trig questions

[uph0]

Not too sure of how to solve these two questions:
Two poles, pole A and pole B, are of the same length and line the edge of a river. They are 100 m apart. On the other side of the edge of the river, stands a person. The person The person's angle of elevation to the top of pole A is 22 degrees. The person's angle of elevation to the top of pole 19 degrees. What is the width of the river? (If you draw a line from the bottom of one of the poles across the width river and to the opposite edge, that's where the person is.)

A survyetor standing on horizontal plain can see a volcano in the distance. The angle of elevation of the top of the volcano is 23. If he moves 750 m closer, the angle of elevation is now 37. What's the height of the volcano?

thanks

 

[mmm4444bot]

If you draw a line from the bottom of one of the poles across the width river and to the opposite edge, that's where the person is.

Hi uph0:

By using the word "width" above, are you trying to say that the two poles and person form a RIGHT triangle?

On the ground, let point A be at the base of pole A, and let point B be at the base of pole B, and let point C be the person.

If the person is across the river from, let's say, pole A, is angle CAB ninety degrees?

Cheers ~ Mark

 

[soroban]

Hello, uph0!

This is a messy one . . .

Two poles, pole A and pole B, are of the same height and line the edge of a river, 100 m apart.
On the other side of the edge of the river, directly opposite pole A, stands a person.
The person The person's angle of elevation to the top of pole A is 22 degrees.
The person's angle of elevation to the top of pole B is 19 degrees.
What is the width of the river?

Looking down at the ground, we have this diagram:

           100
    A * - - - - - * B
      |        *
    w |     * x
      |  *
      *
      P

The person is at P.
The poles are at \(\displaystyle A\) and \(\displaystyle B\!:\;AB = 100\)
The width of the river is \(\displaystyle w\).
Let \(\displaystyle x = PB.\)
\(\displaystyle \text{Pythagorus says: }\:x \:=\:\sqrt{w^2 + 100^2}\;\;[1]\)



The person \(\displaystyle P\), looking at pole \(\displaystyle A\), has this right triangle.

      *
      | *
      |   *
    h |     *
      |       *
      |    22d  *
    A * - - - - - * P
            w

\(\displaystyle \text{We have: }\;\tan22^o \:=\:\frac{h}{w} \quad\Rightarrow\quad h \:=\:w\tan22^p\;\;[2]\)


The person, looking at pole B, has this diagram:

                        *
                     *  |
                  *     |
               *        | h
            *           |
         * 19d          |
      * - - - - - - - - *
      P        x        B

\(\displaystyle \text{We have: }\:\tan19^o \:=\:\frac{h}{x} \quad\Rightarrow\quad h \:=\:x\tan19^o\;\;[3]\)


\(\displaystyle \text{Equate [2] and [3]: }\:w\tan22^o \:=\:x\tan19^o\)

\(\displaystyle \text{Substiutute [1]: }\:w\tan22^o \:=\:\tan19^o\sqrt{w^2+100^2}\)


Now solve for \(\displaystyle w.\)
. . (I'll wait in the car.)

 

[mmm4444bot]

The volcano exercise is not difficult, if you draw a picture.

You've got two right triangles.

The big one has base 750 + x.

The small one has base x.

Both have height h.

You've been given the angles opposite h.

Wrtie two equations using tan(angle) = h/base.

Solve them for h.

Equate the results.

Solve for x.

Substitute this into an equation with h and x.

I get 1,294 meters for the height of the volcano.

You should confirm this.