solutions to csec maths p2 june 2012
TRANSCRIPT
Solutions to CSEC Maths P2 JUNE 2012
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Question 1a
Calculate the exact value of
3
1
5โ
2
3
2 4
5
Numerator: 3 1
5โ
2
3
= 16
5โ
2
3
=48โ10
15
=38
15
Denominator: 2 4
5=
14
5
๐๐ข๐๐๐๐๐ก๐๐
๐ท๐๐๐๐๐๐๐๐ก๐๐ =
38
15รท
14
5
=38
15ร
5
14
=19
21
Question 1b
Data Given: The table below shows the cost price, selling price and profit or loss as a
Percentage of the cost price.
Cost Price Selling Price Percentage
Profit or Loss
$55.00 $44.00 ______
________ $100.00 25% profit
Required to Copy and Complete the table below inserting the missing values.
Cost Price Selling Price Percentage
Profit or Loss
$55.00 $44.00 20% loss
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$80.00 $100.00 25% profit
For the 1st item:
Since the selling price is less than the cost price for the first item,
โน ๐ฟ๐๐ ๐ = ๐ถ๐๐ ๐ก ๐๐๐๐๐ โ ๐๐๐๐๐๐๐ ๐๐๐๐๐
= $55 โ $44
= $11.00
โด ๐๐๐๐๐๐ก๐๐๐ ๐ฟ๐๐ ๐ =๐ฟ๐๐ ๐
๐ถ๐๐ ๐ก ๐๐๐๐๐ ร 100
=11
55ร 100
= 20%
For the 2nd item:
๐ด 25% ๐๐๐๐๐๐ก ๐๐๐๐๐๐๐ ๐กโ๐๐ก ๐กโ๐ ๐ ๐๐๐๐๐๐ ๐๐๐๐๐ ๐๐ 125% ๐๐ ๐กโ๐ ๐๐๐ ๐ก ๐๐๐๐๐.
โด ๐โ๐ ๐๐๐ ๐ก ๐๐๐๐๐ =100
125ร 100
= $80.00
Question 1c part (i)
Data Given: The table below shows some rates of exchange
US $1.00 = EC $2.70
TT $1.00 = EC $0.40
Required to calculate the value of EC $1 in TT $
๐ธ๐ถ $0.40 = ๐๐ ๐1.00
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๐ธ๐ถ $1.00 =๐๐ $1.00
40ร 100
๐ธ๐ถ $1.00 = ๐๐ $2.50
Question 1c part (ii)
Required to calculate the value of US $80 in EC $
๐๐ $1.00 = ๐ธ๐ถ $2.70
๐๐ $80.00 = ๐ธ๐ถ $2.60 ร 80
= ๐ธ๐ถ $216.00
Question 1c part (iii)
Required to calculate the value of TT $648 in US $
๐๐ $1.00 = ๐ธ๐ถ $0.40
๐๐ $648 = ๐ธ๐ถ $0.40 ร 648
๐ธ๐ถ $259.20
Now,
๐ธ๐ถ $2.70 = ๐๐ $1.00
๐ธ๐ถ $1.00 =๐๐ $1.00
๐ธ๐ถ 2.70
๐ธ๐ถ $259.20 =๐๐ $1.00
๐ธ๐ถ 2.70ร ๐ธ๐ถ $259.20
= $96.00
โด ๐๐ $648.00 = ๐๐ $96.00
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Question 2a part (i)
Required to Factorize
2๐ฅ3๐ฆ + 6๐ฅ2๐ฆ2
Step 1: We factor out the common factor in each term
2๐ฅ2๐ฆ is a common factor in each term
Thus, 2๐ฅ2๐ฆ(๐ฅ) + 2๐ฅ2๐ฆ(3๐ฆ)
Step 2: Factorize to get 2๐ฅ2๐ฆ(๐ฅ + 3๐ฆ)
Question 2a part (ii)
Required to Factorize
9๐ฅ2 โ 4
Step 1: We re-write the expression as the difference of two squares:
(3๐ฅ)2 โ (2)2
Step 2: We factorize to get:
(3๐ฅ โ 2)(3๐ฅ + 2)
Question 2a part (iii)
Required to Factorize
4๐ฅ2 + 8๐ฅ๐ฆ โ ๐ฅ๐ฆ โ 2๐ฆ2
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Step 1: Group like terms
4๐ฅ(๐ฅ + 2๐ฆ) โ ๐ฆ(๐ฅ + 2๐ฆ)
Step 2: Factorize to get:
(๐ฅ + 2๐ฆ)(4๐ฅ โ ๐ฆ)
Question 2b
Required to Solve
2๐ฅโ3
3+
5โ๐ฅ
2= 3
Step 1: Find the LCM
2(2๐ฅโ3)+3(5โ๐ฅ)
6= 3
4๐ฅโ6+15โ3๐ฅ
6= 3
Step 2: Simplify
๐ฅ+9
6= 3
Step 3: Make ๐ฅ the subject of the formula
๐ฅ + 9 = 3(6)
๐ฅ + 9 = 18
๐ฅ = 18 โ 9
๐ฅ = 9
Question 2c
Given Data: We are given two equations:
3๐ฅ โ 2๐ฆ = 10
2๐ฅ + 5๐ฆ = 13
Required to solve both equations simultaneously
Using the Elimination Method
Step 1: Let 3๐ฅ โ 2๐ฆ = 10 be Equation 1
Let 2๐ฅ + 5๐ฆ = 13 be Equation 2
Step 2: Multiply Equation 1 by 2 to get Equation 3
2(3๐ฅ โ 2๐ฆ) = 2(10)
6๐ฅ โ 4๐ฆ = 20 [Equation 3]
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Step 3: Multiply Equation 2 by 3 to get Equation 4
3(2๐ฅ + 5๐ฆ) = 3(13)
6๐ฅ + 15๐ฆ = 39 [Equation 4]
Step 4: Equation 4 โ Equation 3
6๐ฅ + 15๐ฆ = 39 โ
6๐ฅ โ 4๐ฆ = 20
19๐ฆ = 19
Step 5: Make y the subject of the formula to find the value of ๐ฆ
๐ฆ =19
19
๐ฆ = 1
Step 6: Substitute ๐ฆ = 1 into Equation 1 to find the value of ๐ฅ
3๐ฅ โ 2๐ฆ = 10
3๐ฅ โ 2(1) = 10
3๐ฅ โ 2 = 10
3๐ฅ = 10 + 2
3๐ฅ = 12
๐ฅ =12
3
๐ฅ = 4
Thus ๐ฅ = 4 and ๐ฆ = 1
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Question 3a part (i)
Data Given: In a survey of 36 students,
30 play tennis ๐ฅ play volleyball only
9๐ฅ play both tennis and volleyball
4 do not play either tennis or volleyball
๐ = { ๐๐ก๐ข๐๐๐๐ก๐ ๐๐ ๐ ๐ข๐๐ฃ๐๐ฆ}
๐ = {๐๐ก๐ข๐๐๐๐ก๐ ๐คโ๐ ๐๐๐๐ฆ ๐ฃ๐๐๐๐๐ฆ๐๐๐๐}
๐ = {๐๐ก๐ข๐๐๐๐ก๐ ๐คโ๐ ๐๐๐๐ฆ ๐ก๐๐๐๐๐ }
Required to Copy and Complete the Venn Diagram showing the number of students in subsets
marked y and z
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Question 3a part (ii)(a)
Required to write an expression in ๐ฅ for the total number of students in the survey
The number of students who play volleyball only = ๐ฅ
The number of students who play tennis only = 30 โ ๐ฅ
The number of students who play both tennis and volleyball = 9๐ฅ
The number of students who do not play tennis nor volleyball = 4
Total number of students in the survey = ๐ฅ + (30 โ 9๐ฅ) + 9๐ฅ + 4
= ๐ฅ + 30 โ 9๐ฅ + 9๐ฅ + 4
= ๐ฅ + 30 + 4
= ๐ฅ + 34
Question 3a part (ii)(b)
Required to write an equation in ๐ฅ for the total number of students in the survey and to solve
for ๐ฅ
Total number of students = 36
Total number of students = ๐ฅ + 34
U
V
T
4
30-9x
9x
x
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Thus ๐ฅ + 34 = 36
๐ฅ = 36 โ 34
๐ฅ = 2
Question 3b part (i)
Data Given: Diagram showing the journey of a ship which started at port ๐ , sailed 15๐๐ due
South to port ๐ and then further 20๐๐ due West to port ๐
Required to Copy and label the diagram to show points ๐ and ๐ and distances 20๐๐ and 15๐๐
Question 3b part (ii)
Required to the shortest distance of the ship from the port to where the journey started
Using Pythagorasโ Theorem
๐๐ 2 = ๐๐2 + ๐ ๐2
๐๐ = โ(15)2 + (20)2
= 25๐๐
Question 3b part (iii)
P
Q R
15km
20km
N
S
W E
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Required to Calculate the measure of angle ๐๐๐ , giving answer to the nearest degree
๐ ๐๐๐๏ฟฝฬ๏ฟฝ๐ =๐๐๐
โ๐ฆ๐
๐ ๐๐๐๏ฟฝฬ๏ฟฝ๐ =20
25
โด ๐๏ฟฝฬ๏ฟฝ๐ = (20
25)
= 53.1ยฐ
53ยฐ [to the nearest degree]
Question 4a part(i)
Data Given: Diagram showing the cross-section of a prism
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Required to calculate the length of the arc ๐ด๐ต๐ถ
Length of the arc ๐ด๐ต๐ถ =270ยฐ
360ยฐร ๐ถ๐๐๐๐ข๐๐๐๐๐๐๐๐ ๐๐ ๐ ๐๐๐๐๐๐๐ก๐ ๐๐๐๐๐๐
=3
4ร 2๐ ร ๐
=3
4ร 2 ร
22
7ร 3.5
= 16.5๐๐
Question 4a part(ii)
Required to calculate the perimeter of the sector ๐๐ด๐ต๐ถ
Perimeter of ๐๐ด๐ต๐ถ = ๐ฟ๐๐๐๐กโ ๐๐ ๐ด๐ + ๐ด๐๐ ๐ฟ๐๐๐๐กโ ๐ด๐ต๐ถ + ๐ฟ๐๐๐๐กโ ๐๐ ๐ ๐๐๐๐ข๐ ๐ถ๐
= 16.5 + 3.5 + 3.5
= 23.5๐๐
Question 4a part (iii)
Required to calculate the area of the sector ๐๐ด๐ต๐ถ
๐ด๐๐๐ ๐๐ ๐ ๐๐๐ก๐๐ ๐๐ด๐ต๐ถ =270ยฐ
360ยฐร ๐ด๐๐๐ ๐๐ ๐ ๐๐๐๐๐๐๐ก๐ ๐๐๐๐๐๐
=3
4ร
22
7ร (3.5)2
= 28.875
= 28.88๐๐2 [to 2 decimal places]
Question 4b part (i)
A
B
C
O 3.5cm
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Data Given: A prism is 20cm long and made of tin
1๐๐3 of tin has a mass of 7.3๐
Required to calculate the volume of the prism
๐๐๐๐ข๐๐ ๐๐ ๐กโ๐ ๐๐๐๐ ๐ = ๐ถ๐๐๐ ๐ โ ๐ ๐๐๐ก๐๐๐๐๐ ๐ด๐๐๐ ร ๐ป๐๐๐โ๐ก
= 28.875 ร 20
= 577.5๐๐3
Question 4b part (ii)
Required to calculate the mass of the prism, to the nearest kg
1๐๐3๐๐ ๐ก๐๐ ๐ค๐๐๐โ๐ 7.3๐
577.5๐๐3๐๐ ๐ก๐๐ = 7.3 ร 577.5๐
= 4215.75๐
=4215.75
1000
= 4.2๐๐
=4 kg [to the nearest kg]
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Question 5a part(i)
Required to Construct triangle ๐๐๐ with ๐๐ = 8๐๐, โ ๐๐๐ = 60ยฐ and โ ๐๐๐ = 45ยฐ
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Question 5a part (ii)
Required to measure and state the length of ๐ ๐
Using a Ruler to measure the length of ๐ ๐
The length of ๐ ๐ = 6.0๐๐ ]
Question 5b part (i)
Data Given: The line ๐ passes through the points ๐(6,6) and ๐(0, โ2)
Required to find the gradient of the line
Let (๐ฅ1, ๐ฆ1) = (0, โ2) and (๐ฅ2, ๐ฆ2) = (6,6)
๐บ๐๐๐๐๐๐๐ก =(๐ฆ2โ๐ฆ1)
(๐ฅ2โ๐ฅ1)
=6โ(โ2)
6โ0
=8
6
=4
3
Question 5b part (ii)
Required to find the equation of the line ๐
The line ๐ is of the form ๐ฆ = ๐๐ฅ + ๐
Since (0, โ2) lies on the line ๐, then the y-intercept is โ2 and the gradient m =4
3
Substituting ๐ =4
3 and ๐ = โ2 into ๐ฆ = ๐๐ฅ + ๐ to get ๐ฆ =
4
3๐ฅ โ 2
Thus the equation of the line ๐ = ๐ฆ =4
3๐ฅ โ 2
Question 5b part (iii)
Required to find the midpoint of the line segment ๐๐
๐๐๐๐๐๐๐๐ก = (๐ฅ1+๐ฅ2
2 ,
๐ฆ1+๐ฆ2
2)
=6+0
2,
6+(โ2)
2
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= (3,2)
Question 5b part (iv)
Required to find the length of the line segment ๐๐
๐โ๐ ๐๐๐๐๐กโ ๐๐ ๐๐ = โ(๐ฅ1 โ ๐ฅ2)2 + (๐ฆ2 โ ๐ฆ1)2
= โ(6 โ 0)2 + (6 โ (โ2))2
= โ62 + 82
= 10 ๐ข๐๐๐ก๐
Question 6a
Data Given: Graph showing triangle ๐ฟ๐๐ and its image ๐๐๐ after an enlargement
Required to Locate the centre of enlargement
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The centre of enlargement was found by finding the point of intersection between the end
points of triangle LNM triangle PRQ.
The centre of enlargement was found to be (1,5)
Question 6b
Required to State the scale factor and the center of enlargement
๐๐๐๐๐๐ ๐น๐๐๐ก๐๐ =๐ฟ๐๐๐๐กโ ๐๐ ๐๐
๐ฟ๐๐๐๐กโ ๐๐ ๐ฟ๐
=6
3
= 2
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Question 6c
Required to Determine the value of ๐ด๐๐๐ ๐๐ ๐๐๐
๐ด๐๐๐ ๐๐ ๐ฟ๐๐
Since ๐ ๐
๐๐=
2
1
Then ๐ด๐๐๐ ๐๐ ๐๐๐
๐ด๐๐๐ ๐๐ ๐ฟ๐๐=
(2)2
(1)2
=4
1
= 4
Question 6d
Required to Draw and Label Triangle ๐ด๐ต๐ถ
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Question 6e
Describe fully the transformation which maps triangle ๐ฟ๐๐ onto triangle ๐ด๐ต๐ถ
Triangle ๐ฟ๐๐ undergo a rotation of 90โ in an anti-clockwise direction about the origin
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(0.0)
Question 7a
Data Given: Table showing the ages of persons who visited the clinic during the week
Required to Copy and Complete the table to show cumulative frequency
Age, ๐ฅ LCL โ UCL
LCB โค ๐ฅ โค UCB Number of
Persons
Cumulative Frequency
Points to be
Plotted
(39.5,0)
40 โ 49 39.5 โค ๐ฅ โค 49.5
4 4 (49.5,4)
50 โ 59 49.5 โค ๐ฅ โค 59.5
11 15 (59.5,15)
60 โ 69
59.5 โค ๐ฅ โค 69.5
20 35 (69.5,35)
70 โ 79
69.5 โค ๐ฅ โค 79.5
12 47 (79.5,47)
80 โ 89
79.5 โค ๐ฅ โค 89.5
3 50 (89.5,50)
Question 7b
Required to Draw the Cumulative Frequency curve to represent the data
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Question 7c part (i)
Required to Estimate the median age for the data
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The Median Age is calculated by finding the x value (years) when the y value is 1
2 ๐๐ 50
on the graph
Thus, the Median Age is found to be 64.75 when y = 25
Question 7c part (i)
Required to Estimate the probability that a person who visited the clinic was 75 years or
younger
๐(๐ด ๐๐๐๐ ๐๐๐ ๐๐ 75 ๐ฆ๐๐๐๐ ๐๐ ๐ฆ๐๐ข๐๐๐๐)
= ๐๐ข๐๐๐๐ ๐๐ ๐๐๐๐ ๐๐๐ 75 ๐ฆ๐๐๐๐ ๐๐ ๐ฆ๐๐ข๐๐๐๐ รท ๐๐๐ก๐๐ ๐๐ข๐๐๐๐ ๐๐ ๐๐๐๐ ๐๐๐
= 42.5 รท 50
=17
20
Question 8a
Given Data: Diagram showing three figures in a sequence of figures
Required to draw the fourth figure in the sequence of figures
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Question 8b
Required to copy and complete the table given
Figure Area of Triangle Number of Pins on Base
1 1 (2 ร 1) + 1 = 3
2 4 (2 ร 2) + 1 = 5
3 9 (2 ร 3) + 1 = 7
4 42 = 16 (2 ร 4) + 1 = 9
โ100 = 10 100 (2 ร 10) + 1 = 21
20 202 = 400 (2 ร 20) + 1 = 41
๐ ๐2 2๐ + 1
Question 9a part (i)
Required to Solve ๐ฆ = 8 โ ๐ฅ and 2๐ฅ2 + ๐ฅ๐ฆ = โ16
Let ๐ฆ = 8 โ ๐ฅ [Equation 1]
2๐ฅ2 + ๐ฅ๐ฆ = โ16 [Equation 2]
Using the Method of Substitution
Step 1: Substitute Equation 1 into Equation 2
2๐ฅ2 + ๐ฅ(8 โ ๐ฅ) = โ16
2๐ฅ2 + 8๐ฅ โ ๐ฅ2 = โ16
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๐ฅ2 + 8๐ฅ = โ16
๐ฅ2 + 8๐ฅ + 16 = 0
(๐ฅ + 4)(๐ฅ + 4) = 0
๐ฅ = โ4
Step 2: Substitute ๐ฅ = 4 into Equation 1
๐ฆ = 8 โ (โ4)
= 8 + 4
= 12
Thus ๐ฅ = โ4 and ๐ฆ = 12
Question 9a part (ii)
Required to State: With reason or not, ๐ฆ = 8 โ ๐ฅ is a tangent to the curve with equation
2๐ฅ2 + ๐ฅ๐ฆ = โ16
Solve the Equation 2๐ฅ2 + ๐ฅ๐ฆ = โ16 to obtain the root ๐ฅ = โ4
Since the roots are real and equal, the straight line does not intersect the curve but rather
it touches the curve at the point (โ4, 12)
Question 9b part (i)
Data Given: ๐ฅ roses and ๐ฆ orchids in each bouquet with the following constraints
The number of orchids must be at least half the number of roses
There must be at least 2 roses
There must be no more than 12 flowers in the bouquet
Required to write the three inequalities for the constraints given
๐ฆ โฅ1
2๐ฅ
๐ฅ โฅ 2
๐ฅ + ๐ฆ โค 12
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Question 9b part (ii)
Required to shade the region that satisfies all three inequalities
Question 9b part (iii)
Required to State the co-ordinates of the points which represent the vertices of the region
showing the solution set
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โณ ๐ด๐ต๐ถ is the region which is shaded.
The co-ordinates for โณ ๐ด๐ต๐ถ is ๐ด = (2,10), ๐ต = (2,1) and ๐ถ = (8,4)
Question 9b part (iv)
Data Given: A profit of $3 is made on each rose and $4 on each orchid
Required to Determine the maximum possible profit on the sale of a bouquet
Step 1: Find an expression for the Total Profit made
Total profit made on ๐ฅ roses at $3 each and on ๐ฆ orchids at $4 each = (๐ฅ ร 3) + (๐ฆ ร 4)
Step 2: Find an equation for the Total Profit
๐ = 3๐ฅ + 4๐ฆ
Step 3: Find the Total Profit made at each of the three points.
At point ๐ด = (2,10) ๐ = 3(2) + 4(10)
๐ = $46.00
A point ๐ต = (2,1) ๐ = 3(2) + 4(1)
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๐ = $10.00
A point ๐ถ = (8,4) ๐ = 3(8) + 4(4)
๐ = $40.00
The Maximum Profit occurs at point ๐ด since the Total Profit made at point ๐ด is $46
Question 10a part(i)
Data Given: Diagram showing quadrilateral ๐๐ ๐๐
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Required to Calculate the length of ๐ ๐
Using the Sine Rule
๐๐
๐ ๐๐๐ =
๐ ๐
๐ ๐๐๐
7
๐ ๐๐60ยฐ=
๐ ๐
๐ ๐๐48ยฐ
7๐ ๐๐48ยฐ = ๐ ๐๐ ๐๐60ยฐ
๐ ๐ =7๐ ๐๐48ยฐ
๐ ๐๐60ยฐ
= 6.006
= 6.01 [to 2 decimal places]
Question 10a part(ii)
Required to Calculate the measure of โ ๐๐๐
Using the Cosine Rule
๐๐2 = ๐๐2 + ๐๐2 โ 2(๐๐)(๐๐)๐๐๐ ๏ฟฝฬ๏ฟฝ
72 = 82 + 102 โ 2(8)(10)๐๐๐ ๏ฟฝฬ๏ฟฝ
๐๐๐ ๏ฟฝฬ๏ฟฝ =(72โ82โ102)
(โ2(8)(10)
=โ115
โ160
๏ฟฝฬ๏ฟฝ = (0.7187)
= 44.02
= 44.0ยฐ [to the nearest 0.1ยฐ]
Question 10b part(i)(a)
8cm
10cm 7cm
R
Q
T
S
48
60
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Required to Calculate the measure of angle ๐๐๐
๐๏ฟฝฬ๏ฟฝ๐ = 180ยฐ โ 70ยฐ
= 110ยฐ
In triangle ๐๐๐, ๐๐ = ๐๐ [since they are radii of the same circle]
โ triangle ๐๐๐ is isosceles
๐โ๐ข๐ , ๐๏ฟฝฬ๏ฟฝ๐ =180 ยฐโ110 ยฐ
2
๐๏ฟฝฬ๏ฟฝ๐ = 35ยฐ
Question 10b part(i)(b)
Required to Calculate the measure of angle ๐๐๐
๐๏ฟฝฬ๏ฟฝ๐ =70 ยฐ
2
= 35 ยฐ
[The angle that is subtended by a chord at the circumference of a circle is half of that angle that
the chord subtends at the center of the circle, and standing on the same arc]
๐๏ฟฝฬ๏ฟฝ๐ ๐๐ ๐๏ฟฝฬ๏ฟฝ๐ = 90ยฐ
[The angle made by the tangent ๐๐ to a circle and a radius ๐๐ , at the point of contact, ๐ is a
right angle.
๐๏ฟฝฬ๏ฟฝ๐ = 180ยฐ โ (90ยฐ + 35ยฐ)
= 55ยฐ
[The sum of the interior angles of a triangle = 180ยฐ]
Question 10b part(i)(c)
Required to Calculate the measure of angle ๐๐๐
๐๏ฟฝฬ๏ฟฝ๐ = 70ยฐ
๐๏ฟฝฬ๏ฟฝ๐ = 90ยฐ
[The sum of the three interior angles of a triangle = 180ยฐ]
๐๏ฟฝฬ๏ฟฝ๐ = 180ยฐ โ (90ยฐ + 70ยฐ)
= 20ยฐ
Question 10b part(ii)(a)
Required to Name the triangle which is congruent to triangle ๐๐๐
๐๐ = ๐๐ and ๐๐ = ๐๐
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๐๏ฟฝฬ๏ฟฝ๐ = ๐๏ฟฝฬ๏ฟฝ๐
[Vertically opposite angles are equal when two straight lines intersect]
Therefore, โณ ๐๐๐ is congruent to โณ ๐๐๐
Question 10b part(ii)(b)
Required to Name the triangle which is congruent to triangle ๐๐๐
Consider โณ ๐๐๐ and โณ ๐๐๐
๐๐ = ๐๐
๐๏ฟฝฬ๏ฟฝ๐ = ๐๏ฟฝฬ๏ฟฝ๐ = 90ยฐ
๐๐ is a common side to both triangles and therefore triangle ๐๐๐ and triangle ๐๐๐ have the
same hypotenuse and share a common side.
Thus, Triangle ๐๐๐ is congruent to Triangle ๐๐๐
Question 11a part(i)(a)
Given Data: ๐๐ด = (6 2 ), ๐๐ต = (3 4 ) ๐๐๐ ๐๐ถ = (12 โ 2 )
Required to Express the vector ๐ต๐ด in the form (๐ฅ ๐ฆ )
Using the Vector Triangle Law
๐ต๐ด = ๐ต๐ + ๐๐ด
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= โ(3 4 ) + (6 2 )
= (3 โ 2 )
Question 11a part(i)(b)
Required to Express the vector ๐ต๐ถ in the form (๐ฅ ๐ฆ )
Using the Vector Triangle Law
๐ต๐ถ = ๐ต๐ + ๐๐ถ
= โ(3 4 ) + (12 โ 2 )
= (9 โ 6 )
Question 11a part(ii)
Required to State one geometrical relationship between BA and BC
|๐ต๐ถ| = 3|๐ต๐ด|
The vector ๐ต๐ด is a scalar multiple of the vector ๐ต๐ถ
Question 11a part(iii)
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Question 11b part(i)
Data Given: (๐ โ 4 1 ๐ )(2 โ 4 1 โ 3 ) = (2 0 0 2 )
Required to Calculate the value of ๐ and ๐
Step 1: Ensure that the dimensions of the matrices allows matrix multiplication
In this case, it does, as all three matrices are 2 ร 2 matrices
Thus, the result will be of the form (๐ ๐ ๐ ๐ )
Step 2: Multiply the matrices
((2 ร ๐) + (โ4 ร 1) (โ4 ร ๐) + (โ4 ร โ3) (1 ร 2) + (๐ ร 1) (1 ร โ4) + (๐ ร โ3) ) =
(2 0 0 2 )
(2๐ โ 4 12 โ 4๐ ๐ + 2 โ 4 โ 3๐ ) = (2 0 0 2 )
Step 3: Equating entries to find the value of ๐ and ๐
2๐ โ 4 = 2 12 โ 4๐ = 0
2๐ = 6 12 = 4๐
๐ =6
2
12
4= ๐
๐ = 3 3 = ๐
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๐ + 2 = 0 โ4 โ 3๐ = 2
๐ = โ2 โ4 โ 2 = 3๐
โ6 = 3๐
โ6
3= ๐
โ2 = ๐
Thus, ๐ = 3 and ๐ = โ2
Question 11b part(ii)
Required to Find the inverse of (2 โ 4 1 โ 3 )
Let ๐ด = (2 โ 4 1 โ 3 )
(3 โ 4 1 โ 2 )(2 โ 4 1 โ 3 ) = (2 0 0 2 )
(3 โ 4 1 โ 2 )(2 โ 4 1 โ 3 ) = 2(1 0 0 1 )
This is of the form ๐ด๐ดโ1 = ๐ผ
1
2(3 โ 4 1 โ 2 )(2 โ 4 1 โ 3 ) = (1 0 0 1 )
(3
2 โ
4
2 1
2 โ
2
2 ) (2 โ 4 1 โ 3 ) = (1 0 0 1 )
Therefore, ๐ดโ1 = (3
2 โ 2
1
2 โ 1 )
Question 11b part(iii)
Required to Solve for ๐ฅ and ๐ฆ in (2 โ 4 1 โ 3 ) = (๐ฅ ๐ฆ ) = (12 7 )
Step 1: Multiply the matrix equation by ๐ดโ1
(3
2 โ 2
1
2 โ 1 ) (2 โ 4 1 โ 3 )(๐ฅ ๐ฆ ) = (
3
2 โ 2
1
2 โ 1 ) (12 7 )
(๐ฅ ๐ฆ ) = (3
2 โ 2
1
2 โ 1 ) (12 7 )
(๐ฅ ๐ฆ ) = ((3
2ร 12) + (โ2 ร 7) (
1
2ร 12) + (โ1 ร 7) )
(๐ฅ ๐ฆ ) = (4 โ 1 )
๐ฅ = 4 and ๐ฆ = โ1
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