Pretty Basic Trig Question...HELP!

[Iamadam]

Ok I've been trying to get this out for aaaages, and have arrived at the same answer countless times - the same answer which just doesn't agree with the 'back of the book' Here is the question:

A tower is erected at the base of a road that is inclined at a constant angle of 8 Degrees to the horizontal. Angles of elevation to the top of the tower are measured at two points that are 80m apart, some distance up the slope away. These angles were found to be 25 and 8 degrees respectively. What is the height of the tower?

I just can't seem to get it right!

(I am getting 16.252m.. the answer is supposed to be 41.5m)

If anyone could help me with this I would GREATLY appreciate it.
Thanks in advance
--Adam

 

[stapel]

...a road that is inclined at a constant angle of 8 Degrees to the horizontal....

Just to clarify: Does the road go up at an eight-degree angle from the tower to the sighting points, or down at an eight-degree angle?

Thank you.

Eliz.

 

[soroban]

Hello, Adam!

"Sloped ground" problems are always tricky . . .

A tower is erected at the base of a road that is inclined at a constant angle of 8^o to the horizontal.

Angles of elevation to the top of the tower are measured at two points that are 80m apart,
some distance up the slope. \(\displaystyle \,\)These angles were found to be 25^o and 8^o, respectively.

What is the height of the tower?

First of all, we need a good diagram . . . not an easy task.
\(\displaystyle \;\;\)I assume you can connect-the-dots.

      A
      *
      |   *     *
      | 65°   *           *
      |         x *             8°  *
    h |               *         --------------*D
      |               25° *           *
      |               --------*C       80
      |               *
      | 82°   *  8°
    --*----------------------------------------
      B

Note that all angles of elevation are measured from the horizontal.

From the given information, we know all the angles in the diagram.
\(\displaystyle \angle ACB\,=\,25^o\,+\,8^o\,=\,33^o\)
\(\displaystyle \angle ACD\,=\,147^o\)
\(\displaystyle \angle ADC\,=\,8^o\,+\,8^o\,=\,16^o\)
\(\displaystyle \angle DAC\,=\,17^o\)
\(\displaystyle \angle ABC\,=\,82^o\;\;\) (I hope you followed all that.)

We are given: \(\displaystyle \,CD\,=\,80\)
Let \(\displaystyle x\,=\,AC\)


In \(\displaystyle \Delta ACD\), use the Law of Sines: \(\displaystyle \,\frac{x}{\sin16^o}\,=\,\frac{80}{\sin17^o}\)

\(\displaystyle \;\;\)and we have: \(\displaystyle \:x\:=\:\frac{80\cdot\sin16^o}{\sin17^o}\:\approx\:75.42\)


In \(\displaystyle \Delta ABC\), use the Law of Sines: \(\displaystyle \,\frac{h}{\sin33^o}\,=\,\frac{x}{\sin82^o}\)

Therefore: \(\displaystyle \:h\:=\:\frac{75.42\cdot\sin33^o}{\sin82^o}\:=\:41.48036003\:\approx\;41.5\) m . . . ta-DAA!

 

[Iamadam]

Thanks alot, now I can sleep at ease
-Adam