# 3d problems

• Dec 21st 2010, 09:17 PM
jgv115
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.
• Dec 21st 2010, 09:52 PM
pickslides
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 $\displaystyle \frac{50}{\tan 26^o}$ 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?
• Dec 21st 2010, 11:12 PM
jgv115
Hi pickslides.

I understand the first part:

http://i22.photobucket.com/albums/b3...ntitled-32.png

So x will be the horizontal length you're talking about and C is the thing standing due south.

http://i22.photobucket.com/albums/b3...ntitled-33.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 right-angled triangle and T the height, so it will look like this:

http://i22.photobucket.com/albums/b3...ntitled-34.png

So $\displaystyle \frac{50}{\tan 26^o}$

$x=102.515$

Then we can use pythag... $b= \sqrt {102.515^2+ 120^2}$

$b=157.827$

Now we know two sides of the last triangle...

$\tan\theta = \displaystyle\frac {50}{157.827}$

$\theta = \tan^{-1} (\displaystyle\frac{50}{157.827})$

$\theta = 17.58^o$

:S
• Dec 22nd 2010, 11:45 AM
pickslides
Sounds good to me!