Plane Trig Question?

[yohanson77]

Hello all,
Thank you again for the help on the last one, it helped loads. This one is on plane trig and I'm finding it a bit tricky, please help!

Q. A vertical mast stands on horizontal ground. A man, posioned due west of the mast, finds the elevation of the top to be 51(degrees) 5(mins). He moves due south 20.0m and finds the elevation to be 47(degrees) 19(mins). Calculate the height of the mast?

I've worked with cosine and sine rule, but I think this is Tan.

yours faithfully

Yohanson77

 

[Deleted member 4993]

For shortest way, you have to use Pythagorus and Tan.

However, you can use sine and Cosine since Tan is derived from those two.

Show us what you have done - we can guide you after that.

 

[yohanson77]

Plane Trig

Hi,
The most I've done is scratch my head. I can't think where to start as there is only 1 angle and 1 length:-(

 

[soroban]

Hello, Yohanson77!

Interesting . . . a three-dimensional problem.
It takes some acrobatics to connect the Given to the Punchline.
. . (But then, that's what Math is all about, isn't it?)

A vertical mast stands on horizontal ground.
A man, positioned due west of the mast, finds the elevation of the top to be 51°5'
He moves due south 20.0m and finds the elevation to be 47°19'
Calculate the height of the mast.

The man stands at \(\displaystyle A\).
The mast is: \(\displaystyle \,y \,=\,PQ\).
His distance is: \(\displaystyle \,x \,=\,AQ\).
Let: \(\displaystyle \,\alpha \,=\,\angle PAQ\).

The diagram looks like this:

                              * P
                           *  |
                        *     |
                     *        |
                  *           | y
               *              |
            *                 |
         *  α                 |
    A * - - - - - - - - - - - * Q
                  x

We have: \(\displaystyle \:\tan\alpha\,=\,\frac{y}{x}\;\;\Rightarrow\;\;y \:=\:x\cdot\tan\alpha\;\) [1]


He moves 20 m directly south to point \(\displaystyle B\).
Hence: \(\displaystyle \,AB\,=\,20\)
Let: \(\displaystyle \,z\,=\,BQ\)

Looking down at the ground, the diagram looks like this:

                  x 
    A * - - - - - - - - - - - * Q
      |                   *
      |               *
   20 |           * z
      |       *
      |   *
    B *

Pythagorus says: \(\displaystyle \:z \:=\:\sqrt{x^2\,+\,20^2}\;\) [2]


His new right triangle looks like this:

                              * P
                          *   |
                      *       |
                  *           | y
              *               |
          *  β                |
    B * - - - - - - - - - - - * Q
                    z

Let: \(\displaystyle \,\beta \,=\,\angle PBQ\)

We have: \(\displaystyle \:\tan\beta \,=\,\frac{y}{z}\;\;\Rightarrow\;\;y \:=\:z\cdot\tan\beta\;\) [3]

Now we can solve the problem . . .


Equate [1] and [3]: \(\displaystyle \:x\cdot\tan\alpha \:=\:z\cdot\tan\beta\)


Substitute [2]: \(\displaystyle \:x\cdot\tan\alpha \:=\:\tan\beta\sqrt{x^2\,+\,400}\)

Square both sides: \(\displaystyle \:x^2\cdot\tan^2\alpha \:=\:\tan^2\beta\left(x^2\,+\,400\right)\)

And we have: \(\displaystyle \:x^2\:=\:\frac{400\cdot\tan^2\beta}{\tan^2\alpha\,-\,\tan^2\beta} \;\;\Rightarrow\;\;x \:=\:\frac{20\cdot\tan\alpha}{\sqrt{\tan^2\alpha\,-\,\tan^2\beta}}\)


Substitute into [1]: \(\displaystyle \L\:y \;=\;\frac{20\cdot\tan\alpha\cdot\tan\beta}{\sqrt{\tan^2\alpha\,-\,\tan^2\beta}}\;\;\) . . . There!


Since \(\displaystyle \,\alpha\,=\,51^o5'\) and \(\displaystyle \beta\,=\,47^o19'\)
. . you can crank it through your calculator . . .

 

[Deleted member 4993]

        W \                      /N
             \        P         /
               \     *|       /
                 \  * |     /
                A/\   |   /
               /    \ | /
             /       O\/
        B /          /\
                     /    \
                   /        \
                 /            \

First draw a sketch - somewhat like above.

PO is the pole

A is the first position of the man.

B is the next position of the man.

Then

AB = 20

PO = h

AO = x

BO = sqrt(AO^2 + AB^2)

angle OAP = 51°05'

angle OBP = 47°19'

Use TAN of these angles and solve for 'x' and 'h'