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CSEC MATHEMATICS JANUARY 2020 PAPER 2
SECTION I
1. (a) Using a calculator, or otherwise, calculate the exact value of the following:
(i) 4 "#ร "%โ 1 "
(
SOLUTION:
)415 ร
13, โ 1
14
We first evaluate: 4 "#ร "%
=215 ร
13 =
75 = 1
25
Nowwehave, 1 9#โ 1 "
(. Since 1 โ 1 = 0, we now have:
25 โ
14 =
8 โ 520
= %9<
(in exact form)
(ii)4.1 โ 1.259
0.005 SOLUTION:
4.1 โ 1.259
0.005 4.1 โ (1.25 ร 1.25)
0.005 Using the calculator,
=4.1 โ 1.5625
0.005
=2.53750.005
= 507.5 (in exact form)
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1(b) A stadium currently has a seating capacity of 15 400 seats.
(i) Calculate the number of people in the stadium when 75% of the seats are occupied.
SOLUTION:
75% of 15 400 seats
=75100 ร 15400
=11 550 persons
(ii) The stadium is to be renovated with a new seating capacity of 20 790 seats. After the renovation, what will be the percentage increase in the number of seats?
SOLUTION:
Increase in the number of seats
= 20 790 โ 15 400
= 5 390
So, percentage increase = BCDEFGHFICJKFCLMNFEOPHFGJHQEIRICGSCLMNFEOPHFGJH
ร 100%
= #%U<"#(<<
ร 100%
= 35%
1(c) A neon light flashes five times every 10 seconds. Show that the light flashes 43 200 times in one day.
SOLUTION:
1 day = 24 hours = 24 ร 60 minutes = 24 ร 60 ร 60 seconds = 86 400 seconds
The light flashes five times in each 10-second interval.
The number of 10-second intervals in 86 400 seconds = VW(<<"<
= 8640
If there are 5 flashes in every 10 seconds, then the number of times the light flashes in one day = 8640ร 5
= 43200
Q.E.D.
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2. (a) Factorise the following expressions completely.
(i) 5โ9 โ 12โ๐
SOLUTION:
5โ9 โ 12โ๐
= 5โ. โ โ 12โ๐
= โ(5โ โ 12๐)
(ii)2๐ฅ9 โ ๐ฅ โ 6
SOLUTION:
2๐ฅ9 โ ๐ฅ โ 6
(2๐ฅ + 3)(๐ฅ โ 2)
2(b) Solve the equation
๐ + 3 = 3(๐ โ 5)
SOLUTION:
๐ + 3 = 3๐ โ 15
๐ โ 3๐ = โ15 โ 3
โ2๐ = โ18
๐ = ]"V]9
๐ = 9
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2(c) Make ๐ the subject of the formula
2๐ด = ๐๐9 + 3๐ก
SOLUTION:
2๐ด = ๐๐9 + 3๐ก
๐๐9 = 2๐ด โ 3๐ก
๐9 = 9c]%Jd
๐ = e9c]%Jd
2(d) A farmer plants two crops, potatoes and corn, on a ten-hectare piece of land. The number of hectares of corn planted, ๐, must be at least twice the number of hectares of potatoes, ๐.
Write two inequalities to represent the scenario above.
SOLUTION:
The area of the entire plot of land = 10 hectares
The number of hectares of corn = c
The number of hectares of potatoes = p
Clearly, the total area planted cannot exceed the area of the plot
Hence, ๐ + ๐ โค 10
The number of hectares of corn is at least twice the number of hectares of potatoes
Hence, ๐ โฅ 2๐
The two inequalities are:
๐ + ๐ โค 10โฆโฆโฆโฆ(1)
๐ โฅ 2๐ โฆโฆโฆโฆโฆ.(2)
c 2p โฅ
๐ + ๐ โฏ 10
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3 (a) The diagram below shows a hexagonal prism.
Complete the following statement.
The prism has
______________ faces.
______________ edges and
______________ vertices.
SOLUTION:
Faces
The prism has 6 rectangular faces and 2 hexagonal faces.
Total number of faces = 6 + 2 = 8
Edges
The prism has 6 edges bounding each of the two hexagonal faces and 6 edges bounding the rectangular faces. Total number of edges = (6 ร 2) + 6 = 18
Vertices
The prism has 6 vertices on each of the two hexagonal face.
Total number of vertices = 6 ร 2 = 12
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3(b) A Sports Club owns a field, PQRS, in the shape of a quadrilateral. A scale diagram of this field is shown below. (1 centimetre represents 10 metres)
In the following parts show all your construction lines.
The field is to be divided with a fence from ๐ to the side ๐ ๐ so that different sports can be played at the same time.
Each point on the fence is the same distance from๐๐ as from ๐๐
(i) Using a straight edge and compasses only, construct the line representing the fence.
(ii) Write down the length of the fence , in metres.
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SOLUTION:
(i) Points equidistant from the line ๐๐ and ๐๐ will lie on the line bisecting the angle ๐๐๐.
The bisector of the angle QPS meets RS at F.
(ii) The line BF is 7.4 cm by measurement.
Since 1 centimetre represents 10 metres, the length of the fence is
7.4 ร 10 = 74๐๐๐ก๐๐๐
PF is the line representing the fence.
T
For any point, T, on PF, the two right-angled triangles are congruent
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3 (c) A quadrilateral ๐๐๐ ๐ and its image ๐โฒ๐โฒ๐ โฒ๐โฒ are shown on the grid below.
(i) Write down the mathematical name for the quadrilateral ๐๐๐ ๐ (ii) ๐๐๐ ๐ is mapped onto ๐โฒ๐โฒ๐ โฒ๐โฒ by an enlargement with scale factor, ๐, about the
centre, ๐ถ(๐, ๐). Using the diagram above, determine the values of ๐, ๐ and ๐.
SOLUTION:
(i) The quadrilateral ๐๐๐ ๐ has only one pair of parallel sides and is therefore a trapezium.
(ii) To locate the center of enlargement, join image points to their corresponding object points by straight lines. Produce them to meet at the center of enlargement.
We join ๐๐โฒ and ๐๐y and the point of intersection, ๐ถ(โ4,1)is shown below. It is not necessary to join more than two such lines since all such lines are concurrent.
Hence, ๐ = โ4 and ๐ = 1.
The scale factor of the enlargement is the ratio of image length to object length.
Hence, ๐ = zy{yz{
= %"= 3
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4 (a) The function ๐ is defined as
๐(๐ฅ) =2๐ฅ + 75
(i) Find the value of ๐(4) + ๐(โ4).
SOLUTION:
๐(4) =2(4) + 7
5 =8 + 75 =
155 = 3.
๐(โ4) =2(โ4) + 7
5 =โ8+ 75 = โ
15
Hence, ๐(4) + ๐(โ4) = 3 + )โ15, = 2
45
C
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(ii) a) Calculate the value of ๐ฅ for which ๐(๐ฅ) = 9.
SOLUTION:
๐(๐ฅ) = 9
Therefore,2๐ฅ + 75 = 9
2๐ฅ + 7 = 45
2๐ฅ = 45 โ 7
2๐ฅ = 38
๐ฅ = 19
b) Hence, or otherwise, determine the value of ๐]"(9)
SOLUTION:
From above, when ๐ฅ = 19, ๐(๐ฅ) = 9.
This means that, for the function ๐(๐ฅ), an input of 19 produces an output of 9. We can also say that the function, ๐maps 19 onto 9.
Hence, the inverse function, ๐]"(๐ฅ),will map 9 onto 19. So, ๐]"(19) = 9.
Alternatively, we could find ๐]"(๐ฅ), and substitute ๐ฅ = 19 in this inverse function.
Let.๐ฆ =2๐ฅ + 75
Interchange ๐ฅ and๐ฆ:
๐ฅ =2๐ฆ + 75
5๐ฅ = 2๐ฆ + 7
2๐ฆ = 5๐ฅ โ 7
๐ฆ =5๐ฅ โ 72
๐]"(๐ฅ) =5๐ฅ โ 72
๐]"(9) =5(9) โ 7
2 =382 = 19
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4(b) The graph below shows two straight lines, ๐ฟ"and๐ฟ9. ๐ฟ" intercepts the ๐ฅ and ๐ฆ axes at (4, 0) and (0, 2) respectively. ๐ฟ9 intercepts the ๐ฅ and ๐ฆ axes at (1.5, 0) and (0,โ3) respectively.
(i) Determine the equation of the line ๐ฟ".
SOLUTION:
The gradient of ๐ฟ" can be found by inspection as 9](= โ "
9
OR by using the points (0,2)and(4,0) and the gradient formula, ๏ฟฝ๏ฟฝ]๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ]๏ฟฝ๏ฟฝ
.
Gradient of ๐ฟ" =<]9(]<
= โ 9(= โ "
9
The general equation of any straight line is ๐ฆ = ๐๐ฅ + ๐,where ๐ is the gradient and ๐ is the๐ฆ-intercept. ๐ฟ" has a ๐ฆ-intercept of 2, so ๐ = 2. Hence, the equation of ๐ฟ" is:
๐ฆ = โ12๐ฅ + 2
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OR
Using the point (๐ฅ", ๐ฆ") = (2,0) and ๐ = โ"9,the equation of a line is given as:
๐ฆ โ ๐ฆ"๐ฅ โ ๐ฅ"
= ๐
๐ฆ โ 2๐ฅ โ 0 = โ
12
2(๐ฆ โ 2) = โ๐ฅ
2๐ฆ โ 4 = โ๐ฅ
2๐ฆ = โ๐ฅ + 4
๐ฆ = โ12๐ฅ + 2
(ii) What is the gradient of the line ๐ฟ9, given that ๐ฟ" and ๐ฟ9 are perpendicular?
SOLUTION:
The product of the gradients of perpendicular lines is โ1.
Gradient of ๐ฟ" ร Gradient of ๐ฟ9 = โ1
โ "9ร Gradient of ๐ฟ9 = โ1
Gradientof๐ฟ9 = โ1
โ12
Gradientof๐ฟ9 = 2
OR
By using the points (1.5, 0) and (0,โ3),
Gradient of ๐ฟ9:
=โ3 โ 00 โ 1.5 =
โ3โ1.5 = 2
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5. A group of 100 students estimated the mass, ๐(grams) of a seed. The cumulative frequency curve below shows the results.
(i) (a) Using the cumulative frequency curve, estimate the (i) median (ii) upper quartile
SOLUTION:
(i) The median is the middle score in a distribution and will occur at the 50th percentile. The estimate of the median is approximately 3.2 grams.
(ii) The upper quartile is the 75th percentile. The estimate of the upper quartile approximately 4.3 grams.
[These values are obtained from the curve and shown on the diagram.]
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(iii) Semi-interquartile range
To obtain the semi-interquartile range, we use the formula: {๏ฟฝ]{๏ฟฝ9
, where ๐% is the upper quartile and ๐"the lower quartile.
The lower quartile is at the 25th percentile and shown on the graph.
๐" โ 2.3๐๐๐๐๐% โ 4.3๐
๐% โ ๐"2 =
4.3 โ 2.32 =
22 = 1
The semi-interquartile range is 1.
Medianโ 3.2๐
Upper Quartile โ 4.3๐
Lower Quartile โ 2.3๐
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(iv) Number of students whose estimate is 2.8 grams or less.
SOLUTION:
The cumulative frequency that corresponds to a mass of 2.8 grams is 38.
Hence 38 students had estimates that were less than 2.8 grams.
(b) (i) Use the cumulative frequency curve given to complete the frequency table below.
38
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SOLUTION:
The cumulative frequency that corresponds to ๐ โค 4 is 68. Hence, 68 students had estimates that were less than or equal to 4 grams.
Since 20 students had estimates that were less than or equal to 2 grams, the number having estimates between 2 and 4 grams is 68 โ 20 = 48.
The cumulative frequency that corresponds to ๐ โค 6 is 93. Hence, 93 students had estimates that were less than or equal to 6 grams.
Since 68 students had estimates that were less than or equal to 4 grams, the number having estimates between 4 and 6 grams is 93 โ 68 = 25
The completed table is shown below:
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(b) (ii) A student is chosen at random. Find the probability that the student estimated the mass to be greater than 6 grams.
From the frequency table, the number of students who estimated the mass to be greater than 6 grams is 6 + 1 = 7.
The total number of students is 100.
Probability (estimate is greater than 6 grams)
=๐๐ข๐๐๐๐๐๐๐ ๐ก๐ข๐๐๐๐ก๐ ๐คโ๐๐๐ ๐ก๐๐๐๐ก๐๐๐กโ๐๐๐๐ ๐ ๐ก๐๐๐๐๐๐๐๐ก๐๐๐กโ๐๐6๐
๐๐๐ก๐๐๐๐ข๐๐๐๐๐๐๐ ๐ก๐ข๐๐๐๐ก๐
=7100
48
25
โ๐ = 100
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6. (a) The radius of EACH circle in the rectangle ๐๐๐๐shown below is 7 cm. The circles fit exactly into the rectangle
(ii) Show that the area of the rectangle is 2 352 cm2.
SOLUTION:
The diameter of the circle = 7๐๐ ร 2 = 14๐๐
The length of the rectangle = 14๐๐ ร 4 = 56๐๐
The width of the rectangle = 14๐๐ ร 3 = 42๐๐
The area of the rectangle = 56 ร 42๐๐9
= 2352๐๐9
(iii) Calculate the area of the shaded region.
SOLUTION:
Area of the 12 circles = 12 ร ๐๐9 = 12 ร 99๏ฟฝร 79
= 12 ร 99๏ฟฝร 7 ร 7 = 1848๐๐9
Area of the shaded region = Area of the rectangle โ Area of the 12 circles
= 2352 โ 1848๐๐9
= 504๐๐9
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6 (b) The diagram below, not drawn to scale, shows triangle MNP in which angle MNP = angle PMN = 52< and MN = 12.5 cm.
(i) State the type of triangle shown above.
SOLUTION:
Since angle N = 180< โ 2(52)< โ 52<,only two angles of triangle MNP are equal.
Hence, triangle MNP is isosceles.
(ii) Determine the value of angle PNM.
SOLUTION:
Angle PNM = 180< โ 2(52)< = 76< (sum of angles in a triangle is 180<)
(iii) Calculate the area of triangle MNP.
SOLUTION:
760
12.5 cm
Area of triangle MNP:
=12 ร 12.5 ร 12.5 ร ๐ ๐๐76
<
=12 ร 12.5 ร 12.5 ร 0.970
= 75.80๐๐9
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7. A sequence of figures is made up stars, using dots and sticks of different lengths. The first three figures in the sequence are shown below.
Study the pattern of numbers in each row of the table below. Each row relates to a figure in the sequence of figures stated above. Some rows have not been included in the table.
(a) Complete Rows (i), (ii) and (iii).
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SOLUTION:
Part (a)
(i) The number of sticks is 12 times the Figure Number. So, in Figure 5, there will be 5 ร 12 = 60 sticks.
The number of dots is one more than the number of sticks. Since Figure 5 has 60 sticks it will have (60 + 1) = 61 dots
(ii) The Figure Number will be the number of dots divided by 12. If a figure has 156 sticks then, the Figure Number will be (156 รท 12) = 13
Hence, the number of dots for Figure 13 is 156 + 1 = 157.
(iii) The number of sticks is always 12 times the Figure Number. So, the ๐JKfigure will have 12๐ sticks.
The number of dots is always one more than the number of sticks. So, the ๐JK figure will have [(12 ร ๐) + 1] = 12๐ + 1 sticks.
The completed table is shown below:
60 61
13 157
12๐ 12๐ + 1
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(b) The sum of the number of dots in two consecutive figures is recorded. This information for the first three pairs of consecutive figures is shown in the table below.
Determine the total number of dots in
(i) Figure 7 and 8 (ii) Figure ๐and Figure (๐ + 1)
SOLUTION:
Part (b)
(i) The total number of dots in Figures 7 and 8
The number of dots in Figure 7 = 12(7) + 1 = 85 (๐ = 7)
The number of dots in Figure 8 = 12(8) + 1 = 97 (๐ = 8)
Therefore the total number of dots in Figures 7 and 8 = 85 + 97 = 182
(iii) The total number of dots in Figure ๐and Figure (๐ + 1) The number of dots in Figure ๐ is12๐ + 1 The number of dots in Figure ๐ + 1 = 12(๐ + 1) + 1
Therefore, the total number of dots in Figure ๐and Figure (๐ + 1)
= [12๐ + 1] + [12(๐ + 1) + 1]
= 12๐ + 1 + 12๐ + 12 + 1
= 24๐ + 14
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SECTION II
ALGEBRA, RELATIONS, FUNCTIONS AND GRAPHS
8. (a) Solve the pair of equations:
๐ฆ9 + 2๐ฆ + 11 = ๐ฅ
๐ฅ = 5 โ 3๐ฆ
SOLUTION:
Let
๐ฆ9 + 2๐ฆ + 11 = ๐ฅ Eq (1)
๐ฅ = 5 โ 3๐ฆ Eq (2)
Substitute ๐ฅ from Equation 2 in Equation 1:
๐ฆ9 + 2๐ฆ + 11 = 5 โ 3๐ฆ
๐ฆ9 + 2๐ฆ + 11 โ 5 + 3๐ฆ = 0
๐ฆ9 + 5๐ฆ + 6 = 0
(๐ฆ + 3)(๐ฆ + 2) = 0
๐ฆ + 3 = 0๐๐๐ฆ + 2 = 0
๐ฆ = โ3๐๐๐ฆ = โ2
When ๐ฆ = โ3, ๐ฅ = 5 โ 3(โ3) = 5 + 9 = 14
When ๐ฆ = โ2, ๐ฅ = 5 โ 3(โ2) = 5 + 6 = 11
Solutions are: ๐ฅ = 14๐๐๐๐ฆ = โ3๐๐ ๐ฅ = 11๐๐๐๐ฆ = โ2
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(b) The function f is defined as follows:
๐(๐ฅ) = 4๐ฅ9 โ 8๐ฅ โ 2
(i) Express ๐(๐ฅ) in the form ๐(๐ฅ + โ)9 + ๐, where ๐, โ and ๐are constants.
SOLUTION:
4๐ฅ9 โ 8๐ฅ โ 2
= 4(๐ฅ9 โ 2๐ฅ) โ 2
= 4[(๐ฅ โ 1)9 โ 1] โ 2
= 4(๐ฅ โ 1)9 โ 4 โ 2
= 4(๐ฅ โ 1)9 โ 6
OR
4๐ฅ9 โ 8๐ฅ โ 2
= 4(๐ฅ9 โ 2๐ฅ) โ 2
= 4(๐ฅ โ 1)9+?
4(๐ฅ โ 1)(๐ฅ โ 1) = 4(๐ฅ9 โ 2๐ฅ + 1) = 4๐ฅ9 โ 8๐ฅ + 4
The constant in the original expression is โ2. Hence, 4+?= โ2 and ? = โ6
So, 4๐ฅ9 โ 8๐ฅ โ 2 = 4(๐ฅ โ 1)9 โ 6
OR
Expanding and equating coefficients;
๐(๐ฅ + โ)9 + ๐ = ๐(๐ฅ9 + 2๐ฅโ + โ9) + ๐
= ๐๐ฅ9 + 2๐ฅ๐โ + ๐โ9 + ๐
4๐ฅ9 โ 8๐ฅ โ 2 = ๐๐ฅ9 + 2๐ฅ๐โ + ๐โ9 + ๐
Equating coefficients:
๐ = 4 2๐โ = โ8 ๐โ9 + ๐ = โ2
2(4) ร โ = โ8 4(โ1)9 + ๐ = โ2
โ = โ1 ๐ = โ6
So, ๐(๐ฅ + โ)9 + ๐ = 4(๐ฅ โ 1)9 โ 6
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(ii) State the minimum value of ๐(๐ฅ)
๐(๐ฅ) = 4(๐ฅ โ 1)9 โ 6
SOLUTION:
4(๐ฅ โ 1)9 โฅ 0, for all values of ๐ฅ.
๐(๐ฅ)MIC = 0 โ 6 = โ6
(iii) Determine the equation of the axis of symmetry.
The axis of symmetry passes through the minimum point.
This occurs when ๐ฅ โ 1 = 0, or๐ฅ = โ1
OR
The equation of the axis of symmetry is ๐ฅ = ]N9G
, for a quadratic of the form ๐(๐ฅ) = ๐๐ฅ9 + ๐๐ฅ + ๐
For the function, ๐(๐ฅ) = 4๐ฅ9 โ 8๐ฅ โ 2, ๐ = 4, ๐ = โ8
๐ฅ = ](]V)9(()
= โ1
The equation of the axis of symmetry is ๐ฅ = โ1.
(c) The speed-time graph below, not drawn to scale, shows the three-stage journey of a car over a period of 40 seconds.
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Determine the acceleration of the car for EACH of the following stages of the journey
Stage 2
Stage 3
SOLUTION:
Acceleration of the car in Stage 2 of the journey
Stage 2 of the journey is represented by a horizontal line. This indicates that the velocity is constant.
Since the velocity is not changing, the rate of change of velocity is zero.
Therefore the acceleration, ๐ = 0๐/๐ 9.
OR
The acceleration can be calculated by finding the gradient of the line. A horizontal line has a gradient of zero. By calculation,
Gradient = "9]"9%9]"W
= 0.
(16, 12) (32, 12)
Acceleration of the car in Stage 3 of the journey
(32, 12) Gradient = "9]<%9](<
= "9]V= โ %
9๐/๐ 9
The car is decelerating (slowing down). The acceleration is โ1.5๐/๐ 9 or the deceleration is 1.5๐/๐ 9.
(40, 0)
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9. (a) The circle shown below has center O and the points A, B, C and D lying on the circumference. A straight line passes through the points A and B. Angle ๐ถ๐ต๐ท = 49< and angle ๐๐ด๐ต = 37<.
(i) Write down the mathematical names of the straight lines BC and OA.
SOLUTION: The straight line, BC, joins two points on the circle and it is therefore a chord of the circle. The straight line OA joins the center of the circle to a point on the circle and is therefore a radius of the circle.
(ii) Determine the value of EACH of the following angles. Show detailed working where necessary and give a reason to support your answer. a) x b) y SOLUTION: Consider triangle OAB ๐๐ด = ๐๐ต (radii)
Hence, triangle OAB is isosceles. Angle ๐๐ต๐ด = 37< (Base angles of an isosceles triangle are equal) ๐ฅ< = 180< โ (37 + 37)< (Sum of angles in a triangle is 180<) ๐ฅ< = 180< โ 74< ๐ฅ = 104
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Consider triangle BCD Angle ๐ต๐ถ๐ท = 90< (Angle in a semi-circle is a right angle) ๐ฆ< = 180< โ (90< + 49<) (Sum of angles in a triangle is 180<) ๐ฆ< = 180< โ 139< ๐ฆ = 41
(b) The diagram below, not drawn to scale, shows the route of a ship cruising from
Palmcity (๐) to Quayton (๐) and then to Rivertown (R). The bearing of ๐from ๐ is 133< and the angle ๐๐๐ is 56<.
(i) Calculate the value of the angle ๐ค. SOLUTION:
๐ฅ
๐ฅ
Let this angle be x. ๐ฅ = 180< โ 133< = 47<
(Angles on a straight line) The two North lines are parallel, hence, this angle is also equal to x (alternate angles). Therefore, ๐ค< = 180< โ (56 + 47)< ๐ค = 77
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(ii) Determine the bearing of P from Q.
The angle that represents the bearing of P from Q is show in the diagram. It is calculated as
360< โ 47< = 313<
(iii) Calculate the distance RP.
Applying the cosine rule to triangle PQR we have:
๐ ๐9 = 2109 + 2909 โ 2(210)(290)๐๐๐ 56< = 44100 + 84100 โ 121800 ร 0.559 = 128200โ 68109.70 = 60090.30 ๐ ๐ = โ60090.30 ๐ ๐ = 245.13๐๐
47<
Bearing of P from Q = 3130
P
Q
N
N
P
Q
R
210 km
290 km
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10. (a) The transformation, ๐ = )0 ๐๐ 0, maps the point ๐ onto ๐ โฒ as shown in the diagram
below.
(i) Determine the value of p and of q.
SOLUTION:
The transformation, ๐ = )0 ๐๐ 0, maps ๐ (2,โ5) onto ๐ y(5,2). Hence,
)0 ๐๐ 0, ยฉ
2โ5ยช = ยฉ52ยช
2ร2 2ร1 2ร1
)0 ร 2 + ๐ ร โ5๐ ร 2 + 0 ร โ5, = ยฉ52ยช
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)โ5๐2๐ , = ยฉ52ยช
Equating corresponding elements, โ5๐ = 5and2๐ = 2 So, ๐ = โ1and๐ = 1
(ii) Describe fully the transformation, M.
SOLUTION: The transformation, ๐ = ยฉ0 โ1
1 0 ยช, is an anticlockwise (or positive) rotation about the origin through 90<.
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(b) ๐๐๐ ๐ is a parallelogram in which ๐๐ยซยซยซยซยซโ = ๐ขand ๐๐ยซยซยซยซโ = ๐ฃ.
M is a point on QS such that ๐๐:๐๐ = 1: 2
(i) Write in terms of ๐and ๐ an expression for
a) ๐๐ยซยซยซยซยซโ
SOLUTION: ๐๐ยซยซยซยซยซโ = ๐๐ยซยซยซยซยซโ + ๐๐ยซยซยซยซโ = โ(๐) + ๐
๐๐ยซยซยซยซยซโ = โ๐ + ๐
b) ๐๐ยซยซยซยซยซยซโ
SOLUTION: ๐๐:๐๐ = 1: 2 Therefore, ๐๐ = "
%๐๐
๐๐ยซยซยซยซยซยซโ =13 (โ๐ + ๐)
๐๐ยซยซยซยซยซยซโ = โ13๐ +
13๐
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(ii) Show that ๐๐ ยซยซยซยซยซยซโ = "%(๐ + 2๐)
SOLUTION: ๐๐ ยซยซยซยซยซยซโ = ๐๐ยซยซยซยซยซยซโ + ๐๐ ยซยซยซยซยซโ ๐๐ ยซยซยซยซยซยซโ = โ๐๐ยซยซยซยซยซยซโ + ๐๐ ยซยซยซยซยซโ = โยฉโ"
%๐ + "
%๐ยช + ๐ ยฒ๐๐ ยซยซยซยซยซโ = ๐๐ยซยซยซยซโ = ๐ยณ
=13๐ โ
13๐ + ๐
=13๐ +
23๐
=13 (๐ + 2๐)
Q.E.D.
(iii) T is the mid-point of PQ. Prove that R, M and T are collinear.
SOLUTION: To prove that R, M and T are collinear, consider the vectors ๐๐ ยซยซยซยซยซโ and๐๐ ยซยซยซยซยซยซโ .
๐๐ ยซยซยซยซยซโ = ๐๐ยซยซยซยซยซโ + ๐๐ ยซยซยซยซยซโ
๐๐ ยซยซยซยซยซโ =12๐ + ๐
๐๐ ยซยซยซยซยซโ =12 (๐ + 2๐)
T
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We know from part (ii) that
๐๐ ยซยซยซยซยซยซโ =13 (๐ + 2๐)
๐๐ ยซยซยซยซยซยซโ =13(๐ + 2๐) =
23ยด12 (๐ + ๐๐)
ยถ
๐๐ ยซยซยซยซยซยซโ =23๐๐ ยซยซยซยซยซโ
The vector ๐๐ ยซยซยซยซยซยซโ is a scalar multiple of ๐๐ ยซยซยซยซยซโ (The scalar multiple being 9
%).
Hence, ๐๐ ยซยซยซยซยซยซโ is parallel to ๐๐ ยซยซยซยซยซโ . In both line segments, R is a common point. Therefore, R, M and T are collinear.
Q.E.D.
13(๐+
2๐)
12(๐+
2๐)
T
R
M
R