The smokestake and the Tower

[browncutie]

Use the following information to determine an approximate height for a smock stake.
*The camera is 440 m from the smokestack
* The smokestack is 3200 m from the CN tower
* The height of the part of the tower hidden behind the smokestake is 330 m

b) How accurate is your response in part a? What factors affect the accuracy of your responsse? Explain.

 

[soroban]

Hello, browncutie!

Sorry, I don't trust the wording/accuracy of this problem . . .

Use the following information to determine an approximate height for a smokestack.
* The camera is 440 m from the smokestack
* The smokestack is 3200 m from the CN tower . . . . over 3 kilometers away?
* The height of the part of the tower hidden behind the smokestake is 330 m
Is that hidden from view, or hidden "horizontally"?

I suspect that the second distance is 320 meters
. . and that the last clue tells us that the smokestack is 330 m high.

If all of that is true, there is a simple solution . . .

   T  *
      |   *
    y |       *
      |           *   S
    D * - - - - - - - *
      |      320      |   *
  330 |            330|       *
      |               |           *
      * - - - - - - - * - - - - - - - *
      A      320      B      440      C

\(\displaystyle \text{The tower is: }\;TA \:=\:y + 330\)

\(\displaystyle \text{The smokestack is: }\:SB \:=\A\:=\:330\)

. . \(\displaystyle \text{and: }\:AB \:=\S \:=\:320\)

\(\displaystyle \text{The camera is at }C:\;CB \:=\:440\)


\(\displaystyle \text{We have right triangles: }\Delta TDS \sim \Delta TAC\)

\(\displaystyle \text{Hence: }\;\frac{TD}{DS} \:=\:\frac{TA}{AC}\quad\Rightarrow\quad \frac{y}{320} \:=\:\frac{y+330}{760}\)


\(\displaystyle \text{Solve for }y\text{, and the height of the tower is: }y + 330.\)