NATIONAL

GRADE 10
SENIOR CERTIFICATE

MEMORANDUM MATHEMATICS PAPER 2

MID-YEAR 2017

MARKS: 75

TIME: 1,5 hours

This marking guideline consists of 8 pages

 N0. QUESTION 1 [08] COMMENTS
1.1 sin( K  L  M )
1
cos2 7K 
sin( L  40 )
sin( 30  50  70)
= 🗸substitution
1
cos2 7(30) 
sin( 50  40) 🗸answer
= -2
(2)
1.2 3 sin 60  sin 45  cos 45 
  

3

3

3

1

1 🗸
2
1 2 2 2
1
3 1 🗸
  2
2 2
1
1 🗸
2
🗸answer

(4)

1.3
4 tan2 288,2 cos164,6
sin 199,4 🗸Simplification
37,00330944  0.9640954042
= 🗸Answer
 0,3321611319
= 107,40 Answer only(full
marks)
(2)

[08]

NO. QUESTION 2 [14] COMMENTS

2.1.1 sin( 2  40) 
 0,175
3
 multiplying by 3
sin( 2  40) 
(3) :  0,525
3 🗸 2  4  31,67
2  4  31,67
2  71,67 🗸answer
  35,83 (3)
2.2 cos 40  sec 40  cos 30 3 1 3
x  sin 60 ∘   🗸 🗸 🗸
 
tan 45  cos 0 2 cos 40 2
1 3
cos 40    🗸 tan 45 ∘  1
x
3
 cos 40  2
2 11 🗸 cos 0  1
2 1
( ): x 
1 🗸answer
3
(6)
x 1

2.3
y

4
O A 🗸Labeled diagram
x & correct quadrant
3

3 tan A  4  0
4
tan A  🗸 tan A 
4
3 3
r  x  y2
2 2

 32  42
 9 16
 25
r  5 🗸r 5

10 sin 2 A  5cos2 A
2 2 🗸correct subst. into
4 3 expression
10   5 
5  5 

 16   9 
10    5 
 25   25 

205

25
1
8 🗸answer
5

(5)

[14]
NO. QUESTION 3 [15] COMMENTS
3.1.1
PR 3,2
 sin Q̂ 🗸 sin 33,45 ∘ 
PQ PQ
3,2 sin 33,45 ∘ 3,2
 🗸 PQ 
PQ 1 sin 33,45 ∘
PQ sin 33,45∘  3,2
🗸Answer
3,2
PQ 
sin 33,45 ∘ (3)
 PQ  5,81cm
3.1.2 P̂  180 ∘  90 ∘  33,45 ∘
 P̂  56,55 ∘ 🗸 P̂  180 ∘  90 ∘  33,45 ∘
🗸Answer
(2)
3.1.3 PR
 tan Q̂
QR 3,2
🗸 tan 33,45 ∘ 
3,2 tan 33,45 ∘ QR

QR 1 3,2
🗸 QR 
QR tan33,45∘  3,2 tan 33,45 ∘
3,2 🗸Answer
QR 
tan 33,45 ∘
(3)
 QR  4,84cm

3.2.1 A

7,3m

22,3 ∘ 48,2 ∘
B C D
 7,3
InACD : tan 48 ∘ 
CD
7,3
 CD 
7,3
tan 48,2∘ 🗸 CD 
 6,5m tan 48,2 ∘
🗸 6,5m
7,3
In ∆ ABD : tan 22,3 
∘
BD 7,3
7,3 🗸 BD 
 BD  tan 22,3∘
tan 22,3∘ 🗸17,8m
 17,8m
BC  BD  CD
 17,8  6,5 🗸answer
 11,3m (5)
3.2.2 EA// BD
 EÂB  AB̂C (alt.s) 🗸S/R
🗸answer
 EÂB  22,3 ∘
(2)
[15]
Question 4 [07]

4.1

f : domain
 range
🗸Asymptotes

g : 🗸 x  intercept
🗸 y  intercept
🗸 Asymptotes
(6)

4.1.1 y 0;2 🗸Answer

(1)

[7]
NO. QUESTION COMMENTS
5.1
A C E
130 ∘ 1 2
85 ∘

1 x
1
B D
 🗸 S/R
Ĉ  B̂ (opp = sides)
1 1
🗸 S/R
  B̂1  Ĉ1  180 ∘ (int. s of  )
2Ĉ1  180 ∘  130 ∘ 🗸 Ĉ1  25 ∘
Ĉ  25 ∘
1

Ĉ1  Ĉ2  85 ∘  180 ∘ (str.line)

Ĉ2  180 ∘  85 ∘  25 ∘
🗸 Ĉ2  70 ∘
Ĉ2  70 ∘
 D1  70 ∘ (alt.s)
🗸 D1  70 ∘ and reason
  D̂1  x  130 ∘ (opps. parm)
x  D̂1  130 ∘ (opps. parm)
x  130 ∘  70 ∘
🗸answer
 x  60 ∘ (6)
5.2.1
P

1
Q S R

In PTS : PS 2  TS 2  PT 2 (Pyth) 🗸S/R

PT 2  PS 2  TS 2    (1)
🗸Equa.2
In STR : SR2  TS 2  TR 2    (2)
(1)+(2): PT 2  SR 2  PS 2  TR 2 🗸Equa.1+equa.2
QS  SR (given)
 PT 2  QS 2  PS 2  TR 2 🗸 QS  SR and reason

(4)
5.2.2 In QRP & TRS : 🗸S/R
Q̂  Tˆ (both  90 ∘ ) 🗸S/R
1 🗸S/R
Rˆ  Rˆ (common)
P̂  Ŝ (3rd <)
PQR ||| STR
(3)
[13]

NO. Question 6 [14] COMMENTS

6.1.1

P T
y1 2 3 4 x S
y x

O
U W

Q R
V

QV  VR;QU  UP(given) 🗸S/R

UV // PR(midptthrm) 🗸S/R
🗸S/R
 ST  TP; SW  WR(given) 🗸S/R
TW // PR (midptthrm) (4)
TW //UV
6.1.2 Tˆ1  Tˆ2 Tˆ 3 Tˆ 4 180 ∘ (str. Line)
🗸S/R
y  y  x  x  180 ∘
2 y  2x  180 ∘ 🗸S
x  y  90∘
Û 1  T̂2  T̂3  180 ∘ TW //UV 🗸 x  y  90 ∘
Û1  90 ∘  180 ∘ 🗸S/R

(4)
6.1.3 1 🗸S/R
TW  PR(midptthrm)
2
1
UV  PR(midptthrm) 🗸S/R
2
TW  UV
🗸 TW  UV
Û 1  90 ∘ ;T̂  T̂  90 ∘ ( proven)
2 3
TW //UV 🗸 TW //UV
TUVW
is a rectangle
(4)
6.2.1 OB  2cm (diagonals of kite bisect each other) 🗸2cm
🗸correct reason
(2)
6.2.2 AÔB  90∘ (diagonals of kite intersect at right s ) 🗸 90 ∘
🗸correct reason
(2)
6.2.3 BÂO  20∘ (diagonals of kite bisect s of kite) 🗸S/R
BÂD  20∘  20∘ BÂD  40∘
 40 ∘ (2)
[14]
END OF MEMO