Download 12_Control_WBHS_2016-07_FinTrig_Memo

GR 12 TUTORIAL MEMO (BG)
27 JULY 2016
For the purposes of this test, a ‘slip’ is defined as an accidental mistake, such as adding two numbers
incorrectly (e.g. 5 + 2 = 8) or writing a digit incorrectly (e.g. 1000 instead of 10000).
An incorrect formula for financial maths results in a breakdown.
1.1
1.2
𝑥 = 1000 (1 +
0.1 −4×10
)
4
✓present value ✓subs
= R372.43
✓ans (c.a. only for slips)
0.09 12𝑛
18 732 = 10 000 (1 +
)
12
0.09 12𝑛
1.8732 = (1 +
)
12
12𝑛 = log (1+0.09) 1.8732
✓equation ✓all correct values
𝑛 = 6.9999 ≈ 7
✓logs
i.e. 𝑛 = 7 years or 84 months.
✓value of 𝑛.
(Note: There are two options here. Either define n as
the number of years, or n as the number of months in
which case one wouldn’t use 12n.)
Accept 7 years, or 7 years and no
months, or 84 months. (Answer has
to be rounded to nearest month, not
given in months…)
12
✓simplify
2.1
Book value = 750 000(1 − 0.2)5
= R245 760
✓form (depr on dim bal.) ✓subs
2.2
Future cost = 750 000(1 + 0.055)5
= R980 220.00
✓formula ✓substitution
2.3
2.4.1
Shortfall = 980 220 − 245 760
= R734 460
Let payment = 𝑥
3
✓answer (c.a. only for slips)
5
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3
✓answer (c.a. only for slips)
3
✓ca: answer
1
0.07 12×5
0.07 12×3
734 460 = 𝑥 (1 +
)
+ 𝑥 (1 +
)
12
12
60
36
0.07
0.07
734 460 = 𝑥 [(1 +
) + (1 +
) ]
12
12
✓LHS ✓first pmt ✓second pmt
Divide both sides by the coefficient of 𝑥, and plug the
monster into a calculator:
✓ca: rearranging/simplifying
∴ 𝑥 = R277 097.12
✓ answer.
(only ca where appropriate)
5
2.4.2
Let payment = 𝑥
✓LHS
3.1
3.2
0.07 12×5
𝑥 [(1 +
)
− 1]
12
734 460 =
0.07
12
0.07
734 460 ×
12
∴𝑥=
12×5
0.07
(1 +
)
−1
12
∴ 𝑥 = 10 258.80
✓correct annuity formula
ℎ
𝐴𝑇
ℎ
∴ 𝐴𝑇 =
sin 𝛼
𝐴𝑇 = 𝑇𝐶
✓trig statement
sin 𝛼 =
✓subs all values correctly
✓rearr
✓ans
✓a: ans (no ca)
5
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2
(given)
𝐴𝐶 2 = 𝐴𝑇 2 + 𝑇𝐶 2 − 2 ⋅ 𝐴𝑇 ⋅ 𝑇𝐶 ⋅ cos 2𝛼
ℎ2
ℎ2
ℎ2
(1 − 2 sin2 𝛼)
=
+
−
2
sin2 𝛼 sin2 𝛼
sin2 𝛼
2ℎ2
2ℎ2
=
− 2 + 4ℎ2
2
sin 𝛼 sin 𝛼
= 4ℎ2
∴ 𝐴𝐶 = 2ℎ
✓cos rule
✓subs
✓expansion of cos 2𝛼
✓4ℎ2
✓ans.
This is a “show that” question, so the
answer must be fully preceded by
logical working.
5
3.3
1
Area of △ 𝐴𝑇𝐶 = 𝐴𝑇 ⋅ 𝑇𝐶 ⋅ sin 2𝛼
2
1 ℎ
ℎ
= ⋅
⋅
⋅ 2 sin 𝛼 cos 𝛼
2 sin 𝛼 sin 𝛼
ℎ2 cos 𝛼
=
sin 𝛼
✓Area rule
✓subs ✓sin 2𝛼 expansion
✓ans
This is a “show that” question, so the
answer must be fully preceded by
logical working.
4.1
Radius 𝐶𝑃 ⊥ tangent 𝑦-axis
✓a
4
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1
4.2
𝑦𝐶 = 2
✓𝑦
(See 4.1)
At C:
3𝑥 + 4(2) + 7 = 0
∴ 3𝑥 = −15
∴ 𝑥 = −5
𝑟 = 𝐶𝑃 = 5
✓sub
✓𝑥
(By inspection)
✓𝑟
∴ Equation of circle:
(𝑥 + 5)2 + (𝑦 − 2)2 = 52
4.3
𝐷𝐸 = 10 units
(Diameter)
4.4
Let the equation of the perpendicular bisector be
given by 𝑦 = 𝑚𝑥 + 𝑐
−1 + 0 −1 + 2
;
)
2
2
1 1
= (− ; )
2 2
✓equation
5
✓a
1
𝑀𝑃𝐸 = (
𝑚𝑃𝐸 =
✓midpoint
2 − (−1)
0 − (−1)
✓𝑚𝑃𝐸
=3
∴𝑚=−
1
3
✓𝑚
Substitute in (−1; −1):
1
−1 = − (−1) + 𝑐
3
4
∴𝑐=−
3
1
4
∴𝑦=− 𝑥−
3
3
4.5
✓sub
✓𝑐
Perpendicular bisector of a chord passes through the
centre C, and DE is a diameter, so both intersect at C.
TOTAL: 50
The equation must be given or
otherwise deduct a mark.
✓perp bisector of chord goes
through centre
✓C lies on DE / DE is diameter
5
2
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