 7.21Length JK = 14.42The length of midsegment MN is one half the length of the parallel side. Each midsegment is half thelength of the parallel side.It has been proved that each midsegment is parallel to the opposite side and is half the length of theparallel side.

Chapter 3 Test1. Expand and simplify.

(a) (x + 4)(x – 7) (b) (2x – 9)(3x – 5) (c) –4(x + 6)2

2. Factor each expression.

(a) x2 – 3x – 28 (b) 81x2 – 25 (c) 25x2 – 20x + 4 (d) 24x2 + x – 10

3. Sketch the graph of each quadratic relation. Clearly label the zeros and the vertex. (a) y = (x – 6)(x + 2) (b) y = –(x – 1)(x – 9)

4. Solve each of the following.

(a) x2 + 7x + 12 = 0 (b) x2 – 2x = 48 (c) 5x2 + 4x = 2 – 5x

5. A model rocket is equipped with a motion detector that measures the height of the rocket at 30 sintervals. On a recent flight, the detector recorded the following data.

Time (s) Height (m)0 0

30 31560 54090 675

120 720150 675180 540210 315240 0

(a) By determining first and second differences, decide whether the data displays a linear, quadratic, orother relationship.

(b) Create a scatter plot of the data and graph the line or curve of best fit.(c) Determine the equation of the axis of symmetry and the maximum height reached by the rocket.(d) Derive an equation that models the relationship.(e) The first zero corresponds to the point at which the rocket leaves the ground. Explain the

significance of the second zero.(f) How high is the rocket after 1 min 40 s?(g) At what point in time is the rocket 400 m above the ground?

6. Accountants for the Hitech running shoe company have determined that the relation P = –2x2 + 20x – 42models the company’s profit for the next quarter. P represents the profit (in $ hundred thousands), and xrepresents the number of pairs of shoes sold (in hundred thousands).(a) Draw a graph that represents this relationship.(b) Determine the range in the number of pairs of shoes the company must sell to yield a profit.(c) How many pairs of shoes must they sell to maximize the profit?(d) What is the maximum profit the company can expect to earn in the quarter?(e) Determine the number of pairs of shoes sold if a profit of $600 000 is realized.

Chapter 3 Test: Solutions

1. (a) (x + 4)(x – 7) (b) (2x – 9)(3x – 5) (c) –4(x + 6)2

= x2 – 7x + 4x – 28 = 6x2 – 10x – 27x + 45 = –4(x + 6)(x + 6)

= x2 – 3x – 28 = 6x2 – 37x + 45 = –4(x2 +12x + 36)

= –4x2 – 48x – 144

2. (a) x2 –3x – 28 (b) 81x2 – 25 = (x – 7)(x + 4) = (9x – 5)(9x + 5)

(c) 25x2 – 20x + 4 (d) 24x2 + x – 10 = (5x – 2)(5x – 2) = (3x + 2)(8x – 5)

3. (a) y = (x – 6)(x + 2) (b) y = –(x – 1)(x – 9) zeros (x-intercepts) at 6 and –2 zeros (x-intercepts) at 1 and 9

axis of symmetry x = axis of symmetry x = x = 2 x = 5

at the vertex, x = 2 at the vertex, x = 5 y = (2 – 6)(2 + 2) y = –(5 – 1)(5 – 9) y = (–4)(4) y = –(4)(–4) y = –16 y = 16 (2, –16) are the coordinates of (5, 16) are the coordinates of the vertex. the vertex

4. (a) x2 +7x + 12 = 0 (b) x2 –2x = 48 (c) 5x2 + 4x = 2 – 5x

(x + 3)(x + 4) = 0 x2 – 2x – 48 = 0 5x2 + 4x + 5x – 2 = 0

x = –3 and x = –4 (x – 8)(x + 6) = 0 5x2 + 9x – 2 = 0 x = 8 and x = –6 (5x –1)(x + 2) = 0

x = and x = –2

5. (a) Time (s) Height (m) FirstDifference

SecondDifference

0 030 315 31560 540 225 –9090 675 135 –90

120 720 45 –90150 675 –45 –90180 540 –135 –90210 315 –225 –90240 0 –313 –90

The second differences are constant, so the relationship is quadratic.

b)(c) The rocket reaches its maximum height at the vertex. In thiscase, the graph has zeros at 0 and 240. The axis of symmetry is x =or x = 120. From the table of values, y = 720 when x = 120. After120 s have elapsed, the rocket reaches its maximum height of 720m.

(d) From the zeros, h = a(t – 0)(t – 240). And (120, 720) is a known point on the graph.

720 = a(120 – 0)(120 – 240) 720 = a(120)(– 120) 720 = –14 400a

–0.05 = a

The equation h = –0.05t(t – 240) or h = –0.05t2 + 12t models this relationship.

(e) The second zero corresponds to the rocket’s return to the ground. This happens at 240 s.

(f) 1 min 40 s = 100 s

h = –0.05t2 + 12t

= –0.05(100)2 + 12(100)

= –500 + 1200= 700The rocket is 700 m highafter 1 min 40 s.

(g) When the rocket is 400 m high, h = 400. Substitute and solve for t.

h = –0.05t2 + 12t

400 = –0.05t2 + 12t

0 = –0.05t2 + 12t – 400

0 = –0.05(t2 – 240t +8000)

0 = –0.05(t – 200)(t – 40)t = 200 and t = 40

The rocket is 400 m high after 40 s and again after 200 s.

6. (a) P = –2x2 + 20x – 42

= –2(x2 – 10x + 21) = –2(x –7)(x – 3)The parabola opens down and has zeros at 7 and 3.The axis of symmetry is x = or x = 5.When x =5,

P = –2(5)2 + 20(5) – 42

= –50 + 100 – 42 = 8The vertex is at (5, 8).

(b) Since the zeros are at 3 and 7, the company only generates a profit if it sells between 300 000 and700 000 pairs of shoes.

(c) The maximum profit occurs when x = = 500 000.(d) When x = 500 000, the profit is $800 000.(e) A profit of $600 000 is represented by P = 6.

P = –2x2 + 20x – 42

6 = –2x2 + 20x – 42

0 = –2x2 + 20x – 48

0 = x2 – 10x + 24 0 = (x – 6)(x – 4) x = 6 and x =4The company makes a profit of $600 000 if they sell either 400 000 or 600 000 pairs of shoes.

Chapter 4 Test1. Find the quadratic relation in vertex form that

(a) has its vertex at (5, 8) and passes through the point (11, –46)(b) has zeros –3 and 7 and passes through the point (5, –8)

(c) is defined by y = 2x2 + 12x – 17

2. Describe the transformations applied to the graph of y = x2 to obtain the graph of each quadratic relation.Sketch the graph of the relation by hand on graph paper.

(a) y = 2(x + 3)2 – 6 (b) y = –x2 + 11

3. Find the roots of each of the following quadratic equations using the most appropriate algebraic method.Express your answers to two decimal places, where appropriate.

(a) 3(t – 7)2 – 15 = 0 (b) 2x2 + 4x – 30 = 0 (c) 3x2 – 2x – 11 = x2 – 5x

4. A model rocket is launched from a 5 m high pad straight up into the air with an initial velocity of 150

m/s. The height of the rocket h, in metres, is modelled by h = –5t2 + 150t + 5, where t is the elapsed timein seconds.(a) Use the method of completing the square to determine the maximum height attained by the rocket.(b) For what length of time was the rocket above 730 m?

5. The Sudbury Transit System carries 15 000 bus riders per day for a fare of $1.90. The city wishes toincrease ridership in an effort to reduce car pollution, while at the same time maximizing the Transitrevenues. The city conducted a survey and the results indicated that the number of riders would increaseby 500 for every $0.05 decrease in fare.(a) Write an equation to represent the daily revenue.(b) What fare will produce the greatest revenue? What is the expected daily ridership?(c) The city calculates that it must take in $24 675 per day to break even. What ticket prices would

enable the city to earn a profit?

6. A ball is thrown into the air from the top of a bridge. Its height above the ground is tracked by a stop-motion camera from the instant the ball is released. The table below shows some of the recorded data.

Time since releasein seconds

Height above theground in metres

0.25 1311.0 1501.5 1604.5 1655.0 1586.5 1207.0 1037.5 83

(a) Create a scatter plot of the data and draw a curve of best fit(b) Can you use a table of first differences to determine the type of relationship that exists? Explain.(c) Estimate the maximum height of the ball and the time at which it occurred.(d) Find an algebraic model of the relation between height and time.(e) Use your model to predict the height of the ball after 8.5 s.(f) Use your model to predict the time when the ball will hit the ground.(g) Use your model to predict the height of the bridge.(h) Determine a quadratic regression model using technology and compare it to your algebraic model. (i) Use the regression model to answer questions (c), (f) and (g).

Chapter 4 Test: Solutions

1. (a) Since the vertex is (5, 8), the relation expressed in vertex form is y = a(x – 5)2 + 8.

Substitute the coordinates of the point (11, –46) into the relation and solve for a.

–46 = a(11 – 5)2 + 8

–46 = 36a + 8–54 = 36a

a = –36

54 = –

2

3

The equation of the relation is y = –2

3 (x – 5)

2 + 8

(b) Since the zeros are –3 and 7, the relation can be expressed as y = a(x + 3)(x – 7).Substitute the coordinates of the point (5, –8) into the relation and solve for a.–8 = a(5 + 3) (5 – 7)–8 = a(8)(–2)

a = 2

1

The equation of the relation in factored form is y = 2

1 (x + 3)(x – 7).

The x-coordinate of the vertex is located at the midpoint of the zeros, x = 2

)73( +− = 2.

Substitute x = 2 into the relation and solve for the y-coordinate of the vertex,

y = 2

1(5)(–5) = –

2

25.

The equation of the relation is y = 2

1 (x –2)

2 +

2

25

(c) The relation does not factor completely. The vertex can be found using partial factors or bycompleting the square.

y = 2x2 + 12x – 17

y = 2(x2 + 6x) – 17

y = 2(x2 + 6x + 9 – 9) – 17

y = 2(x2 + 6x + 9) – 18 – 17

y = 2(x + 3)2 – 35

y = x(2x + 12) – 17 or y = 2x(x + 6) – 17The points (0, –17) and (–6, –17) are on the graph.Using symmetry, find the vertex at x = –3.Substitute x = –3 and solve for the y-coordinate.

y = 2(–3)2 + 12(–3) – 17 = –35

The equation is y = 2(x + 3)2 – 35.

2. (a) Translate the graph 3 units to the left, vertically stretch the graph by a factor of 2, and thentranslate the graph down 6 units.(b) Vertically compress the graph by a factor of 2, reflect the graph about the x-axis, and then translate

the graph upwards 11 units.

Obtain the graphs by applying the transformations, or by locating the vertex and using first differences to locate

other points on the graph.

3. (a) 3(t – 7)2 – 15 = 0

3(t – 7)2 = 15

(t – 7)2 = 5

t =̇ 7 ±2.24

t = 7.24 and t = 4.76

(b) 2x2 + 4x – 30 = 0

2(x2 + 2x –15) = 0

2(x + 5)(x – 3) = 0

x = –5 and x = 3

(c) 3x2 – 2x – 11 = x

2 – 5x

2x2

+ 3x – 11 = 0

x =̇ 1.71 and x =̇ –3.21

4. (a) h = –5t2 + 150t + 5

h = –5(t2 – 30t) + 5

h = –5(t2 – 30t + 225 – 225) + 5

h = –5(t2 – 30t + 225) +1125 + 5

h = –5(t – 15)2 + 1130

After 15 s, the rocket will reach its maximum height of 1130 m.

(b) Solve (b) by completely factoring the quadratic formula or by isolating the perfect square and solving

for t.

730 = –5(t – 15)2 + 1130

–400 = –5(t – 15)2

80 = (t – 15)2

t =̇ 15 ± 8.94

t = 23.94 and t = 6.06

The rocket was above 730 m between 6.06 s and 23.94 s for a total time of 17.88 s.

5. Let x represent the number of $0.05 decreases.(a) R = (15 000 + 500x)(1.90 – 0.05x)

(b) To find the greatest revenue, locate the vertex. The easiest way to do this is to find the zeros of Rabove because it is in factored form. Alternatively, expand R and use partial factoring to locate twopoints on the graph. The midpoint of these points will give the x-coordinate of the vertex.

Let R = 0 and find the zeros by letting 15000 + 500x = 0 and 1.90 – 0.05x = 0. This gives zeros at x = –

30 and x = 38. The maximum revenue will occur at x = 2

38)+ (-30 = 4.

For maximize revenue, the number of $0.05 decreases is 4, and the fare is $1.70. The expected ridershipis (15 000 + 500(4)) or 17 000 riders per day.

(c) To calculate the break-even point, let R = 24 675 and solve for x.

(15 000 + 500x)(1.90 – 0.05x) = 24 675

28 500 – 750x + 950x – 25x2 – 24 675 = 0

–25x2

+ 200x + 3825 = 0Solve by factoring or using the quadratic formula.

–25(x2 – 8x – 153) = 0

–25(x – 17)(x + 9) = 0

x = 17 and x = –9

Calculate the fares.

1.90 – 0.05(17) = $1.05 and 1.90 – 0.05(–9) = $2.35. Since the parabola opens downwards, a farebetween $1.05 and $2.35 will enable to city to earn a profit.

6. (a) Students should create a scatter plot with appropriate labels and scales. Their hand drawn curve ofbest fit should resemble the graph produced below.

(b) A table of first differences cannot be used to determine the relationship because the increments in thetimes are not constant.

(c) Answers will vary. Appropriate solutions should have a maximum height near 175 m at a time near 3s.

(d) Given the approximated vertex in (c), one possible algebraic model is h = a(t – 3)2 + 175. Select one

other point on the curve of best fit and solve for a. The point (5, 155) appears to be on the curve ofbest fit.

155 = a(5 – 3)2 + 175

–20 = 4a

a = –5

Answers between –4 and –6 are appropriate.

One possible algebraic model is h = –5(t – 3)2 + 175.

(e) h = –5(8.5 – 3)2 + 175 = 23.75 m

(f) The ball hits the ground when h = 0.

0 = –5(t – 3)2 + 175

35 = (t – 3)2

t =̇ 8.92 and t =̇ –2.92

The negative root is extraneous to the problem. The ball hits the ground about 8.9 s after it is thrown.

(g) Find the height of the bridge at t = 0.

h = –5( –3)2 + 175 = 139 m

(h) The TI-83 Plus calculator can be used to produce the regression equation

Expand the algebraic model to compare it to the algebraic model.

h = –5(t – 3)2 + 175

= –5(t2 – 6t + 9) + 175

= –5t2 + 30t – 130

The algebraic model and the regression model are very similar. The regression model has a very strongcorrelation to the data. The graphs of both models show a very good fit to the data.

(i) You may find the maximum height of 173.7 m at 3.2 s using the

maximum command on the CALCULATE menu, as shown in the

screen. Adjust the WINDOW settings and use the zero command on the

CALCULATE menu to find the time when the ball hits the ground.

This occurs at t = 9.15 s, as shown below. The height of the bridge

corresponds to the height at time t = 0. Use the Trace feature to find

that the height of the bridge is about 123.6 m.

Chapter 5 Test

1. Determine the values of the three primary trigonometric ratios for ∠BAC in each of the followingtriangles.(a) (b)

2. Find the measures of the indicated sides or angles in each of the following diagrams.(a) (b) (c)

(d) (e)

3. Consider this diagram.

(a) Explain why ∆ABZ ~ ∆XBY.

(b) Explain why ∆ABC is also similar to the triangles in (a). What is the order in which the vertices of

∆ABC match the vertices of ∆XBY?

(c) Show how to use the fact that all three triangles are similar to compute the length of AB.

(d) Show how to determine the ratios of the area and the perimeter of ∆ABC to the area and the

perimeter of ∆ABZ without actually computing any of the values.

4. An underground parking lot has an access ramp that slopes down at a 20° angle from the horizontal. Thebuilding code for such a ramp indicates that the slope must not be greater than 1 : 5 for safety reasons.Prove whether or not the ramp meets the code requirements.

5. Triangle ABC is an isosceles triangle with equal sides of length 20 cm. The angle between the equalsides is 40°.(a) Find the length of the altitude of this triangle.(b) Find the length of the base of this triangle.

6. To determine the height of a bridge support column, a surveyor uses a transit mounted on a tripod. Thetransit sits vertically 1.5 m above ground level and is placed 20 m from the base of the column. Theangle to the top of the column is 60° from the horizontal. (a) Determine the height of the column.(b) Determine the transit’s angle of declination from the horizontal to the base of the column.

7. ABCD is a parallelogram in which AB || DC and AD || BC. ∠A = 50°, AB = 5 cm, and BC = 6.8 cm.(a) Compute the area of ABCD.(b) Determine the length of the diagonal BD.

Chapter 5 Test: Solutions1. (a) Using the Pythagorean theorem,

AB2 = AC2 + BC2

= 162 + 242

= 832Therefore, AB = 832 28.84

sin ∠BAC =

16

28.84 0.555

cos ∠BAC =

24

28.84 0.832

tan ∠BAC =

16

24 0.667

(b) AC = 7 and BC = 8.

Thus, AB = 72

+ 82

= 113 10.63

sin ∠BAC = 63.10

8 0.753

cos ∠BAC = 63.10

7 0.659

tan ∠BAC = 7

8 1.143

2. (a)

a

10 = sin 50° 0.766

a = 10 × 0.766 7.66

10

b = cos 50° 0.643

b = 643.0

10 15.552

(b) tan c = 16

20 = 1.25

c = tan–1(1.25) 51.3°

(c) e = 502

− 322

= 1476 38.42

sin d = 50

32 = 0.64

d = sin–1(0.64) 39.8°

(d) tan f = 2

5 = 2.5

f = tan–1(2.5) 68.2°

(e) Label the diagram to make it easier to discuss angles.

g = ∠DCE

∠DCE is the slope angle of the line AC.

Therefore, slope of AC = tan ∠DCE.

slope angle ∠DCE = tan–1(slope of AC)

= tan–1

−−

−

04

08

= tan–1(–2)

–63.4°

While the slope angle is negative, the angle g is positive. Therefore, g is about 63.4°.

Similarly, ∠ABC is the slope angle of the line AB.

∠ABC = tan–1(slope of AC)

= tan–1

−−−

−

)6(4

08

= tan–1(4)

76.0°

Therefore, h 180° – (63.4° + 76.0°) 40.6°.

3. (a) ∠ABZ = ∠XBY (same angle)

∠XYB = ∠AZB = 90°

Therefore, ∆ABZ ~ ∆XBY (AA~).

(b) ∠ABC = ∠YBX (same angle)

∠BAC = ∠BYX = 90°

Therefore, ∆ABC ~ ∆YBX (AA~).

(c) Since ∆ABZ ~ ∆XBY,

AB

XB =

AZ

XY =

8

4

Calculate XB using the Pythagorean theorem.

XB = 52

+ 42

= 41 6.40

AB

6.40 =

8

4 = 2

Therefore, AB 12.80.

(d) In similar triangles, the ratio of the perimeters equals the ratio of the corresponding sides.

Since ∆ABC ~ ∆YBX ~∆ZBA,

ZBA

ABC

∆

∆

ofperimeter

ofperimeter =

ZB

AB

YB

ZB =

XY

AZ =

4

8, so ZB = 2(YB) = 10.

ZBA

ABC

∆

∆

ofperimeter

ofperimeter

10

80.12 = 1.28

The ratio of the areas is the square of 1.28, or approximately 1.64.

4. The following diagram shows that the slope of the ramp is the same as the angle of declination from thehorizontal.

The slope of the ramp should not exceed 1 : 5. That means the maximum slope angle is

slope angle = tan–1

5

1

= tan–1(0.2)

11.3°

Since the slope of the ramp is 20°, it exceeds the safety limit of the building code by about 8.7°.

5. As shown in the diagram, the altitude of an isosceles triangle bisects the angle at the top vertex and also isthe median to the base.

(a) In ∆CAD,

cos ∠CAD =

AD

AC

cos 20° =

AD

20

Therefore, AD = 20 cos 20° 18.79. The length of the altitude of this triangle is approximately 19 cm.

(b) Using the same triangle,

sin ∠CAD =

CD

AC

sin 20° =

CD

20

Therefore, CD = 20 sin 20° 6.84.

The length of the base of this triangle is 2 × CD 14 cm.

6. Draw a labeled diagram.

(a)In ∆ATB,

tan ∠ATB = TB

AB

tan 60° = 20

AB

AB = 20 tan 60° 34.6 The height of the column is about (34.6 + 1.5) m, or 36.1 m.

(b) In ∆TBC,

tan ∠BTC = BT

BC =

20

5.1 = 0.075

∠BTC = tan–1(0.075) 4.3°

The angle of declination of the transit from the horizontal is

about 4.3°.

7. Draw a labeled diagram showing what needs to be found.

(a) The area formula for a parallelogram is area A = bh, where b isthe length of the base and h is the altitude of the parallelogram.

In the diagram, b = AD = 6.8 cm and h = BX.

In ∆BXA,

sin A = BA

BX

sin 50° = 5

BX

Therefore, BX = 5 sin 50° 3.8

The area of ABCD is approximately (6.8 × 3.8) 25.8 cm2. (b) The length BX 3.8 cm was found in (a). If the length of AX is also known, then the length of XD can

be computed. The Pythagorean theorem can then be used in ∆BXD to compute the length of BD.

In ∆BXA,

cos A = BA

AX

cos 50° = 5

AX

Therefore, AX = 5 cos 50° 3.2 DX = (6.8 – 3.2) 3.6 cmUsing the Pythagorean theorem, BD2 = BX2 + DX2

= 3.62 + 3.82

= 27.4

BD = 27.4 5.2

The length of BD is about 5.2 cm.

Chapter 6 TestPlease note that, in addition to being marked according to an objective scale, specific questions will beassessed for demonstrating evidence of having attained achievement levels.

1. In each case, find the value of the indicated unknown and give reasons for your choice of strategy.

(a) In ∆ABC, ∠B = 90°, a = 15 m, b = 40 m, ∠C = n

(b) In ∆XYZ, x = 12 cm, y = 15 cm, z = 20 cm, ∠Z = n

(c) In ∆DEF, find the measure of the shortest side, if ∠D = 63°, ∠C = 47°, and d = 8 cm.

2. In a book about Navajo crafts, the diagram of a hand made rug shows triangular patterns with side

lengths 5 cm and 12 cm surrounding an angle of 50°. A large wall hanging for an art gallery is to bemade using the pattern but the triangles in the hanging will have an area 25 times as large as the diagram.What will be the measurement of

(a) the third side of the diagram?

(b) the corresponding side of the actual hanging?

(c) the opposite angle in the actual hanging?

3. Solve ∆ABC if a = 13 cm, ∠ABC = 66°, and c = 9 cm.

4. A ship in distress sends up flares that are visible to two other ships. The Athabasca notes that the flares

are at N41°E of its position. The Britannia, which is 12.7 nautical miles due east of the Athabasca sees

the flares at a bearing of 310°. Will the Athabasca or the Britannia reach the ship in distress first if theytravel at equal speeds?

5. Jose has a summer job working at a conservation site with an ecologically sensitive bog area. A floatingwalkway is to be constructed so visitors can see the plant and animal life without causing damage. Aviewing platform already exists directly north of the visitors’ centre and the walkway is planned tostretch from the viewing platform to the other side of the bog area. From a position 300 m directly east

of the visitors’ centre, Jose records that the viewing platform is located N35° W and the other end of the

proposed walkway is 500 m from his position at a bearing of 30°. Find the length of the walkway and thedirection visitors would face as they leave the viewing platform to use the walkway.

Chapter 6 Test: Solutions

1. (a)

Because the triangle contains a right angle, primary trigonometric ratios can beused.

cos C = = 0.375

∠C =̇ 68°

(b)

Because no angles are known, the cosinelaw is used. Because the required angle is Z, the version of thecosine law selected begins and ends with z.

z2 = x2 + y2 – 2xy cos Z

202 = 122 + 152 – 2(12)(15) cos Z cos Z = 0.157 41

∠Z =̇ 81°

(c)Because two angles are known there is no possible

ambiguity. Because an angle and its opposite side andone other fact are known, the sine law can be applied.

= =

°63sin

8=

°47sin

e

e =̇ 6.6 cm

2. Diagram in book

(a) Using the cosine law in ∆DEF,

e2 = d2 + f2 – 2df cos E

e2 = 122 + 52 – 2(12)(5) cos 50°

e =̇ 9.6 cmThe third side of the diagram is 9.6 cm.

(b) If the ratio of the areas is 25 : 1, the ratio of the corresponding sides will be 5 : 1.

DEF

ABC

∆

∆

area

area= (

e

b)2

= ()2 or = b =̇ 48 cm

The length of the corresponding side in the actual hanging will be 48 cm.

(c) Because angles do not change in similar figures and the diagram is a scale drawing of the actual hanging

(i.e. a similar figure) the opposite angle will be 50°.

3. Using the cosine law,

b2 = a2 + c2 – 2ac cos B

b2 = 132 + 92 – 2(13)(9) cos 66° b =̇ 12.4 cmUsing the sine law,

= =

= °66sin

4.12=

∠A =̇ 73°

∠C =̇ 42°

4. Because the two rescue ships travel at the same speed, the distances will determine the first arrival.

∠SAB = 90° – 41°= 49°

∠SBA = 360° – 310°= 50°

Because the shorter side is opposite the smaller angle, SA <SB, so the Athabasca can reach the ship in distress first.

Note: Students could determine the interior angles and thenapply the sine law to actually calculate the distances.

5. In ∆VCJ,

∠VJC = 90° – 35°= 55°

Using a trigonometric ratio in ∆VCJ,

cos 55° = VJ

300

VJ =̇ 523 m

In ∆EVJ,

∠VJE = 35° + 30°= 65°

Using the cosine law in ∆EVJ,

VE2 = 5232 + 5002 – 2(523)(500) cos 65° VE =̇ 550The walkway is about 550m long.

In ∆VCJ,

∠CVJ = 180° – (90° + 55°)= 35°

Using the sine law in ∆VJE,

JVΕ∠ sin

500=

°65sin

550

∠JVE =̇ 55°

180° – 55° – 35° = 90°The visitors would be facing directly east.