
3d problems
Can someone help me with this question?
Standing due south of a tower 50 m high, the angle of elevation of the top is 26◦.What is the angle of elevation after walking a distance 120 m due east?
The question has me confused so I can't draw a diagram.

Ok this problem has to be done in a few parts. A diagram will really help here.
Firstly find the horizontal length from the tower before moving eastward.
You get $\displaystyle \displaystyle \frac{50}{\tan 26^o}$ Do you know why?
Now using pythagora's thm, find the horizontal distance from the tower with the 120m and the value you just found. What do you get?

Hi pickslides.
I understand the first part:
http://i22.photobucket.com/albums/b3...ntitled32.png
So x will be the horizontal length you're talking about and C is the thing standing due south.
mm.. After thinking about your post I drew this diagram:
http://i22.photobucket.com/albums/b3...ntitled33.png
T is tower, x is the horizontal distance in the previous diagram and b will be the distance from the tower to the location after moving 120m due east.
After that we can make b the base of a rightangled triangle and T the height, so it will look like this:
http://i22.photobucket.com/albums/b3...ntitled34.png
Theta will be our answer
So $\displaystyle \displaystyle \frac{50}{\tan 26^o}$
$\displaystyle x=102.515 $
Then we can use pythag... $\displaystyle b= \sqrt {102.515^2+ 120^2} $
$\displaystyle b=157.827 $
Now we know two sides of the last triangle...
$\displaystyle \tan\theta = \displaystyle\frac {50}{157.827} $
$\displaystyle \theta = \tan^{1} (\displaystyle\frac{50}{157.827}) $
$\displaystyle \theta = 17.58^o $
:S
