This memorandum consists of 21 pages. MATHEMATICS P2. FEBRUARY/
MARCH 2013. MEMORANDUM. NATIONAL. SENIOR CERTIFICATE. GRADE 12
...
NATIONAL SENIOR CERTIFICATE
GRADE 12
MATHEMATICS P2 FEBRUARY/MARCH 2013 MEMORANDUM
MARKS: 150
This memorandum consists of 21 pages.
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2 NSC – Memorandum
DBE/Feb.–Mar. 2013
NOTE: • If a candidate answers a question TWICE, only mark the FIRST attempt. • If a candidate has crossed out an attempt of a question and not redone the question, mark the crossed out version. • Consistent Accuracy applies in ALL aspects of the marking memorandum. QUESTION 1 1.1
Scatter plot of exchange rate versus oil price 82 81 80
any 4 points correctly plotted any 9 points correctly plotted all points correctly plotted
79 78 77
Oil price (in $)
76 75 74 73 72 71 70 69 68 67 66 65 6.7
6.8
6.9
7
7.1
7.2
7.3
7.4
7.5
7.6
7.7
Exchange rate (in R/$) 1.2
1.3
1.4 1.5
As the exchange rate (R/$) increases the oil price ($) decreases. OR There is a negative correlation between the exchange rate and oil price. 852,6 Mean = 12 = 71,05 Standard deviation is: σ = 4,09 2 standard deviations from the mean = 71,05+ 2(4,09) = 79,23 The public will be concerned in December 2010
7.8
(3) reason (2) 852,6 71,05 (2) 4,09 (2) 79,23 Dec 2010 (2) [11]
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QUESTION 2 2.1
94 – 68 26
Range of Peter’s scores is 94 – 68 = 26
(2) 2.2
76
Vuyani’s minimum score is 76
(1) 2.3
Vuyani was more consistent during the year because the range of his scores is more clustered about the median value OR the range and inter-quartile range are smaller than Peters.
Vuyani reason (2) [5]
QUESTION 3 3.1
Cumulative Frequency Graph 160 150
plotting points at cumulative frequencies
140 130
Cumulative frequency
120 110
plot against upper limits
100 90 80
grounded at (0 ; 0)
70 60 50 40
smooth curve
30 20 10 0 0
10
20
30
40
50
60
Percentage interval
3.2.1
3.2.2
(85 ; ± 144 ) ± 144 learners that scored below 85% (Accept: 144 to 146) Q 1 = 25 or 27 or 26 Q 3 = 61 or 62 or 64 Interquartile range = 36 or 35 or 38
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70
80
90
100
(4)
(85 ; ± 144) ± 144 learners (2) lower quartile upper quartile IQR (3) [9]
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QUESTION 4 4.1
m AD =
y 2 − y1 x 2 − x1
substitution
7 − (−3) = 1 − (−4) =2
2 (2)
4.2
m AD = 2
AD//BC m AD = m BC = 2
substitute into formula
y − y1 = m( x − x1 ) y − (−8) = 2( x − (−2) ∴ y = 2x − 4
y = 2x − 4 (3)
4.3
4.4
At F: y = 0 0 = 2x – 4 x=2 F(2 ; 0) D is translated C according to the rule: D( x ; y ) → C ( x + 2 ; y − 5) A must also be translated according to this rule to B/. ∴A(1 ; 7) → B/ (3 ; 2)
y=0 x=2 (2) x=3 y=2 (2)
OR x=3 y=2
x B / = −2 + (1 + 4) = 3 y B / = −8 + (7 + 3) = 5 4.5
(2)
m BC = 2 tan θ = 2 β θ = 63,43° D(–4 ; –3) − 8 − (−3) 5 =− m DC = − 2 − (−4) 2 C(–2 ; –8) 5 tan β = − 2 β = 180° − 68,20° = 111,80° α = 111,80° − 63.43° = 48,37°
θ F(2 ; 0)
63,43° tan β = −
5 2
111,8° 48,37° (4)
OR
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DC = (−4 + 2) 2 + (−3 + 8) 2 = 29 CF = (−2 − 2) 2 + (−8 − 0) 2 = 80 Subst in cosformula cos α subject 0,6643... 48,37° (4)
DF = (2 + 4) 2 + (0 + 3) 2 = 45 29 + 80 − 45 cos α = 2( 29 )( 80 ) = 0,6643...
α = 48,37° OR DC = (−4 + 2) 2 + (−3 + 8) 2 = 29 DB = (3 + 4) 2 + (2 + 3) 2 = 74 BC = (3 + 2) 2 + (2 + 8) 2 = 125 29 + 125 − 74 cos α = 2( 29 )( 125 ) = 0,6643...
Subst in cosformula cos α subject 0,6643... 48,37° (4)
α = 48,37° 4.6
DC = (−4 + 2) 2 + (−3 + 8) 2 = 29 CF = (−2 − 2) 2 + (−8 − 0) 2 = 80 1 .DC.CF. sin α 2 1 = ( 29 )( 80 ) sin 48,37° 2 = 18 units 2
substitution into formula 29 substitution into formula 80
Area ∆DCF =
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substitution into the area rule 18 (6)
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OR 6
F
(1) Area=9
3
establishing rectangle and area
D 8 5
(3) Area =5
(2) Area = 16 relationship of areas (1) = 9 (2) = 16 (3) = 5 18 units2 (6)
2 C 4 Area ∆DCF = Area of rectangle – (1) – (2) – (3) = 48 – 9 – 5 – 16 = 18 sq units
OR
G(–4 ; 0)
H(–2 ; 0)
F (2 ; 0)
drawing perpendiculars
D (–4 ; –3)
C (–2 ; –8) Area CDF = Area CHF + Area CDGH - Area DGF 1 1 1 × 4 × 8 + 2 × (3 × 8) − × 6 × 3 2 2 2 = 16 + 11 − 9 = 18 =
relationship of areas 16 11 9 18 units2 (6) [19]
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QUESTION 5 5.1.1
x 2 + y 2 + 2x + 6 y + 2 = 0 x 2 + 2 x + 1 + y 2 + 6 y + 9 = −2 + 10 ( x + 1) 2 + ( y + 3) 2 = 8 M (−1 ; − 3)
5.1.2 5.2
radius of circle C1 = 8
x 2 + ( x − 2) 2 + 2 x + 6( x − 2) + 2 = 0
( x + 1) 2 + ( y + 3) 2 = 8 –1 –3 (3) 8 (1) substitution
x 2 + x 2 − 4 x + 4 + 2 x + 6 x − 12 + 2 = 0 2x 2 + 4x − 6 = 0 x 2 + 2x − 3 = 0 (x + 3)( x − 1) = 0 x = −3 or x ≠ 1 y = −3 − 2 = −5
standard form factors value of x value of y (5)
∴ D(−3 ; − 5) OR
( x + 1) 2 + ( y + 3) 2 = 8 subst. y = x − 2
substitution
( x + 1) 2 + ( x − 2 + 3) 2 = 8 ( x + 1) 2 + ( x + 1) 2 = 8 x 2 + 2x − 3 = 0 (x + 3)( x − 1) = 0 x = −3 or x ≠ 1 y = −3 − 2 = −5 OR ( x + 1) 2 + ( y + 3) 2 = 8 subst. y = x − 2
standard form factors value of x value of y (5)
substitution
( x + 1) 2 + ( x − 2 + 3) 2 = 8 ( x + 1) 2 + ( x + 1) 2 = 8
(x + 1)2
= x +1 = x= y=
4 ±2 −3 or x ≠ 1 −3 − 2 = −5
simplification square root of both sides value of x value of y
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PM makes 45° with the x-axis.
8 = 22 + 22 Therefore: x D = x M − 2 = −1 − 2 = −3
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8 = 2 2 + 2 2 value of x value of y (5)
y D = −3 − 2 = −5 5.3
MD ⊥ DB
(tangent ⊥ radius)
MB 2 = MD 2 + DB 2
(Pythagoras)
= ( 8 ) 2 + (4 2 ) 2 = 40 MB is the radius of C 2
MB = 40
tangent ⊥ radius substitution into Pythagoras
40
(3) 5.4
5.5
( x + 1) + ( y + 3) = 40 2
2
(
)
Distance from 2 5 ; 0 to centre =
(2
)
5 + 1 + (0 + 3) 2
2
= 6,24
(
)
6,24 < 6,32 40 Distance from 2 5 ; 0 to centre < radius of circle. 2 5 ; 0 lies inside the circle.
(
)
(
)
LHS RHS (2) substitution into distance formula 6,24 6,24 < 6,32 conclusion (4) [18]
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QUESTION 6 6.1.1
A( – 5 ; 3) A / (−5 + 4 ; 3 − 3) = (−1 ; 0)
6.1.2
A (−5 ; − 3)
–1 0 (2) –5 –3
/
(2) 6.2.1
/
Scale factor of enlargement is
/
KM 15 3 = = KM 10 2
/
KM KM 3 2
/
OR 3 3 K (−4 ; 2) → K / (−6 ; 3) = K / × −4 ; × 2 2 2 3 Scale factor is 2
y
3 3 × −4 ; × 2 2 2 3 2 (2) 3 x 2 3 y 2 (2) 9 2 3 (2)
6.2.2
3 3 ( x ; y) → x ; 2 2
6.2.3
3 3 P/ ×3 ; 2× 2 2 9 = P / ; 3 2
6.2.4
a=1
a=1
6.2.5
K (4 ; − 2)
4 –2
6.2.6
K K =5
K K = 5 K / M /// = 15
(2)
//
(2) ///
/
K / M /// = 15 K / K /// 5 1 = = / /// 15 3 K M
///
/
1 3 (3) [17]
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QUESTION 7 b –a
7.1
K / (b ; − a )
7.2
K (b cos θ − a sin θ ; − a cos θ − b sin θ ) //
OR K (a cos(90 + θ ) + b sin(90° + θ ) ; b cos(90° + θ ) − a sin(90° + θ ) //
= K // (−a sin θ + b cos θ ; − b sin θ − a cos θ )
7.3
T // (−(−4) sin θ + (−2) cos θ ; − (−2) sin θ − (−4) cos θ ) = T // (4 sin θ − 2 cos θ ; 2 sin θ + 4 cos θ )
OR
T // (−2 cos θ − (−4) sin θ ; − (−4) cos θ − (−2) sin θ ) = T // (−2 cos θ + 4 sin θ ; 4 cos θ + 2 sin θ )
7.4
2 3 + 1 = 4 sin θ − 2 cos θ ........(1) 3 − 2 = 2 sin θ + 4 cos θ ........(2) (2) × 2 : 2 3 − 4 = 4 sin θ + 8 cos θ ....(3) (1) − (3) : 5 = −10 cos θ 1 − = cos θ 2 ∴ θ = 180° − 60° = 120° OR
2 3 + 1 = 4 sin θ − 2 cos θ ........(1) 3 − 2 = 2 sin θ + 4 cos θ ........(2) (1) × 2 : 4 3 + 2 = 8 sin θ − 4 cos θ ....(3) (2) + (3) : 5 3 = 10 sin θ
(2) b cos θ − a sin θ − a cos θ − b sin θ (2) 4 sin θ − 2 cos θ 2 sin θ + 4 cos θ (2) 4 sin θ − 2 cos θ 2 sin θ + 4 cos θ (2) substitution to form equation substitution to form equation 5 = −10 cos θ 1 − = cos θ 2 120° (5) substitution to form equation substitution to form equation 5 3 = 10 sin θ
3 = sin θ 2 ∴ θ = 180° − 60° = 120°
3 = sin θ 2 120°
(5) OR
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mOT =
mOT / =
11 NSC – Memorandum
1 ˆT = 1 ⇒ tan XO 2 2 ˆ XOT = 206,565...° 3−2
⇒ tan XOˆ T // =
2 3 +1 ˆ T = −3,434° XO
3−2 2 3 +1
= −0,06....
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ˆT = tan XO
1 2
206.565...° – 0,06...
90° + θ = 209,99...° ≈ 210° θ = 120°
– 3.434° 120° (5) OR
(TT/)2 = OT2 + (OT/)2 – 2(OT)(OT/).cos (90° + θ) 40 + 20 3 = 40 − 40. cos(90° + θ ) 3 cos(90° + θ ) = − 2 90° + θ = 150° θ = 60°
(TT/)2 = 40 + 20 3 substitution in cos-rule simplification 150° 60° (5) [11]
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QUESTION 8 8.1
1 − sin 2 θ + 3 − cos 2 θ
simplification
= 4 − (sin 2 θ + cos 2 θ ) =3
3 (2) OR
cos θ + 3 − cos θ =3 2
substitution with identity 3
2
(2) 8.2
4
sin 150°
.2
rewrite using reduction formula substituting special angles simplification
3 tan 225°
= 4 sin 30°.2 3 tan 45° =
(2 ) .2 1 2 2
3
= 16 =4
4 (4) OR
1 sin 150° = 2 tan 225° = 1
1 2 tan 225° = 1
sin 150° =
4 sin 150° 2 3 tan 225° 1
= 4 2 23 = 2.2
3
= 16 =4
8.3
cos 2 x(sin 2 x + cos 2 x) 1 − sin x 2 cos x . (1) = 1 − sin x (1 − sin 2 x) = 1 − sin x (1 + sin x)(1 − sin x) = 1 − sin x =1 + sin x
LHS =
= RHS
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1
42 = 2 4 (4) factorisation 1 1 − sin 2 x factors
(4)
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8.4
13 NSC – Memorandum
cos 3θ = cos(2θ + θ ) = cos 2θ . cos θ − sin 2θ . sin θ = (2 cos θ − 1). cos θ − 2 sin θ . cos θ . sin θ 2
= 2 cos 3 θ − cos θ − 2 sin 2 θ . cos θ = 2 cos 3 θ − cos θ − 2(1 − cos 2 θ ). cos θ = 2 cos θ − cos θ − 2 cos θ + 2 cos θ 3
3
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expansion 2 cos 2 θ − 1 2 sin θ . cos θ 1− cos 2 θ
= 4 cos 3 θ − 3 cos θ
8.5
cos 3θ = 4 cos θ − 3 cos θ
(4)
3
cos 3(20°) = 4 cos 3 (20°) − 3 cos(20°) 1 = 4 x 3 − 3x 2 8x 3 − 6 x − 1 = 0
θ = 20° cos 60° =
1 2 (2) [16]
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QUESTION 9 9.1
cos160°. tan 200° 2 sin( −10°) (− cos 20°)(tan 20°) = 2(− sin 10°)
− cos 20° tan 20° − sin 10°
sin 20° (− cos 20°) cos 20° = − 2 sin 10° 2 sin 10° cos10° = 2 sin 10° = cos10° 9.2.1
2 sin 10° cos10° cos10°
LHS = cos( x + 45°). cos( x − 45°) = (cos x. cos 45° − sin x sin 45°)(cos x cos 45° + sin x sin 45°) = cos 2 x. cos 2 45° − sin 2 x. sin 2 45° 2 2 − sin 2 or cos 2 = cos x 2 2 1 1 = cos 2 x − sin 2 x 2 2 1 = (cos 2 x − sin 2 x) 2 1 = cos 2 x 2 2
sin 20° cos 20°
2
2
2
1 x − sin 2 2
2 1 x 2
(6) expand cos( x + 45°) expand cos( x − 45°) substitute special angles simplification
(4) OR 2 cos α cos β = cos(α + β ) + cos(α − β ) 1 cos α cos β = (cos(α + β ) + cos(α − β ) ) 2 Let α = x + 45° and β = x − 45° ∴ cos( x + 45°) cos( x − 45°)
deriving identity
1 (cos(( x + 45° + x − 45°) + cos( x + 45° − x + 45°) ) 2 1 = (cos 2 x + cos 90°) 2 1 = cos 2 x 2
substitution
=
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simplification
(4)
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9.2.2
15 NSC – Memorandum
cos( x + 45°) cos( x − 45°) has a minimum when
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1 cos 2 x has a 2
minimum. The minimum value of cos 2 x is –1
cos 2 x = −1 2 x = 180°
minimum value of –1 2x = 180° x = 90°
x = 90°
(3) [13]
QUESTION 10 10.1
Range = [−1 ; 1]
[−1 ; 1] (2)
10.2
3 3 f x = sin 2 x 2 2 = sin 3 x 360° ∴ Period = = 120° 3
sin 3x 120° (2)
OR 3 3 f x = sin 2 x 2 2 = sin 3 x = sin(3 x + 360°) = sin 3( x + 120°) ∴ Period = 120°
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sin 3x
120° (2)
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10.3 1.5
f
1
(30° ; 1) 0,87
0.5
(90° ; 0,5) -180
-120
-60
60
x intercepts turning points shape
-0.5
(-180° ; -0,87)
g
-1
(4)
(-150° ; -1) -1.5
10.4
(−180° ; − 90°) or (−60° ; 0°)
> – 180° < –90° > –60° < 0°
OR − 180° < x < −90° or − 60° < x < 0°
(4) translation 30° to the left
10.5
y = sin 2(x + 30°) ∴ translation of 30° to the left
10.6
sin 2 x = cos( x − 30°) sin 2 x = sin[90° − ( x − 30°)] = sin(120° − x) 2 x = 180° − (120° − x) + 360°k 2 x = 120° − x + 360°k ; k ∈ Z or 2 x − x = 60° + 360°k 3 x = 120° + 360°k x = 60° + 360°k ; k ∈ Z x = 40° + 120°k ; k ∈ Z
(2) using co-function 2 x = 120° − x + 360°k x = 40° + 120°k 2 x = 180° − (120° − x) + 360°k x = 60° + 360°k k ∈Z (6)
OR sin 2 x = cos( x − 30°) cos(90° − 2 x) = cos( x − 30°) 90° − 2 x = x − 30° + 360°k or 90° − 2 x = 360° − ( x − 30°) + 360°k
− 3 x = −120° + 360°k x = 40° − 120°k ; k ∈ Z
− x = 300° + 360°k x = −300° − 360°k ; k ∈ Z
∴ x = 40° + 120°k or x = 60° + 360°k ; k ∈ Z
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cos(90° − x) = cos( x − 30°) 90° − 2 x = x − 30°
+ 360°k x = 40° − 120°k 90° − 2 x = 360° − ( x − 30°) + 360°k x = −300° − 360°k k ∈Z (6) [20] Please turn over
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QUESTION 11 11.1
AB = sin 2 x r AB = r sin 2 x
AB = sin 2 x r AB = r sin 2 x
(2) 11.2
ˆ C = 90° + x AK
AKˆ C = 90° + x (1)
11.3 C
sine rule
x x
substitution
r
B
90°- x 90°+ x K
A
making AK subject of the formula cos x
In ΔAKC : ˆ C sin ACˆK sin AK = AC AK sin(90° + x) sin x = r AK r sin x r sin x AK = = sin(90° + x) cos x AK 2 = AB 3 r sin x cos x = 2 r sin 2 x 3 sin x 2 cos x = 2 sin x cos x 3 sin x 1 2 × = cos x 2 sin x cos x 3 1 2 = 2 2 cos x 3 4 cos 2 x = 3
2sinx.cosx
cos x =
3 cos x = 2 x = 30°
3 2
x = 30° OR
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1 2 cos 2 x
(8)
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C x x r
90°- x 90°+ x K Using the sine-formula in ∆CBK and ∆CKA: sin x sin(90° − x) sin x sin(90° + x) = = and KA AC BK BC BK KA ∴ = BC AC 1 2 ∴ = BC r 1 ∴ BC = r 2 BC 12 r 1 ∴ cos 2 x = = = 2 AC r ∴ 2 x = 60° ∴ x = 30° B
A
sin x sin(90° − x) = BK BC sin x sin(90° + x) = KA AC BK KA = BC AC 1 2 = BC r 1 BC = r 2 1 cos 2 x = 2 2x = 60°
x = 30°
OR
(8)
C x x r
90°- x 90°+ x K c
B ∆CBK: KC =
90°- 2 x 2c
A
c sin x
sin x sin(90° − 2 x) sin(90° − 2 x). sin x = = 2c KC c
∆CKA:
1 = sin 30° 2 90° - 2x = 30° x = 30°
∴ sin (90° - 2x) = ∴
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KC =
c sin x
sin x sin(90° − 2 x) = 2c KC substitution 1 sin (90° - 2x) = 2 90° - 2x = 30° x = 30°
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(8)
OR
C x x
r
B
90°- x 90°+ x K c
90°- 2 x 2c
A
∆CBK:
3c = 2 sin x. cos x r 3c .......................(1) ∴ r sin x = 2 cos x sin 2 x =
∆CKA: 2c r = sin x cos x ∴ r sin x = 2c cos x ..........................(2) Equate (1) and (2): 3c 2c. cos x = 2 cos x 3 ∴ cos 2 x = 4 3 ∴ cos x = 2 x = 30° ∴ OR
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3c r 2 sin x. cos x 3c r sin x = 2 cos x sin 2 x =
2c r = sin x cos x r sin x = 2c cos x
equating
cos x = 30°
3 2
(8)
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AK 2 = =2 KB 1 2=
1 2 1 2
AK .BC BK .BC
multiplying by 12 BC area of triangles
area AKC area ABC 1 rCK sin x = 12 2 BC .CK sin x =
area formula in triangles r =2 BC BC 1 = r 2 1 cos 2 x = 2 2x = 60°
r BC BC 1 ∴ = r 2 =
∴ cos 2 x =
1 2
∴ 2 x = 60° ∴ x = 30°
x = 30° (8) OR C x x r
B
c
K
By the Internal Bisector Theorem: CB BK 1 = = CA KA 2 1 cos 2 x = 2 2 x = 60° x = 30°
2c
A For stating Internal Bisector Theorem CB BK 1 = = CA KA 2
cos 2 x =
1 2
2x = 60° x = 30° (8)
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OR D x
r
C x x r x B
0°- x 90°+ x K
Produce BC to D and draw CK parallel to DA. CAˆ D = KCˆ A and BCˆ K = Dˆ ∴ DC = CA = r ∴ ∆BKC ||| ∆BAD BK BC = =3 ∴ BA BD ∴ BD = 3BC = BC + r
∴ BC =
1 r 2
∴ cos 2 x = ∴ 2 x = 60° ∴ x = 30°
1 2
r 1 = r 2
A
DC = CA = r ∆BKC ||| ∆BAD BK BC = =3 BA BD BD = BC + r 1 BC = r 2 1 cos 2 x = 2 2x = 60° 30° (8) [11] TOTAL: 150
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