Altitude of a plane when given 2 angles and speed.

[lynx720]

I am having extreme trouble with this problem. "A plane is observed approaching your home and you assume its speed is 550 miles per hour. If the angle of elevation of the plane is 16 degrees at one time and 57 degrees 1 minute later, approximate the altitude of the plane."

I figured out that the plane is moving approximately 9.17 miles per minute, so I know it went that far. However, unlike the other problems before it, in which you were given at least one clear cut side to calculate a trigonometric function based on an angle and a side, there is not one of those here. This leaves me very stuck. I've drawn multiple pictures and have tried to browse for this problem online. No one has posted this except for one other board, and in that case, the guy who responded didn't give a way to arrive at the problem. I know from the answer in the back of the book that it is 17,054 ft in the air ( 3.23 miles) but I want to know how you would arrive at that answer.

 

[Loren]

Name the distance from your position to the plane's starting position "a" and the altitude of the plane "x". Assuming your calculation in arriving at 9.17 is correct...

\(\displaystyle \frac{9.17}{\sin(57^o-16^o)}=\frac{a}{sin(180^o-41^o-16^o)}\) <<< Solve for a.

\(\displaystyle \sin 16^o = \frac{x}{a}\) <<< Solve for x.

I leave you to figure out where all the numbers came from.

 

[soroban]

Hello, lynx720!

A plane is observed approaching your home and you assume its speed is 550 mph.
If the angle of elevation of the plane is 16 degrees at one time and 57 degrees 1 minute later,
approximate the altitude of the plane.

At its closest approach, the plane is situated like this:

                  * P
                * |
              *   |
            *     | h
          *       |
        * 57d     |
      * - - - - - *
           x

\(\displaystyle \text{And we have: }\;\tan 57^o \:=\:\frac{h}{x} \quad\Rightarrow\quad x \:=\:\frac{h}{\tan57^o}\) .[1]


\(\displaystyle \text{One minute before, the plane was }9\tfrac{1}{6} \,=\,\tfrac{55}{6}\text{ miles further away.}\)

\(\displaystyle \text{Its position looked like this:}\)

                              * P
                          *   :
                      *       : 
                  *           : h
              *   :           :
          * 16d   :           :
      * - - - - - * - - - - - * 
      :     x     :    55/6   :

\(\displaystyle \text{And we have: }\;\tan16^o \:=\:\frac{h}{x+\frac{55}{6}} \quad\Rightarrow\quad x \:=\:\frac{6h-55\tan16^o}{6\tan16^o}\) .[2]


Equate [2] and [1]:

. . \(\displaystyle \frac{6h-55\tan16^o}{6\tan16^o} \:=\:\frac{h}{\tan57^o} \quad\Rightarrow\quad 6h\tan57^o - 55\tan16^o\tan57^o \;=\;6h\tan16^o\)

. . \(\displaystyle 6h\tan57^o - 6h\tan16^o \;=\;55\tan16^o\tan57^o \quad\Rightarrow\quad 6h(\tan57^o - \tan16^o) \;=\;55\tan16^o\tan57^o\)


\(\displaystyle \text{Therefore: }\;h \;=\;\frac{55\tan16^o\tan57^o}{6(\tan57^o-\tan16^o)} \;=\;3.229966363 \;\approx\;3.23\text{ miles.}\)

 

[lynx720]

Thanks a bunch Soroban. That answer made perfect sense to me, although much more complicated than the previous problems in the chapter. Thanks for the help.