An oil well was drilled 3400 ft before it hit a ledge of....

[Apcanton]

An oil well at A was drilled 3400 ft before it hit a ledge of shale. The same ledge was only 1950 ft deep when drilled from a well at B, which is 8000 ft directly downhill from A. The hill is inclined at 15 degrees to the horizontal.

a. Find how far down the hill from B until shale ledge emerges.
b. What is the angle between the ledge and the hill?

 

[Denis]

Hints:

Make 3400 line from A meet ledge at E;
make hill and ledge meet at D;
extend AE to F, such that AFD = right triangle.

Make line BC with C on AE, such that ACB = right triangle.

You now have 2 similar right triangles ACB and AFD;
with triangle ACB, you have AB = 8000 and angle at B = 15 degrees,
so you can calculate lengths of AC and BC.

Make 1950 line from B meet ledge at G.
You now have 2 similar non-right triangles: AED and BGD.

Hope you can carry on...

 

[soroban]

Re: An oil well was drilled 3400 ft before it hit a ledge of

Hello, Apcanton!

Another approach . . .

An oil well at A was drilled 3400 ft before it hit a ledge of shale.
The same ledge was only 1950 ft deep when drilled from a well at B,
which is 8000 ft directly downhill from A.
The hill is inclined at 15° to the horizontal.

a. Find how far down the hill from B until shale ledge emerges.
b. What is the angle between the ledge and the hill?

                                          *A
                                       *  |
                               8000 *     |
                                 *        |3400
                             C*           |
                           *  |           |
                     x  *     |1950       *B
                     *        |     *     :
                  *           *D          :
               *        *     :           :
            *     *           :           :
         *  *                 :           :
    - * - - - - - - - - - - - + - - - - - + -
      O                       F           E

Let $\displaystyle x \,=\,OC.$

We have: $\displaystyle \,\Delta CDO\,\sim\,\Delta ABO$

. . Hence: $\displaystyle \:\frac{x}{1950} \:=\:\frac{x\,+\,8000}{3400}\;\;\Rightarrow\;\;x \:=\:\frac{312,000}{29}\:\approx\:\fbox{10,758.62\text{ ft}}$

                             C*
                           *  |
                        *     |1950
                 x   *        |
                  *           *D
               *        *     :
            *     *           :
         *  *                 :
      * - - - - - - - - - - - *
      O                       F

$\displaystyle OF\:=\:x\cos15^o\:\approx\:10,392.03$
$\displaystyle CF \:=\:x\sin15^o \:\approx\:2784.54$

Then: $\displaystyle \,DF\:=\:CF\,-\,CD \:=\:2784.54\,-\,1950\:=\:834.54$

And: $\displaystyle \,\tan(\angle DOF) \:=\:\frac{834.54}{10,392.03} \:=\:0.080305773$

Therefore: \(\displaystyle \angle DOF \:\approx\:\fbox{4.6^o}\)