HONG KONG DIPLOMA OF SECONDARY EDUCATION
EXAMINATION
MATHEMATICS
Compulsory Part
SCHOOL-BASED ASSESSMENT
Sample Assessment Task
Angle of Taking a Scenery Photograph
Marking Guidelines
教育局 課程發展處 數學教育組
Mathematics Education Section, Curriculum Development Institute
The Education Bureau of the HKSAR
Assessment Scale
The assessment scale for tasks on Mathematical Investigation is shown in the following table.
Level of
Performance
Very good
Good
Fair
Weak
Marks
Mathematical Knowledge and Investigation Skills
13–16
The student demonstrates a complete understanding of
the underlying mathematical knowledge and
investigation skills which are relevant to the task, and is
consistently competent and accurate in applying them in
handling the task. Typically, the student recognises
patterns, and states a conjecture which is justified by a
correct proof.
9–12
The student demonstrates a substantial understanding of
the underlying mathematical knowledge and
investigation skills which are relevant to the task, and is
generally competent and accurate in applying them in
handling the task. Typically, the student recognises
patterns, draws inductive generalisations to arrive at a
conjecture and attempts to provide a proof for the
conjecture.
5–8
The student demonstrates a basic understanding of the
underlying mathematical knowledge and investigation
skills which are relevant to the task, and is occasionally
competent and accurate in applying them in handling
the task. Typically, the student recognises patterns and
attempts to draw inductive generalisations to arrive at a
conjecture.
1–4
The student demonstrates a limited understanding of the
underlying mathematical knowledge and investigation
skills which are relevant to the task, and is rarely
competent and accurate in applying them in handling
the task. Typically, the student manages to recognise
simple patterns and organises results in a way that helps
pattern identifications.
Marks Mathematical Communication Skills
4
The student communicates ideas in
a clear, well organised and logically
true manner through coherent
written/verbal accounts, using
appropriate and correct forms of
mathematical
presentation
to
present conjectures and proofs.
3
The student is able to communicate
ideas
properly
through
written/verbal accounts, using
appropriate and correct forms of
mathematical presentation such as
algebraic
manipulations
and
geometric deductions.
2
The student is able to communicate
ideas with some appropriate forms
of mathematical presentation such
as
algebraic
formulae
and
geometric facts.
1
The
student
attempts
to
communicate
ideas
using
mathematical
symbols
and
notations,
diagrams,
tables,
calculations etc.

The full mark of a SBA task on Mathematical Investigation submitted should be scaled to 20 marks, of which 16 marks
are awarded for the mathematical knowledge and investigation skills while 4 marks are awarded for the communication
skills.

Teachers should base on the above assessment scale to design SBA tasks for assessing students with different abilities
and to develop the marking guidelines.
Angle of Taking a Scenery Photograph_Marking Guidelines
2
Marking Guidelines
Solution
Performance
Part A
1.
(a)
(i)
(ii)
tan x = 10
5
x = 63.4°
Evidence:
The solutions
AC =
52  22  29
BC =
(10  2)2  52  89
AC 2  BC 2  AB 2
29  89  100

2  AC  BC
2  29  89
y = 79.8°
Weak:
Attempt the question but
all answers are incorrect
Fair:
Some mistakes are found
in the working and not all
answers are correct
Good:
All answers are correct but
minor mistakes are found
in the working
cos y 
y  180  tan 1
Alternative method:
(iii) AC =
5
5
 tan 1
 79.8
2
10  2
52  122  13
Very good: All answers and working
are correct
5  12  10  29
2
2
BC =
AC 2  BC 2  AB 2 169  29  100

2  AC  BC
2  13  29
z = 45.6°
cos z 
Alternative method:
(b)
z  90  tan 1
5
12  10
 tan 1
 45.6
12
5
Let DC = x cm .
AC =
x 2  52 , BC =
10  x 2  52
AC 2  BC 2  AB 2 x 2  25  10  x   25  100

2
2  AC  BC
2 x 2  25 10  x   25
Evidence:
The solutions
2
cos ACB 


2  x  5
Weak:
Attempt the question but
all answers are incorrect
Fair:
Some mistakes are found
in the working and not all
answers are correct
Good:
All answers are correct
but minor mistakes are
found in the working.
2
2
2 x 2  25 10  x   25
DC = 5 cm when ACB is 90°.
Very good: All answers and working
are correct
Angle of Taking a Scenery Photograph_Marking Guidelines
3
Marking Guidelines
Solution
Performance
Evidence:
1.
The answers
2.
The proof
3.
The explanation
Part B
2.
(a)
The maximum lens view angle is 90° and C is (50, 0).
(b)
(i)
Let AC cuts the circle at the Point E.
If C is any point on the line L but not the Point D,
 ADB = AEB
(angle in the same segment)
= ACB + EBC
(ext  of triangle)
> ACB
If C is the Point D, then ADB = ACB

(ii)
(c)
ADB
Weak:
All the answers are
incorrect. Proof and
explanation are not
complete
Fair:
Some answers are correct
but the proof and
explanation are not
complete
Good:
All answers are correct but
the proof and explanation
are not complete
 ACB
Let F be the foot of perpendicular from D onto AB.
 DF = AB
2
DF = FB = FA
 F is the centre of the circle,
we have ADB = 90°
( in semi-circle)
Very good: All answers are correct and
the proof and explanation
are complete
Yes, I agree. From (b) (i), we see that ADB is the largest angle.
Part C
3.
(a)
(i)
In ∆BOD and ∆DOA ,
BOD = DOA = 60°
OBD = ODA
ODB = 180°  BOD  OBD
= 180°  DOA  ODA
= OAD

(ii)
(Common angles)
(angle in alt segment)
( sum of triangle)
Weak:
The proof is not done and
the answer is incorrect
Fair:
The proof is not complete
and the answer is incorrect
Good:
The proof is not complete,
but the answer is correct
∆BOD ~ ∆DOA (A.A.A.)
Let OD = x m .

Evidence:
1.
The answer
2.
The proof
∆BOD ~ ∆DOA ,
OB OD

OD OA
40  100 x

x
40
Very good: The proof is complete and
the answer is correct
x  5600  20 14

OD is 20 14 m .
Angle of Taking a Scenery Photograph_Marking Guidelines
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Marking Guidelines
Solution
3.
(b)
(i)
Performance
Let C be any point on Road M (but is not the Point D) and BC cuts the
circle at E .
ADB = AEB
( in the same segment)
= ACB + EAC
(ext  of triangle)
> ACB
If C is the Point D , then ADB = ACB

(ii)
ADB
Evidence:
1.
The answer
2.
The proof
Weak:
The proof is not done and
the answer is incorrect
Fair:
The proof is not complete
and the answer is
incorrect
Good:
The proof is not
complete, but the answer
is correct
 ACB
AD2 = OA2 + OD2  2OAODcos60°
= 402 + 5600  2(40)( 20 14 )(0.5)
AD = 64.8589 m
Very good: The proof is complete and
the answer is correct
DB2 = OB2 + OD2  2OBODcos60°
= (40+100)2 + 5600  2(140)( 20 14 )(0.5)
DB = 121.3399 m
AD 2  DB 2  AB 2
2  AD  DB
 64.8589 2  121.3399 2  100 2

2  64.8589 121.3399 
ADB  55.4
cos  ADB 
(c)
The maximum lens view angle is 55.4 .
Part D
4.
Yes, I agree.
Let BC cuts the circle at F.
If C is any point on Road M but not the Point D ,
ADB = AFB
= ACB + FAC
> ACB
Evidence:
1.
The proof
2.
The explanation
( in the same segment)
(ext  of triangle)
Weak:
Attempt to give a proof
Fair:
The proof is partly
correct
Good:
The proof is correct but
incomplete
If C is the Point D , then ADB = ACB .

ADB
 ACB
Hence the maximum lens view angle is ADB.
Angle of Taking a Scenery Photograph_Marking Guidelines
Very good: The proof is correct and
complete
5
Marking Guidelines
Solution
Performance
Part E
5.
(a)
(i)
Let F be the foot of perpendicular from E onto the y-axis.
AF = AB  b  a (line from centre perp. to chord bisects chord)
2
2
ba
2
ab

2
2
2
2
ba   ab ba 
x 2  EF 2  r 2  


 
 

 2   2   2 
r  OF  OA  AF  a 
(ii)
 ab
Evidence:
1.
The answer
2.
The explanation
Weak:
All answers are incorrect
Fair:
One of the answers is
correct and the
explanation is relevant
Good:
All the answers are
correct. The explanation
is correct but incomplete
 x  ab
(b)
Very good: All the answers are
correct. The explanation
is correct and complete
Yes, I agree.
  ADB =  ODB   OD A
tan  ADB  tan (  ODB   ODA )
= tan ODB  tan ODA
1  tan ODB  tan ODA
b a

 x x
ab
1 2
x
(b  a ) x
 2
x  ab
ba

2 ab
 ba
 ADB  tan 1 
 2 ab




Angle of Taking a Scenery Photograph_Marking Guidelines
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