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Exemplar for internal assessment resource Mathematics and Statistics for Achievement Standard 91575
© NZQA 2014
Exemplar for Internal Achievement Standard
Mathematics and Statistics Level 3
This exemplar supports assessment against:
Achievement Standard 91575
Apply trigonometric methods in solving problems
An annotated exemplar is an extract of student evidence, with a commentary, to explain key aspects of the standard. It assists teachers to make assessment judgements at the grade boundaries.
New Zealand Qualifications Authority
To support internal assessment
Exemplar for internal assessment resource Mathematics and Statistics for Achievement Standard 91575
© NZQA 2014
Grade Boundary: Low Excellence
1. For Excellence, the student needs to apply trigonometric methods, using extended abstract thinking, in solving problems. This involves one or more of: devising a strategy to investigate or solve a problem, identifying relevant concepts in context, developing a logical chain of reasoning or proof, or forming a generalisation, and also using correct mathematical statements or communicating mathematical insight. This evidence is a student’s response to the TKI task ‘Maths End Ferris Wheels’. This student has devised a strategy to investigate a problem by finding a model for the Kiddy-wheel (1) and a model for the Flying-high wheel (2), although the latter model is incorrect and it does not simplify the problem. The student has used these to solve the problem (3). Correct mathematical statements have been used throughout the response. For a more secure Excellence, the student could have found the correct equation for the Flying-high wheel. The student could also consider and explain that one of the intervals is not a sensible solution, and have chosen to discard it.
Jade – Kiddy Wheel Height 0.5m to 8m 2 revolutions per minute 8 – 0.5 = 7.5
75.325.7
DCtBAy )(sin A = 3.75 D = 3.75 + 0.5 = 4.25 1530
2 B
So by a process of elimination
Kiddy Wheel is 25.4)5.7(15
sin75.3)( tth
Manu – Flying - High Ferris Height 3m to 43m 3 revolutions per minute 43 – 3 = 40
20240
A = 20 D = 23 1020
2 B C = 5
So 23)5(10
sin20)( tth
Looking at the two graphs together
Where can Jade see Manu
Jade 5 525.4)5.7(15
sin75.3 t
t is between 8.461 & 21.5385 and 38.461 & 51.5385 GC Manu – going up, 5 and 20
2023)5(10
sin205 t
t is between 1.44 & 4.52 and 21.44 & 24.52 and 41.44 & 44.52 GC so the intersection of these solutions is the time when Jade can see Manu in the first 60 seconds t is between 21.44 to 21.54 sec and 41.44 to 44.52 sec this will happen every 60 seconds for the duration of the ride.
Exemplar for internal assessment resource Mathematics and Statistics for Achievement Standard 91575
© NZQA 2014
Grade Boundary: High Merit
2. For Merit, the student needs to apply trigonometric methods, using relational thinking in solving problems. This involves one or more of: selecting and carrying out a logical sequence of steps, connecting different concepts or representations, demonstrating understanding of concepts, or forming and using a model, and also relating findings to a context or communicating thinking using appropriate mathematical statements. This evidence is a student’s response to the TKI task ‘Exact Values’. This student has demonstrated an understanding of concepts by finding the value of the reciprocal trigonometric functions for angles found using the compound angle formulae and double angle formulae (1). The values for tan120° and cot120° are incorrect (2). In finding general solutions, the student has found all the angles that have a sine and cosine of 1/√2 and a tangent of 1, but has not clearly communicated what the angles represent (3). The statements for the reciprocal ratios are incorrect (4). To reach Excellence, the student could correct the errors, and the generalisations need to be extended to exact values other than those for the angles in the special triangles.
Special angles
Using these triangles and
H
Osin and
H
Acos and
A
Otan
I can determine that sin cos tan
300, 6
2
1
2
3
3
1
450, 4
2
1
2
1 1
600, 3
2
3
2
1 3
because
tan
1cot
cot 300 = 3 , cot 450 = 1, cot 600 = 3
1
because
cos
1sec
sec 300 = 3
2, sec 450 = 2 , sec 600 = 2
because cosec
sin
1
cosec 300= 2, cosec 450 = 2 , cosec 600 =
3
2
Compound angles
22
1
22
3)
2
1
2
1()
2
1
2
3(45sin60cos45cos60sin)4560sin(
12,15
22
1315sin
therefore cosec
13
2215
22
3
22
1)
2
1
2
3()
2
1
2
1(45sin60sin45cos60cos)4560cos(
22
3115cos
therefore sec
31
2215
13
13
)13(1
13)4560tan(
13
1315tan
therefore cot
13
1315
Double angles
24
32)
22
3(2)
2
1
2
3(260cos60sin2)60(2sin
cossin22sin
3
2,120
AAA
24
32120sin and cosec
32
24120
5.012
11)
2
1(2)60(2cos
60
1cos22cos
2
2
A
AA
cos 1200 = -0.5 and sec 1200 = 5.0
1
2
2
32
31
2
32
)1
3(1
1
32
120tan)60(2tan
60
tan1
tan22tan
2
2
A
A
AA
cot 1200 =
2
32
2
General solutions
2
1sin x
2
1sin 1x x=450 x=n1800 + (-1)n450
2
1cos x
2
1cos 1x x=450 x=2(1800)n +/- 450
1tan x 1tan 1x x=450 x=n1800+450
Therefore
cosec45
1)1(
180
1 nnx
45
1)
360
1(sec nx
45
1
180
1cot nx
Exemplar for internal assessment resource Mathematics and Statistics for Achievement Standard 91575
© NZQA 2014
Grade Boundary: Low Merit
3. For Merit, the student needs to apply trigonometric methods, using relational thinking in solving problems. This involves one or more of: selecting and carrying out a logical sequence of steps, connecting different concepts or representations, demonstrating understanding of concepts, or forming and using a model, and also relating findings to a context or communicating thinking using appropriate mathematical statements. This evidence is a student’s response to the TKI task ‘Exact Values’. This student has connected different concepts in finding the value of the reciprocal trigonometric functions for the angles found using the compound angle formulae (1). For a more secure Merit, the student could use the unit circle or graphs to investigate other angles.
deg rad 300
6
900 2/
450
4
600
3
H
Osin
2
130sin
2
145sin
2
360sin
H
Acos
2
330cos
2
145cos
2
160cos
A
Otan
3
130tan
145tan 360tan
cosecO
H cosec 2
1
230 cosec 2
1
245 cosec
3
260
A
Hsec
3
230sec 2
1
245sec 2
1
260sec
O
Acot 3
1
330cot
145cot
3
160cot
22
13
22
1
22
3
2
1
2
1
2
3
2
130sin45cos30cos45sin)3045sin(75sin
22
31
22
3
22
1
2
3
2
1
2
1
2
160sin45cos60cos45sin)6045sin(105sin
14
3
4
1
2
3
2
3
2
1
2
160sin30cos60cos30sin)6030sin(90sin
22
13
22
1
22
3
2
1
2
1
2
3
2
130sin45sin30cos45cos)3045cos(75cos
22
31
22
3
22
1
2
3
2
1
2
1
2
160sin45sin60cos45cos)6045cos(105cos
04
3
4
3
2
3
2
1
2
1
2
360sin30sin60cos30cos)6030cos(90cos
3
11
3
11
30tan45tan1
30tan45tan)3045tan(75tan
31
31
60tan45tan1
60tan45tan)6045tan(105tan
0
3
13
11
3
13
60tan30tan1
60tan30tan)6030tan(90tan
= unidentified
22
13
22
1
22
3
2
1
2
1
2
3
2
130sin45cos30cos45sin)3045sin(15sin
22
13
22
1
22
3
2
1
2
1
2
3
2
130sin45sin30cos45cos)3045cos(15cos
3
11
3
11
30tan45tan1
30tan45tan)3045tan(15tan
cosec 75 = 13
22
13
2275sec
3
11
3
11
75cot
cosec 105 = 31
22
31
22105sec
31
31105cot
cosec 90 = 1 090sec cot 90 = unidentified
cosec 15 = 13
22
31
2215sec
3
11
3
11
15cot
Exemplar for internal assessment resource Mathematics and Statistics for Achievement Standard 91575
© NZQA 2014
Grade Boundary: High Achieved
4. For Achieved, the student needs to apply trigonometric methods in solving problems. This involves selecting and using methods, demonstrating knowledge of concepts and terms and communicating using appropriate representations. This evidence is a student’s response to the TKI task ‘Maths End Ferris Wheels’. This student has selected and used properties of trigonometric functions by identifying the correct equation of the Kiddy-wheel (1) and finding the correct equation of the Flying-high wheel (2). This student has solved trigonometric equations to find the correct intervals for both Ferris wheels (3). To reach Merit, the student could link the solutions back to the problem to find an interval when Jade can see Manu and show a contextual understanding of the problem.
Jade
Drew on GC 5.475.3)( tth
60 rotations in 60 seconds - unbelievable
25.45.715
sin75.3)( tth
Drew on GC Looks good
25.45.715
4)( tth
Drew on GC -period not correct
25.41530
cos4)( tth
Drew on GC 60 rotations - unbelievable
So 25.45.715
sin75.3)( tth
is the correct equation.
Manu:
40
2
231020
sin20)( tth
Jade:
5385.51461.38
5385.21461.8
525.45.715
sin75.3
t
t
t
Manu:
20231020
sin205 t
going up
04.8987.82
04.4987.42
04.987.2
t
t
t
Exemplar for internal assessment resource Mathematics and Statistics for Achievement Standard 91575
© NZQA 2014
Grade Boundary: Low Achieved
5. For Achieved, the student needs to apply trigonometric methods in solving problems. This involves selecting and using methods, demonstrating knowledge of concepts and terms and communicating using appropriate representations. This evidence is a student’s response to the TKI task ‘Maths End Ferris Wheels’. This student has selected and used properties of trigonometric functions in finding the correct equation of the Kiddy-wheel (1) and solved a trigonometric equation to find an interval when Jade is above 5 m (2). As above. For a more secure Achieved, the student could complete the equation for the Flying-high wheel and find an interval for Manu.
Kiddy wheel: 25.4)5.7(15
sin75.3 th
A B C?? D
2)343( Flying High 23)(sin20 CtBh moved up 23 (3+23) B: 3 revolutions in 2 minutes?
Jade > 5m t between 8.46 and 21.54 (GC) Manu < 20m and going up…?????
Exemplar for internal assessment resource Mathematics and Statistics for Achievement Standard 91575
© NZQA 2014
Grade Boundary: High Not Achieved
6. For Achieved, the student needs to apply trigonometric methods in solving problems. This involves selecting and using methods, demonstrating knowledge of concepts and terms and communicating using appropriate representations. This evidence is a student’s response to the TKI task ‘Exact Values’. This student has selected and used reciprocal trigonometric functions to find exact values for some of the reciprocal functions for the angles in the special triangles (1). The use of the double angle formula to find sin90° is correct, but this is a known angle (2). To reach Achieved, the student could determine the exact value for sin135°.
2
130sin cosec 230
sin(2x30)=2sin30xcos30
2
160cos 260sec
2
145sin cosec 245
2
145cos 245sec
135sin
45sin045cos90sin)4590sin(
090cos
190sin
2
12
2
1
2
1245cos45sin2)245sin(