October 10th, 2016, 01:11 PM  #1 
Member Joined: Mar 2015 From: uk Posts: 33 Thanks: 1  Height of a water tower problem
This question is at the end of a section on limits and differentiation, A landmark on a distant hill is 'x' metres from a water tower. The angle of elevation from the top of the tower is observed to be '0' (0 is theta) degrees whereas the angle from the foot of the tower is observed to be 0+h degrees a)how high is the water tower b)show that if 'h' is small then the height of the water tower is approx (Pi)xh / 180(cosn)^2 I've managed to do the first part. If the height of the landmark is 't' and the height of the water tower is 'n' then, tan (0+h) = t/x and tan 0 = (t  n) / x Eliminating t, t = xtan(0+h) = xtan0 +n So the height of the water tower 'n' is x(tan(0+h)  tan0) Can anyone show me how to do the second part. I don't understand how (Pi) and 180 can become involved?? 
October 11th, 2016, 01:06 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,763 Thanks: 697 
Statement if b) doesn't look right. How can a dumber with dimension of length appear as an argument for cosn? Pi and 180 become involved, since trig functions use radians as a "natural" argument  for example sin(x) ~ x when x is in radians. 
October 11th, 2016, 01:53 PM  #3 
Senior Member Joined: Sep 2015 From: USA Posts: 2,452 Thanks: 1337 
if $h$ is small then $\tan(\theta+h) \approx \tan(\theta)+\left. \dfrac{d}{d\theta}\tan(\theta)\right_{\theta=h}$ $\tan(x+h) \approx \tan(\theta)+\sec^2(h)$ the rest follows 
October 12th, 2016, 01:45 PM  #4 
Global Moderator Joined: May 2007 Posts: 6,763 Thanks: 697  This assumes h is in radians, not in degrees. h in degrees is where $\displaystyle \frac{\pi}{180}$ comes into play.

October 20th, 2016, 02:29 AM  #5 
Member Joined: Mar 2015 From: uk Posts: 33 Thanks: 1 
Thanks for the replies. Having looked at it some more I think this is how it is done n=x(tan(0+h)  tan0) now dy/dx is defined as limit as h tends to zero of [f(x+h)  f(x)] / h therefore n=xh [lim (h tends to zero) (tan(0+h)  tan0 / h)] so n=xh (d/d0 tan0) = xh / (cos0)^2 degrees or (pi)xh / 180(cos0)^2 degrees 

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