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Trigonometry, 2ADV T1 2005 HSC 9b

Trig Ratios, 2UA 2005 HSC 9b
 

The triangle  `ABC`  has a right angle at `B, \ ∠BAC = theta` and  `AB = 6`. The line `BD` is drawn perpendicular to `AC`. The line `DE` is then drawn perpendicular to `BC`. This process continues indefinitely as shown in the diagram.

  1. Find the length of the interval `BD`, and hence show that the length of the interval  `EF`  is  `6 sin^3\ theta`.  (2 marks)

  2. Show that the limiting sum 
     
    `qquad BD + EF + GH + ···`
     
    is given by  `6 sec\ theta tan\ theta`.  (3 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
Show Worked Solution
i.   Trig Ratios, 2UA 2005 HSC 9b Answer

`text(Show)\ EF = 6\ sin^3\ theta`

`text(In)\ ΔADB`

`sin\ theta` `= (DB)/6`
`DB` `= 6\ sin\ theta`
`∠ABD` `= 90 − theta\ \ \ text{(angle sum of}\ ΔADB)`
`:.∠DBE` `= theta\ \ \ (∠ABE\ text{is a right angle)}`

 

`text(In)\ ΔBDE:`

`sin\ theta` `= (DE)/(DB)`
  `= (DE)/(6\ sin\ theta)`
`DE` `= 6\ sin^2\ theta`
`∠BDE` `= 90 − theta\ \ \ text{(angle sum of}\ ΔDBE)`
`∠EDF` `= theta\ \ \ (∠FDB\ text{is a right angle)}`

 

`text(In)\ ΔDEF:`

`sin\ theta` `= (EF)/(DE)`
  `= (EF)/(6\ sin^2\ theta)`
`:.EF` `= 6\ sin^3\ theta\ \ …text(as required)`

 

ii. `text(Show)\ \ BD + EF + GH\ …`

`text(has limiting sum)\ =6 sec theta tan theta`

`underbrace{6\ sin\ theta + 6\ sin^3\ theta +\ …}_{text(GP where)\ \ a = 6 sin theta, \ \ r = sin^2 theta}`
 

`text(S)text(ince)\ \ 0 < theta < 90^@`

`−1` `< sin\ theta` `< 1`
`0` `< sin^2\ theta` `< 1`

 
`:. |\ r\ | < 1`
 

`:.S_∞` `= a/(1 − r)`
  `= (6\ sin\ theta)/(1 − sin^2\ theta)`
  `= (6\ sin\ theta)/(cos^2\ theta)`
  `= 6 xx 1/(cos\ theta) xx (sin\ theta)/(cos\ theta)`
  `= 6 sec\ theta\ tan\ theta\ \ …text(as required.)`

Financial Maths, 2ADV M1 2005 HSC 7a

Anne and Kay are employed by an accounting firm.

Anne accepts employment with an initial annual salary of  $50 000. In each of the following years her annual salary is increased by $2500.

Kay accepts employment with an initial annual salary of  $50 000. In each of the following years her annual salary is increased by 4%.

  1. What is Anne’s annual salary in her thirteenth year?  (2 marks)

  2. What is Kay’s annual salary in her thirteenth year?  (2 marks)

  3. By what amount does the total amount paid to Kay in her first twenty years exceed that paid to Anne in her first twenty years?  (3 marks)

Show Answers Only
  1. `$80\ 000`
  2. `text{$80 052 (nearest dollar)}`
  3. `text{$13 904 (nearest $)}`
Show Worked Solution

i.   `text(Let)\ T_n = text(Anne’s salary in year)\ n`

`T_1 = a = $50\ 000`

`T_2 = a + d = $52\ 500`

`⇒\ text(AP where)\ a = $50\ 000,\ \ d = $2500`

`T_n = a + (n − 1)d`

`T_13` `= 50\ 000 + (13 − 1) xx 2500`
  `=80\ 000`

 

`:.\ text(Anne’s salary in her 13th year is $80 000.)`

 

ii.  `text(Let)\ K_1 =text(Kay’s salary in year)\ n`

`K_1` `= a` `= 50\ 000`
`K_2` `= ar` `= 50\ 000 xx 1.04 = 52\ 000`
`⇒\ text(GP where)\ \ a = 50\ 000, \ \ r = 1.04`
`K_n` `= ar^(n − 1)`
 `K_13` `= 50\ 000 xx (1.04)^12`
  `= $80\ 051.61…`
  `= $80\ 052\ \ \ text{(nearest dollar)}`

 

iii.  `text(Anne)`

`S_n` `= n/2[2a + (n − 1)d]`
 `S_20` `= 20/2[2 xx 50\ 000 + (20 − 1)2500]`
  `= 10[100\ 000 + 47\ 500]`
  `= $1\ 475\ 000`

 

`text(Kay)`

`S_n` `= (a(r^n − 1))/(r − 1)`
`S_20`  `= (50\ 000(1.04^20 -1))/(1.04 − 1)`
  `= $1\ 488\ 903.929…`

 

`text(Difference)`

`= 1\ 488\ 903.929… − 1\ 475\ 000`

`= $13\ 903.928…`

`= $13\ 904\ \ \  text{(nearest $)}`

 

`:.\ text(Kay’s total salary exceeds Anne’s by)\ $13\ 904`

Financial Maths, 2ADV M1 SM-Bank 11

Express the recurring decimal  `0.323232...`  as a fraction.  (2 marks)

Show Answers Only

`32/99`

Show Worked Solution
`0.3232…` `=32/100+32/10^4+32/10^6+…`
  `=32/100(1+1/10^2+1/10^4+…)`
`=>\ text(GP where)\ a=1,\ \ r=1/10^2`
  `=32/100(a/(1-r))`
  `=32/100(1/(1-1/100))`
  `=32/100(1/(99/100))`
  `=32/99`

Financial Maths, 2ADV M1 2014 HSC 14d

At the beginning of every 8-hour period, a patient is given 10 mL of a particular drug.

During each of these 8-hour periods, the patient’s body partially breaks down the drug. Only  `1/3`  of the total amount of the drug present in the patient’s body at the beginning of each 8-hour period remains at the end of that period.  

  1. How much of the drug is in the patient’s body immediately after the second dose is given?    (1 mark)

  2. Show that the total amount of the drug in the patient’s body never exceeds 15 mL.     (2 marks)

Show Answers Only
  1. `13.33\ text{mL  (2 d.p.)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

i.   `text(Let)\ \ A =\ text(Amount of drug in body)`

`text(Initially)\ A = 10`

`text(After 8 hours)\ \ \ A` `=1/3 xx 10`
`text(After 2nd dose)\ \ A` `= 10 + 1/3 xx 10\ text(mL)`
  `=13.33\ text{mL  (2 d.p.)}`

 

ii.   `text(After the 3rd dose)`

`A_3` `= 10 + 1/3 (10 + 1/3 xx 10)`
  `= 10 + 1/3 xx 10 + (1/3)^2 xx 10`

 
`  =>\ text(GP where)\ a = 10,\ r = 1/3`

`text(S)text(ince)\ \ |\ r\ | < 1:`

`S_oo` `= a/(1\ – r)`
  `= 10/(1\ – 1/3)`
  `= 10/(2/3)`
  `= 15`

 

 
`:.\ text(The amount of the drug will never exceed 15 mL.)`

Financial Maths, 2ADV M1 2008 HSC 4b

The zoom function in a software package multiplies the dimensions of an image by 1.2.  In an image, the height of a building is 50 mm. After the zoom function is applied once, the height of the building in the image is 60 mm. After the second application, it is 72 mm.

  1. Calculate the height of the building in the image after the zoom function has been applied eight times. Give your answer to the nearest mm.     (2 marks)

  2. The height of the building in the image is required to be more than 400 mm. Starting from the original image, what is the least number of times the zoom function must be applied?     (2 marks)

Show Answers Only
  1. `text(215 mm)`
  2. `12`
Show Worked Solutions
i.    `T_1` `=a=50`
  `T_2` `=ar^1=50(1.2)=60`
  `T_3` `=ar^2=50(1.2)^2=72`

 
`=>\ text(GP where)\ \ a=50,\ \ r=1.2`

`\ \ vdots` 

MARKER’S COMMENT: Within this GP, note that `T_9` is the term where the zoom has been applied 8 times.
`T_9` `=50(1.2)^8`
  `=214.99`

 

`:.\ text{Height will be 215 mm  (nearest mm)}`

 

ii.    `T_n=ar^(n-1)` `>400`
  `:.\ 50(1.2)^(n-1)` `>400`
  `1.2^(n-1)` `>8`
  `ln 1.2^(n-1)` `>ln8`
  `n-1` `>ln8/ln1.2`
  `n` `>12.405`

 

`:.\ text(The height of the building in the 13th image)`

`text(will be higher than 400 mm, which is the 12th)`

`text(time the zoom would be applied.)`

Financial Maths, 2ADV M1 2009 HSC 4a

A tree grows from ground level to a height of 1.2 metres in one year. In each subsequent year, it grows `9/10` as much as it did in the previous year.

Find the limiting height of the tree.     (2 marks)

Show Answer Only

`12\ text(m)`

Show Worked Solution

`a=1.2, \ \ \ r=9/10`

`text(S)text(ince)\ \ |\ r\ |<1,`

`S_oo` `=a/(1-r)`
  `=1.2/(1-(9/10))`
  `=12\ text(m)`

 

`:.\ text(Limiting height of tree is 12 m.)`

Financial Maths, 2ADV M1 2011 HSC 5a

The number of members of a new social networking site doubles every day. On Day 1 there were 27 members and on Day 2 there were 54 members.

  1. How many members were there on Day 12?     (1 mark)

  2. On which day was the number of members first greater than 10 million?     (2 marks)

  3. The site earns 0.5 cents per member per day. How much money did the site earn in the first 12 days? Give your answer to the nearest dollar.     (2 marks)

Show Answers Only
  1. `55\ 296`
  2. `text(20th)`
  3. `$553`
Show Worked Solutions
MARKER’S COMMENT: Better responses stated the general term, `T_n` before any substitution was made, as shown in the worked solutions.

i.   `T_1=a=27`

`T_2=27xx2^1=54`

`T_3=27xx2^2=108`

`=>\ text(GP where)\ \ a=27,\ \ r=2`

`\ \ \ vdots`

`T_n` `=ar^(n-1)`
`T_12` `=27 xx 2^11=55\ 296`

 

`:.\ text(On Day 12, there are 55 296 members.)`

 

ii.   `text(Find)\ n\ text(such that)\  T_n>10\ 000\ 000`

MARKER’S COMMENT: Many elementary errors were made by students in dealing with logarithms. BE VIGILANT.
`T_n` `=27(2^(n-1))`
`27xx2^(n-1)` `>10\ 000\ 000`
`2^(n-1)` `>(10\ 000\ 000)/27`
`ln 2^(n-1)` `>ln((10\ 000\ 000)/27)`
`(n-1)ln2` `>ln(370\ 370.370)`
`n-1` `>ln(370\ 370.370)/ln 2`
`n-1` `>18.499…`
`n` `>19.499…`

 

`:.\ text(On the 20th day, the number of members >10 000 000.)`

 

iii.  `text(If the site earns 0.5 cents per day per member,)`

`text(On Day 1, it earns)\  27 xx 0.5 = 13.5\ text(cents)`

`text(On Day 2, it earns)\  27 xx 2 xx 0.5 = 27\ text(cents)`

`T_1=a=13.5`

`T_2=27`

`T_3=54`

`=>\ text(GP where)\ \ a=13.5,\ \ r=2`
 

`S_12=text(the total amount of money earned in the first 12 Days)`

♦ Mean mark 44%.
NOTE: This question can also be easily solved by making `S_12` the total sum of members (each day) and then multiplying by 0.5 cents.
`S_12` `=(a(r^n-1))/(r-1)`
  `=(13.5(2^12-1))/(2-1)`
  `=55\ 282.5\ \ text(cents)`
  `=552.825\ \ text(dollars)`

 

`:.\ text{The site earned $553 in the first 12 Days (nearest $).}`

Financial Maths, 2ADV M1 2012 HSC 15a

Rectangles of the same height are cut from a strip and arranged in a row. The first rectangle has width 10cm. The width of each subsequent rectangle is 96% of the width of the previous rectangle.
 

2012 15a
 

  1. Find the length of the strip required to make the first ten rectangles.     (2 marks)

  2. Explain why a strip of 3m is sufficient to make any number of rectangles.     (1 mark)

Show Answers Only
  1. `83.8\ text{cm  (1 d.p.)}`
  2. `S_oo=2.5\ text(m)\ \ =>\ \ text(sufficient.)`
Show Worked Solutions
i.    `T_1` `=a=10`
`T_2` `=ar=10xx0.96=9.6`
`T_3` `=ar^2=10xx0.96^2=9.216`

 
`=>\ text(GP where)\ \ a=10\ \ text(and)\ \ r=0.96`

MARKER’S COMMENT: A common error was to find `T_10` instead of `S_10`
`S_10` `=\ text(Length of strip for 10 rectangles)`
  `=(a(1-r^n))/(1-r)`
  `=10((1-0.96^10)/(1-0.96))`
  `=83.8\ text{cm   (to 1 d.p.)}`

 

ii.   `text(S)text(ince)\ |\ r\ |<\ 1`

`S_oo` `=a/(1-r)`
  `=10/(1-0.96)`
  `=250\ text(cm)`

 

`:.\ text(S)text(ince  3 m > 2.5 m, it is sufficient.)`

Probability, 2ADV S1 2013 HSC 15d

Pat and Chandra are playing a game. They take turns throwing two dice. The game is won by the first player to throw a double six. Pat starts the game.

  1. Find the probability that Pat wins the game on the first throw.     (1 mark)

  2. What is the probability that Pat wins the game on the first or on the second throw?     (2 marks)

  3. Find the probability that Pat eventually wins the game.     (2 marks)

Show Answers Only
  1. `1/36`
  2. `(2521)/(46\ 656)\ \ text(or)\ \ 0.054`
  3. `36/71`
Show Worked Solutions

i.   `P\ text{(Pat wins on 1st throw)}=P(W)`

`P(W)` `=P\ text{(Pat throws 2 sixes)}`
  `=1/6 xx 1/6`
  `=1/36`

 

ii.  `text(Let)\ P(L)=P text{(loss for either player on a throw)}=35/36`

`P text{(Pat wins on 1st or 2nd throw)}` 

♦♦ Mean mark 33%
MARKER’S COMMENT: Many students did not account for Chandra having to lose when Pat wins on the 2nd attempt.

`=P(W) + P(LL W)`

`=1/36\ + \ (35/36)xx(35/36)xx(1/36)`

`=(2521)/(46\ 656)`

`=0.054\ \ \ text{(to 3 d.p.)}`

 

iii.  `P\ text{(Pat wins eventually)}`

`=P(W) + P(LL\ W)+P(LL\ LL\ W)+ … `

`=1/36\ +\ (35/36)^2 (1/36)\ +\ (35/36)^2 (35/36)^2 (1/36)\ +…`

 
`=>\ text(GP where)\ \ a=1/36,\ \ r=(35/36)^2=(1225)/(1296)`

♦♦♦ Mean mark 8%!
 COMMENT: Be aware that diminishing probabilities and `S_oo` within the Series and Applications are a natural cross-topic combination.

 
`text(S)text(ince)\ |\ r\ |<\ 1:`

`S_oo` `=a/(1-r)`
  `=(1/36)/(1-(1225/1296))`
  `=1/36 xx 1296/71`
  `=36/71`

 

`:.\ text(Pat’s chances to win eventually are)\  36/71`.

Financial Maths, 2ADV M1 2013 HSC 12c

Kim and Alex start jobs at the beginning of the same year. Kim's annual salary in the first year is $30,000 and increases by 5% at the beginning of each subsequent year. Alex's annual salary in the first year is $33,000, and increases by $1,500 at the beginning of each subsequent year.

  1. Show that in the 10th year, Kim's annual salary is higher than Alex's annual salary.     (2 marks)

  2. In the first 10 years how much, in total, does Kim earn?     (2 marks)

  3. Every year, Alex saves `1/3` of her annual salary. How many years does it take her to save $87,500?     (3 marks)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `$377\ 336.78`
  3. `text(7  years)`
Show Worked Solutions

i.    `text(Let)\ \ K_n=text(Kim’s salary in Year)\  n`

`{:{:(K_1=a=30\ 000),(K_2=ar^1=30\ 000(1.05^1)):}}{:(\ =>\ GP),(\ \ \ \ \ \ a=30\ 000),(\ \ \ \ \ \ r=1.05):}`

`vdots`

`:.K_10=ar^9=30\ 000(1.05)^9=$46\ 539.85`

 

`text(Let)\ \ A_n=text(Alex’s salary in Year)\ n`

`{:{:(A_1=a=33\ 000),(A_2=33\ 000+1500=34\ 500):}}{:(\ =>\ AP),(\ \ \ \ \ \ a=33\ 000),(\ \ \ \ \ \ d=1500):}`

`vdots`

`A_10=a+9d=33\ 000+1500(9)=$46\ 500`

`=>K_10>A_10`
 

`:.\ text(Kim earns more than Alex in the 10th year)`

 

ii.    `text(In the first 10 years, Kim earns)`

`K_1+K_2+\  ….+ K_10`

`S_10` `=a((r^n-1)/(r-1))`
  `=30\ 000((1.05^10-1)/(1.05-1))`
  `=377\ 336.78`

 

`:.\ text(In the first 10 years, Kim earns $377 336.78)`

 

iii.   `text(Let)\ T_n=text(Alex’s savings in Year)\ n`

`{:{:(T_1=a=1/3(33\ 000)=11\ 000),(T_2=a+d=1/3(34\ 500)=11\ 500),(T_3=a+2d=1/3(36\ 000)=12\ 000):}}{:(\ =>\ AP),(\ \ \ \ a=11\ 000),(\ \ \ \ d=500):}`
 

`text(Find)\ n\ text(such that)\ S_n=87\ 500`

♦ Mean mark 45%.
IMPORTANT: Using the AP sum formula to create and then solve a quadratic in `n` is challenging and often examined. Students need to solve and interpret the solutions.
`S_n` `=n/2[2a+(n-1)d]`
`87\ 500` `=n/2[22\ 000+(n-1)500]`
`87\ 500` `=n/2[21\ 500+500n]`
`250n^2+10\ 750n-87\ 500` `=0`
`n^2+43n-350` `=0`
`(n-7)(n+50)` `=0`

 
`:.n=7,\ \ \ \ n>0`

`:.\ text(Alex’s savings will be $87,500 after 7 years).`