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#1 2018-07-14 02:26:09

Tompson Lee
Member
Registered: 2018-07-05
Posts: 2

A difficult problem from 1957

Here is a problem I found which is from a math class in 1957:

A man is standing due East of a tower and notes it subtends an angles of 45 degrees with the tower.
He walks South 42.4 feet and the subtended angle is 30 degrees.
How tall is the tower?

(You are only allowed to use only pencil(pen) and a paper)

Credits to MindYourDecision: https://www.youtube.com/user/MindYourDecisions

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#2 2018-07-14 22:41:52

Alg Num Theory
Member
Registered: 2017-11-24
Posts: 339
Website

Re: A difficult problem from 1957

Why can’t they use metric?

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#3 2018-07-16 06:19:15

bob bundy
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Registered: 2010-06-20
Posts: 8,407

Re: A difficult problem from 1957

Hi Thomson Lee,

Did you want me to provide a solution or what?  No square brackets (to hide) on my Kindle so you just have to close your eyes.

Also no diagram making, sorry.

This is 3D so I'll describe the diagram.

Let TB be the tower in the 'z' direction and BE be an easterly direction with the first observation at E (x direction).

Draw an oblique line NES to indicate the North/South direction (y) with ES= 42.4 ft.

As TEB = 45, BE = TB = h, the height of the tower.

And as TSB= 30, BS = root 3 h.

So in triangle BES we have BE = h, BS = root 3 h and ES = 42.4 with a right angle at E.

So it would be easy to use Pythag to get h (approx) 30 ft I think.  But here's a construction for it.  I have no paper nor instruments handy so the scale might not work.

Draw a verticle line BE' in the top right corner of your page. I'll make it 5cm to represent 10 ft. Construct a right angle at E' and extend the line right across the paper.

On a fresh sheet, construct an equilateral triangle sides 5 cm.  Construct a perpendicular bisector so the height = 5 root 3 and set this as a radius.  On the first diagram, centre B, make an arc to cut the horizontal line at S'

The triangle BE'S' is the right shape for BES but needs to be scaled up.

On E'S' produced Mark X so that E'X = 21.2 cm.

Draw XS parallel to BE' with BS' extended crossing it at S. SE parallel to S'E' will fix E and hence the correct size for triangle BES.  h = BE.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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