Mathematics/P2
RBHS
September 2016
Gr 12 Mathematics
September Paper 2
MEMORANDUM
GRADE 12
Sept Paper 2 2016
MARK DISTRIBUTION
PAPER 2(150 MARKS)
Q
Data
Handling
(20±3)
1
2
Analytical
Geometry
(40±3)
Trigonometry
(40±3)
19
7
14
9
10
11
12
13
150
QUESTIONS
Routine
procedures(35%)
Complex
procedures(30%)
Problem
Solving(15%)
150
9
11
20
14
16
10
40
19
7
14
40
12
8
12
15
3
50
1
2
DH
3
4
5
Analyt
6
7
8
Trig
9
10
11
12
12
Euclid
7
10
85%
12
12
0
60%
10
2
8
50%
9
8
5
6
0
56%
2
1
15%
2
4
5
27.5%
9
3
2
35%
3
0
7
5
0
30%
0
0
0%
0
0
5
12.5%
0
2
4
15%
0
0
0
4
3
14%
9
11
20
14
16
10
40
19
7
14
40
12
8
12
15
3
50
(50±3)
14
16
10
6
7
8
TOTAL
PAPER 2(150 MARKS)
Euclidean
Geometry
9
11
3
4
5
COGNITIVE LEVELS
TOTAL
12
8
12
15
3
TOTAL
20
40
40
50
150
TOTAL
89
43
18
150
%
13
27
27
33
100
%
59
29
12
100
Ideal
55
30
15
Page 1 of 14
Mathematics/P2
RBHS
September 2016
QUESTION 1
1.1
𝑥̅ =
𝑥̅ =
1.2
4+5+8+13+19+22+25+26+23+17+14+7
12
183 
12 
𝑥̅ = 15,25 
(3)R
Standard deviation: 7,59522 ≈ 7,60 (R)
(2)R
1.3.1 Increase in mean =
(3×5)+(9×1)
12
= 2°C per month 
(2)R
1.3.2 Standard deviation decreases  The maximum value increases by 1°C and the minimum
value increases by 5°C i.e the range decreases  (new SD = 6,27).
NB they must reference range to get second mark. Esp if they write new SD
(2)C
[9]
Page 2 of 14
Mathematics/P2
RBHS
September 2016
QUESTION 2
2.1.1
Time (in min)
Frequency
11 ≤ 𝑡 < 15
15 ≤ 𝑡 < 19
19 ≤ 𝑡 < 23
23 ≤ 𝑡 < 27
27 ≤ 𝑡 < 31
6
7
15
18
2
Cumulative
frequency
6
13
28
46
48

(1)R
2.1.2
grounded
plotting points at
upper limits
−1 𝑝𝑒𝑟 𝑚𝑖𝑠𝑡𝑎𝑘𝑒
 smooth curve
(4)R
2.1.3 a)
b)
median = 22 (𝑢𝑠𝑒 𝑐𝑢𝑟𝑣𝑒)
(1)R
lower quartile 18
(1)R
2.1.4 48 − 37 = 9 𝑝𝑝𝑙
(𝑢𝑠𝑒 𝑐𝑢𝑟𝑣𝑒)
(𝑢𝑠𝑒 𝑐𝑢𝑟𝑣𝑒)
(2)R
2.2.1 C
𝑦 = 58,28 − 4,48𝑥
(1)R
2.2.2 C
late-coming and success
(1)C
[11]
Page 3 of 14
Mathematics/P2
RBHS
QUESTION 3
September 2016
y
T
S
63,43º
P(-5:0)
O
R
x
3.1.1 𝑚𝑃𝑇 = tan 63,43° 
= 2  (answer only full marks)
(2)R
3.1.2 𝑦 = 2𝑥 + 𝑐
0 = 2(−5) + 𝑐
𝑐 = 10 (for substituting gradient and point)
𝑃𝑇: 𝑦 = 2𝑥 + 10 
(2)R
3.1.3 𝑃𝑆 = √ (−5 − 0)2 + (0 − 10)2  OR
𝑃𝑆 2 = 102 + 52  (𝑝𝑦𝑡ℎ𝑎𝑔)
𝑃𝑆 = 5√5 
= 5√5
3.1.4 T(5; 20) 
3.2
(2)R
𝑃𝑂: 𝑂𝑅 = 2: 3
5 ∶ 𝑂𝑅 = 2 ∶ 3 
2 × 𝑂𝑅 = 15
𝑂𝑅 = 7,5
𝑅 (7,5; 0)  (answer only full marks)
3.3
(2)R
1
(2)C
1
𝐴𝑟𝑒𝑎 ∆𝑃𝑇𝑅 = 2 × 𝑃𝑅 × ℎ  = 2 × 12,5  × 20 
= 125 𝑢𝑛𝑖𝑡𝑠 2 
𝑶𝑹
1
𝐴𝑟𝑒𝑎 ∆𝑃𝑇𝑅 = 2 (10√5)(12,5) sin 63,43°   (𝑓)
= 124,99
(4)R
[14]
Page 4 of 14
Mathematics/P2
RBHS
September 2016
QUESTION 4
C(-1;26)
y
20
A

B(-1;1)
R

O
x
4.1
𝐵𝐶 = 26 − 1 = 25 𝑢𝑛𝑖𝑡𝑠
𝐶𝐴̂𝐵 = 90° (𝑡𝑎𝑛 ⊥ 𝑟𝑎𝑑)
𝐴𝐵 = 15
(3)R
4.2
(𝑥 + 1)2  + (𝑦 − 1)2  = 225
(3)R
4.3
𝜃 = 53,13° 
𝑚𝐶𝐴 = 𝑡𝑎𝑛 53,13°
4
= 3  (or 2 marks if no theta calc)
𝑆𝑢𝑏𝑠𝑡 𝐶(−1; 26): 26 =
4
y = 3𝑥 +
4.4
82
3
4
3
4
(−1) + 𝑐
1
 𝑜𝑟 𝑦 = 3 𝑥 + 27 3
𝑜𝑟 3𝑦 = 4𝑥 + 82
(4)C
3
𝑚𝐴𝐵 = − 4 
𝑆𝑢𝑏 𝐵(−1; 1)
3
1 = − 4 (−1) + 𝑐
3
1
y = −4𝑥 + 4
4.5
3
1
−4𝑥 + 4 =
4
3
(3)R
𝑥+
x = -13
82
3

y = 10
A(-13;10) 
(3)R
[16]
Page 5 of 14
Mathematics/P2
RBHS
QUESTION 5
September 2016
y
O
x
B
5.1
𝑂𝐵 = √(0 − 2𝑝)2 + (0 + 𝑝)2 =√45 +√20
√4𝑝2 + 𝑝2 = 3√5 +2√5
= 5√5 
√5𝑝2
5𝑝
5.2
2
= 125
𝑝2
= 25
𝑝
= 5
(5)C
𝑥 2 + 4𝑥 cos 𝜃 + 4 cos2 𝜃 + 𝑦 2 + 8𝑦 sin 𝜃 + 16 sin2 𝜃 = −3 + 4 cos 2 𝜃 + 16 sin2 𝜃
(𝑥 + 2 cos 𝜃)2 
+ (𝑦 + 4 sin 𝜃)2  = −3 + 4(1 − sin2 𝜃) + 16 sin2 𝜃
𝑟 2 = 1 + 12 𝑠𝑖𝑛2 𝜃
Max value of sin 𝜃 = 1, so max value of 1 + 12 𝑠𝑖𝑛2 𝜃 = 13
Max value of 𝑟 = √13
(5)P
[10]
Page 6 of 14
Mathematics/P2
RBHS
September 2016
QUESTION 6
5
6.1.1 cos 2𝐴 = − 13 
12
(quad)
sin 2𝐴 = 13
(values)
(3)R
13
5
12
6.1.2 cos 2𝐴 = 13
2A
5
2 cos 2 𝐴 − 1 = 13
-5
4
cos2 𝐴 = 13 
4
cos 𝐴 = √13 =
6.2
2
√13
=
2√13
13

(3)R
cos 70°. cos 10° + cos 20°. cos 80°
= cos 70°. cos 10° + sin 70°. sin 10°
= cos(70° − 10°) = cos 60°
1
= 2
(4)R
6.3.1 3 cos 𝑥 + 4 sin 𝑥
𝐿𝐻𝑆
3
4
3
= 5 (5 cos 𝑥 + 5 sin 𝑥) 
tan 𝐵 = 4  𝑝𝑖𝑐
5
= 5(sin 𝐵. cos 𝑥 + cos 𝐵. sin 𝑥) 
B
= 5 sin(𝐵 + 𝑥)
OR
𝑅𝐻𝑆
4
= 5(sin 𝐵. cos 𝑥 + cos 𝐵. sin 𝑥) 
3
3
pic
4
= 5 (5 . cos 𝑥 + 5 . sin 𝑥) 
= 3 cos 𝑥 + 4 sin 𝑥
(3)C
5
6.3.2 3 cos 𝑥 + 4 sin 𝑥 = 2
5
∴ 5 sin(𝐵 + 𝑥) = 2 
1
∴ sin(𝐵 + 𝑥) = 2 
𝑟𝑒𝑓 ∠ = 30°
3
𝐺𝑖𝑣𝑒𝑛 tan 𝐵 = 4
∴ 𝐵 = 36,87°
𝐵 + 𝑥 = 30° + 𝑛. 360°
𝐵 + 𝑥 = 180° − 30° + 𝑛. 360
or
𝑥 = −6,87° + 𝑛. 360°
𝑥 = 113,13° + 𝑛. 360° (𝑛 ∈ 𝑍)
(If they don’t sub in B value max 4/6)
Page 7 of 14
(6)C
[19]
Mathematics/P2
RBHS
September 2016
QUESTION 7
7.1
𝑎 = 30°
𝑏=2 
(2)R
7.2
−180° ≤ 𝑥 < 120° ; 𝑥 = −45° ; 60° < 𝑥 ≤ 180°
(3)C
7.3
𝑦 = −2sin(𝑥 − 30°)
(2)P
[7]
Page 8 of 14
Mathematics/P2
RBHS
September 2016
QUESTION 8
A
8.1.1
h
C
B
D
𝑐𝑜𝑛𝑠𝑡𝑟𝑢𝑐𝑡 𝐴𝐷 ⊥ 𝐵𝐶
𝑙𝑒𝑡 𝐶𝐷 = 𝑥 ∴ 𝐷𝐵 = 𝑎 − 𝑥
ℎ2 = 𝑏 2 − 𝑥 2 (𝑝𝑦𝑡ℎ𝑎𝑔 ∆𝐴𝐵𝐷)
ℎ2 = 𝑐 2 − (𝑎 − 𝑥 )2 (𝑝𝑦𝑡ℎ𝑎𝑔 ∆𝐴𝐷𝐶 )
𝑏 2 − 𝑥 2 = 𝑐 2 − (𝑎2 − 2𝑎𝑥 + 𝑥 2 )
𝑏 2 = 𝑐 2 − 𝑎2 − 2𝑎𝑥
𝑥
cos 𝐶̂ = 𝑏 ∴ 𝑥 = 𝑏𝑐𝑜𝑠 𝐶̂ 
𝑐 2 = 𝑎2 + 𝑏 2 − 2𝑎𝑏. cos 𝐶̂
8.1.2 cos 𝐶 =
𝑎2 +𝑏 2 −𝑐 2
2𝑎𝑏
Cos 𝐶 + 1 =
cos 𝐶 + 1 =
(5)R

𝑎2 +𝑏2 −𝑐 2
2𝑎𝑏
2𝑎𝑏
+ 2𝑎𝑏 
(𝑎+𝑏)2 −𝑐 2 
2𝑎𝑏
(𝑎+𝑏+𝑐)(𝑎+𝑏−𝑐)
1 + cos 𝐶̂ =
2𝑎𝑏
(4)P
8.2.2. 𝐷𝐹 = 41 (𝑝𝑦𝑡ℎ𝑎𝑔)
G
𝐺𝐹 = 20 + 41 − 2(20)(41) cos 15,6°
2
2
2
20m
𝐺𝐹 = 22,39 𝑚
(3)R
h
F
9m
15,6
ℎ
8.2.1 sin 15,6° = 20°
D
ℎ = 5,38𝑚
40 m
E
(2)C
[14]
Page 9 of 14
Mathematics/P2
RBHS
September 2016
QUESTION 9
M
A
1
2
x
C
B
1
1
2
4
3
2
1
S4
2
3
Q
2
1
1
2
3
R
P
9.1
Statement
Reason
𝐵̂1 = 𝑥
𝐴̂1 = 𝑥
𝒕𝒂𝒏 𝒄𝒉𝒐𝒓𝒅 𝒕𝒉𝒆𝒐𝒓𝒆𝒎 
𝑃𝐴 = 𝑃𝐵
𝒕𝒂𝒏𝒈𝒆𝒏𝒕𝒔 𝒇𝒓𝒐𝒎 𝒄𝒐𝒎𝒎𝒐𝒏 𝒑𝒐𝒊𝒏𝒕
̂𝟐 
𝑩
∠′ 𝒔 𝒐𝒑𝒑 = 𝒔𝒊𝒅𝒆𝒔 
=𝑥
𝑃̂2 = 𝑥
̂ 
𝑸
∠′ 𝑠 𝑜𝑝𝑝 = 𝑠𝑖𝑑𝑒𝑠
𝑎𝑙𝑡 ∠′ 𝑠 ; 𝐶𝐵||𝑃𝑄
=𝑥
𝑐𝑜𝑟𝑟𝑒𝑠𝑝  ’𝑠; 𝐶𝐵||𝑃𝑄
(5)R
9.2.1 𝐴̂1 = 𝑃̂2 𝑆 (𝑝𝑟𝑜𝑣𝑒𝑑 𝑎𝑏𝑜𝑣𝑒)
∴𝐴𝐵𝑅𝑃 𝑖𝑠 𝑐𝑦𝑐𝑙𝑖𝑐 (𝑐𝑜𝑛𝑣. ∠′ 𝑠 𝑖𝑛 𝑠𝑎𝑚𝑒 𝑠𝑒𝑔)𝑅
9.2.2 𝐴̂1+2 = 𝐵̂1+2 (𝑝𝑟𝑜𝑣𝑒𝑑 𝑎𝑏𝑜𝑣𝑒) 
∴𝐴𝑃 = 𝑃𝐵 (𝑠𝑖𝑑𝑒𝑠 𝑜𝑝𝑝 = ∠′ 𝑠)  S/R
𝑃̂2 = 𝑄̂ (𝑝𝑟𝑜𝑣𝑒𝑑 𝑎𝑏𝑜𝑣𝑒)
∴𝐵𝑃 = 𝐵𝑄 (𝑠𝑖𝑑𝑒𝑠 𝑜𝑝𝑝 = ∠′ 𝑠)  S/R
∴𝐴𝑃 = 𝐵𝑄
OR
𝑖𝑛 ∆𝑄𝐵𝑅 𝑎𝑛𝑑 ∆𝐴𝑃𝑅:
1. 𝐴̂2 = 𝑄̂ = 𝑥 (𝑝𝑟𝑜𝑣𝑒𝑛 𝑎𝑏𝑜𝑣𝑒)
2. 𝐴𝑅 = 𝑅𝑄 (𝑠𝑖𝑑𝑒𝑠 𝑜𝑝𝑝 = ∠𝑠)
3. 𝐵̂4 = 𝐴𝑃̂𝑅 (𝑒𝑥𝑡 ∠ 𝑜𝑓 𝑐𝑦𝑐𝑙𝑖𝑐 𝑞𝑢𝑎𝑑)  𝑓𝑜𝑟 𝑎𝑙𝑙 3 (−1 𝑝𝑒𝑟 𝑚𝑖𝑠𝑡𝑎𝑘𝑒)
∴∆𝐴𝑃𝑅 ≡ 𝑄𝐵𝑅 (𝐴𝐴𝑆) 
∴𝐴𝑃 = 𝑄𝐵
9.3
𝐴𝐶
=
𝐴𝐵
(2) R
(3)C
(𝑝𝑟𝑜𝑝 𝑡ℎ𝑒𝑜𝑟𝑒𝑚; 𝐶𝐵||𝑅𝑄) 𝑆  𝑅
𝐴𝑅
𝐴𝑄
𝐵𝐶
𝐴𝐵
∴ 𝐴𝑅 = 𝐴𝑄
(𝐴𝐶 = 𝐵𝐶)
(2) R
[12]
Page 10 of 14
Mathematics/P2
RBHS
September 2016
QUESTION 10
10.1
𝑐𝑜𝑛𝑠𝑡𝑟𝑢𝑐𝑡 𝐴𝑂 𝑎𝑛𝑑 𝑒𝑥𝑡𝑒𝑛𝑑 𝑡𝑜 𝐷
𝐿𝑒𝑡 𝐵𝐴̂𝑂 = 𝑥
∴𝐴𝐵̂ 𝑂 = 𝑥 (∠′ 𝑠 𝑜𝑝𝑝 = 𝑠𝑖𝑑𝑒𝑠)
𝐿𝑒𝑡 𝐶𝐴̂𝑂 = 𝑦
∴ 𝐶𝐴̂𝑂 = 𝑦 (∠′ 𝑠 𝑜𝑝𝑝 = 𝑠𝑖𝑑𝑒𝑠)
𝐵𝑂̂𝐷 = 2𝑥 (𝑒𝑥𝑡 ∠ ∆𝐵𝐴𝑂)
𝐶𝑂̂𝐷 = 2𝑦 (𝑒𝑥𝑡 ∠ ∆𝐶𝐴𝑂)
∴𝐵𝑂̂𝐶 = 2𝑥 + 2𝑦 = 2(𝑥 + 𝑦)
= 2(𝐴̂)
∴𝐵𝑂̂𝐶 = 2 × 𝐵𝐴̂𝐶
10.2
A
O
C
B
(4)R
P
Q
2x
O
R
S
𝑃𝑂̂ 𝑅 = 180° − 2𝑥 (𝑐𝑜𝑖𝑛𝑡 ∠′ 𝑠; 𝑃𝑄||𝑂𝑅) 𝑆/𝑅
𝑆̂ = 90° − 𝑥 (∠ 𝑎𝑡 𝑐𝑒𝑛𝑡𝑟𝑒 = 2 × ∠ 𝑎𝑡 𝑐𝑖𝑟𝑐𝑢𝑚𝑓)𝑆/𝑅
∴𝑃𝑄̂ 𝑅 = 90° + 𝑥 (𝑜𝑝𝑝 ∠′ 𝑠 𝑐𝑦𝑐𝑙𝑖𝑐 𝑞𝑢𝑎𝑑) 𝑆 𝑅
(4)R
[8]
Page 11 of 14
Mathematics/P2
RBHS
September 2016
QUESTION 11
P
Given: 𝑃𝑄 = 30 units
𝑄𝑅 = 20 units
𝑃𝑅 = 40 units
𝑄𝑆 = 10 units
𝑅𝑆 = 15 units
40
30
20
Q
10
11.1
𝑃𝑄
𝑆𝑄
𝑃𝑅
3.
11.2
11.3
30
𝑅𝑆
𝑄𝑅
2.
15
S
In ∆𝑃𝑄𝑅 𝑎𝑛𝑑 ∆𝑅𝑆𝑄:
1.
R
𝑅𝑄
= 15 = 2
20
= 10 = 2
=
40
20
=2
 (S)
∴∆𝑃𝑄𝑅|||∆𝑅𝑆𝑄 (𝑠𝑖𝑑𝑒𝑠 𝑜𝑓 ∆ 𝑖𝑛 𝑝𝑟𝑜𝑝) 𝑅
(3)R
𝑃𝑅̂ 𝑄 = 𝑅𝑄̂ 𝑆 (∆𝑃𝑄𝑅|||∆𝑅𝑆𝑄) 𝑆
∴ 𝑄𝑆||𝑃𝑅 (𝑎𝑙𝑡 ∠′ 𝑠 =) 𝑅
(2)R
In ∆𝑄𝑋𝑆 and ∆𝑃𝑋𝑅:
1.
𝑄𝑋𝑆 = 𝑃𝑋𝑅 (𝑣𝑒𝑟𝑡 𝑜𝑝𝑝 ∠′ 𝑠 =) 𝑆
2.
𝑄𝑆𝑋 = 𝑋𝑃𝑅 (𝑎𝑙𝑡 ∠′ 𝑠; 𝑄𝑆||𝑃𝑅)S
∴ ∆𝑄𝑋𝑆|||∆𝑃𝑋𝑅 (𝐴𝐴𝐴) 𝑅
𝑄𝑋
𝑄𝑆
∴𝑅𝑋 = 𝑅𝑃 (∴ ∆𝑄𝑋𝑆|||∆𝑃𝑋𝑅)
𝑄𝑋
10
∴20−𝑄𝑋 = 40  S
∴𝑄𝑋 = 4 𝑆
11.4
(5)C
∆𝑃𝑄𝑅: ∆𝑄𝑆𝑅 = 4: 1  𝑆
𝑎𝑟𝑒𝑎 ∆𝑃𝑄𝑅
𝑎𝑟𝑒𝑎 𝑃𝑄𝑆𝑅
4
=5 𝑆
(2)C
[12]
Page 12 of 14
Mathematics/P2
RBHS
September 2016
C
QUESTION 12
B
12.1
5
T
x
3
A
12.1.1
12.1.2
D
In ∆𝐴𝐵𝑇 and ∆𝐶𝐷𝑇
1.
𝐵𝑇̂𝐴 = 𝐷𝑇̂𝐶 (𝑣𝑒𝑟𝑡 𝑜𝑝𝑝 ∠′ 𝑠 =) 𝑆
̂ 𝐶 (𝑎𝑙𝑡∠′ 𝑠; 𝐴𝐵||𝐷𝐶) 𝑆
2.
𝐴𝐵̂ 𝑇 = 𝑇𝐷
∴ ∆𝐴𝐵𝑇|||∆𝐶𝐷𝑇 (𝐴𝐴𝐴) 𝑅
(3)R
𝐴𝐵 𝑖𝑠 𝑎 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 (𝑐𝑜𝑛𝑣. ∠′ 𝑠 𝑖𝑛 𝑠𝑒𝑚𝑖 𝑐𝑖𝑟𝑐𝑙𝑒) S R
𝐴𝐵
𝐴𝑇
= 𝐶𝑇 (∆𝐴𝐵𝑇|||∆𝐶𝐷𝑇)
𝐶𝐷
𝐴𝐵
3
 𝑆
∴𝑥 =5
3
𝐴𝐵 = 5 𝑥
3
𝑆
𝑟 = 10 𝑥
3
𝐴𝑟𝑒𝑎 = 𝜋 (10 𝑥)
2
9
𝐴𝑟𝑒𝑎 = 100 𝜋𝑥 2 𝑢𝑛𝑖𝑡𝑠 2
S
(5)C
P
12.2
R
Q
12.2.1
12.2.2
𝐿𝑒𝑡 𝑃̂ = 𝑥
∴ 𝑥 + 4𝑥 + 4𝑥 = 180°  𝑆 (∠ 𝑠𝑢𝑚 𝑜𝑓 ∆) 𝑅
𝑃̂ = 20°  𝑆
(3)R
𝑄𝑂̂𝑅 = 40°  𝑆 (∠𝑎𝑡 𝑐𝑒𝑛𝑡𝑟𝑒 = 2 × ∠𝑎𝑡 𝑐𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑐𝑒) 𝑅
360°
(∠′ 𝑠 𝑎𝑟𝑜𝑢𝑛𝑑 𝑎 𝑝𝑜𝑖𝑛𝑡) 𝑆/𝑅
𝑛=
40°
𝑛 =9𝑆
(4)P
[15]
Page 13 of 14
Mathematics/P2
RBHS
September 2016
QUESTION 13
A
x y
B
x
a
M
N
P
C
ab
b
y D
𝐶𝑜𝑛𝑠𝑡𝑟𝑢𝑐𝑡 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑠 𝑎𝑡 𝑎𝑙𝑙 𝑝𝑜𝑖𝑛𝑡𝑠 𝑜𝑓 𝑐𝑜𝑛𝑡𝑎𝑐𝑡, 𝑙𝑎𝑏𝑒𝑙 𝑝𝑜𝑖𝑛𝑡𝑠 𝑀, 𝑁 𝑎𝑛𝑑 𝑃 𝑎𝑠 𝑝𝑒𝑟 𝑑𝑖𝑎𝑔𝑟𝑎𝑚
𝑀𝐵 = 𝑀𝐴 (𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑠 𝑓𝑟𝑜𝑚 𝑐𝑜𝑚𝑚𝑜𝑛 𝑝𝑜𝑖𝑛𝑡)
𝑁𝐵 = 𝑁𝐶 (𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑠 𝑓𝑟𝑜𝑚 𝑐𝑜𝑚𝑚𝑜𝑛 𝑝𝑜𝑖𝑛𝑡)
𝑃𝐶 = 𝑃𝐷 (𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑠 𝑓𝑟𝑜𝑚 𝑐𝑜𝑚𝑚𝑜𝑛 𝑝𝑜𝑖𝑛𝑡) 
∴ 𝑏𝑎𝑠𝑒 𝑎𝑛𝑔𝑙𝑒𝑠 𝑜𝑓 𝑎𝑙𝑙 𝑟𝑒𝑙𝑒𝑣𝑎𝑛𝑡 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒𝑠 𝑎𝑟𝑒 𝑒𝑞𝑢𝑎𝑙  (∠′ 𝑠 𝑜𝑝𝑝 = 𝑠𝑖𝑑𝑒𝑠)
𝑠𝑒𝑒 𝑑𝑖𝑎𝑔𝑟𝑎𝑚.
2𝑥 + 2𝑦 + 2𝑎 + 2𝑏 = 360° (∠𝑠𝑢𝑚 𝑖𝑛 𝑞𝑢𝑎𝑑 𝐴𝐵𝐶𝐷) 
∴ 𝑥 + 𝑦 + 𝑎 + 𝑏 = 180°
𝐴̂ + 𝐶̂ = 𝑥 + 𝑦 + 𝑎 + 𝑏 = 180°
∴𝐴𝐵𝐶𝐷 𝑖𝑠 𝑐𝑦𝑐𝑙𝑖𝑐 (𝑐𝑜𝑛𝑣. 𝑜𝑝𝑝 ∠′ 𝑠 𝑐𝑦𝑐𝑙𝑖𝑐 𝑞𝑢𝑎𝑑)
[3]P
[TOTAL 150]
Page 14 of 14