A pilot has just started on the glide path for landing...

[TVigz42]

Could someone please help me using the information given below:

A pilot has just started on the glide path for landing at an airport with a runway of length 9000 feet. The angles of depression from the plane to the ends of the runway are 17.5 degrees and 18.5 degrees.

(a) Draw a diagram that visually represents the problem
(b) Find the air distance the plane must travel until touching down on the near end of the runway.
(c) Find the ground distance the plane must travel until touching down
(d) Find the altitude of the plane when the pilot begins the descent

 

[soroban]

Re: A pilot has just started on the glide path for landing..

Hello, TVigz42!

A pilot has just started on the glide path for landing at an airport
with a runway of length 9000 feet. .The angles of depression from the plane
to the ends of the runway are 17.5° and 18.5°.

(a) Draw a diagram that visually represents the problem

(b) Find the air distance the plane travels to touch down on the near end of the runway.
(c) Find the ground distance the plane must travel until touching down
(d) Find the altitude of the plane when the pilot begins the descent

    P * - - - - - - - - - - - - - - - - - Q
      |  *   * 17.5°
      |     * 1°  *
      |        *       *
    y |         z *         *
      |             *           *
      |                 *              *
      |              18.5° *          17.5°  *
      * - - - - - - - - - - - * - - - - - - - - - - *
      R          x            A        9000         B

The plane is a \(\displaystyle P.\;\) The runway is \(\displaystyle AB\,=\,9000.\)

\(\displaystyle \angle QPB\,=\,17.5^o.\;\)Then \(\displaystyle \angle PBA\,=\,17.5^o\)
\(\displaystyle \angle QPA\,=\,18.5^o.\;\)Then \(\displaystyle \angle PAR\,=\,18.5^o\)
. . And: \(\displaystyle \,\angle BPA\,=\,1^o\)
Let: \(\displaystyle x\,=\,RA,\;y\,=\,PR,\:z\,=\,PA\)

In triangle \(\displaystyle PAB\), use the Law of Sines: \(\displaystyle \L\:\frac{z}{\sin17.5^o} \:=\:\frac{9000}{\sin1^o}\)

. . Therefore: (b)\(\displaystyle \:z \:=\:\frac{9000\cdot\sin17.5^o}{\sin1^o} \:\approx\:155,070.4\) feet


In right triangle \(\displaystyle PRA\), we have: \(\displaystyle \angle A \,=\,18.5^o\) and the hypotenuse.

You can now solve for \(\displaystyle x\) and \(\displaystyle y.\)

 

[TVigz42]

Thanks!

Thank you!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!