Please help me solve: A skier sees the top of a tower....

[Joyce]

Hi, I've been trying to solve this problem for awhile but I'm having a problem determining what exactly you have to do to solve it.

Questions: A skier sees the top of a radio tower, due south of him, at an angle of elevation of 32 (degrees). He then skies on a bearing of 130 (degrees) for 560 m. and finds himself due east of the tower. Calculate:

a) the height of the radio tower
b) the distance from the tower to the skier
c) the angle of elevation of the top of the tower from the new position

For a) I'm confused about what the angle of elevation is (where it would be located when I draw a diagram) and the 130 (degress). Would I substract 180-130-32 ? . I was also wondering if I should use the cosine law.......

 

[galactus]

Here's a drawing. I believe this is what you mean.

Let's put the skier at the origin before he/she sets out.

Since the skier travels 560m at 50 degrees east of the tower, you can

use 560cos(50) or 560sin(40) to find the distance from the origin to the tower.

Since the skier eyed the vertical angle from the origin, you now have the distance form the origin to the tower and the 32 degree angle. You can figure the tower height using tangent.

 

[Joyce]

o o.k thanks for your help

 

[soroban]

Re: Please help me solve: A skier sees the top of a tower...

Hello, Joyce!

A skier sees the top of a radio tower, due south of him, at an angle of elevation of 32°.
He then skies on a bearing of 130° for 560 m. and finds himself due east of the tower. Calculate:

a) the height of the radio tower
b) the distance from the tower to the skier
c) the angle of elevation of the top of the tower from the new position.

This is a tricky problem . . . it is three-dimensional.

Galactus gave you an aerial view of the situation.

      A
      * 
      |\
      | \
      |  \ 560
      |50°\
      |    \
      |      \
      * - - - *
      T       B

The skier starts at A, where he first sights the tower (T),
\(\displaystyle \;\;\)then he skis 560 meters to B.

We have enough information to solve this right triangle.

\(\displaystyle \cos50^o\,=\,\frac{AT}{560}\;\;\Rightarrow\;\;AT\,=\,560\cdot\cos50^o\:=\:359.9610614\:\approx\:360\) m

\(\displaystyle \sin50^o\,=\,\frac{TB}{560}\;\;\Rightarrow\;\;TB\,=\,560\cdot\sin50^o\,=\,428.9848881\,\approx\,429\) m \(\displaystyle \,\) (b)


This is a side view of the first sighting.

                  *
                / |
              /   |
            /     |h
          /       |
        / 32°     |
      * - - - - - *
      A    360    T

The skier is at \(\displaystyle A\); the base of the tower is at \(\displaystyle T\).
He looks up at an angle of 32° to the top of the tower.

We have: \(\displaystyle \,\tan32^o\,=\,\frac{h}{360}\;\;\Rightarrow\;\;h\,=\,360\cdot\tan32^o\,=\,224.9529667\:\approx\:225\) m \(\displaystyle \,\) (a)


This is a side view of the second sighting.

      *
      | \
      |   \
      |     \
  225 |       \
      |       θ \
      * - - - - - *
      T    429    B

The skier has moved to point \(\displaystyle B.\)
He looks up to the top to the 225-meter tower.

We have: \(\displaystyle \,\tan\theta \.= \.\frac{225}{429}\,=\,0.524475524\)

Therefore: \(\displaystyle \:\theta \:=\:27.67591046\:\approx\:28^o\,\) (c)