trigonometry? speed of an airplane

marie54

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An airbus A320 aircraft is cruising at an altitude of 10000 m. The aircraft is flying in a straight line away from Rachel, who is standing on the ground. If she sees the angle of elevation of the aircraft change from 70 degrees to 33 degrees in one minute, what is its cruising speed, to the nearest kilometer per hour?
 
Hello, marie54!

An airbus A320 aircraft is cruising at an altitude of 10,000 meters (10 kilometers).
The aircraft is flying in a straight line away from Rachel, who is standing on the ground.
She sees the angle of elevation of the aircraft change from 70 degrees to 33 degrees in one minute.
What is its cruising speed, to the nearest kilometer per hour?
Code:
                        A        d        B
                        * - - - - - - - - *
                      * :             *   :
                    *   :         *       :
                  *     :     *           :
                *       : *               : 10 km
              *       * :                 :
            *     *     :                 :
          *   *         :                 :
        * *             :               - :
      * - - - - - - - - * - - - - - - - - *
      R        x        P        d        Q

Rachel is at R.

The plane is sighted at A.\displaystyle A.
The point on the ground directly below A\displaystyle A is P.\displaystyle P.
. . AP=10,  ARP=70o\displaystyle AP = 10,\;\angle ARP = 70^o.

One minute later, the plane is at B.\displaystyle B.
The point on the ground directly below B\displaystyle B is Q.\displaystyle Q.
. . BQ=10,  BRQ=33o\displaystyle BQ = 10,\;\angle BRQ = 33^o

Let d=AB=PQ,    x=RP\displaystyle \text{Let }\,d = AB = PQ,\;\;x = RP


\(\displaystyle \text{In right triangle }APR\!:\;\;\tan70^o \,=\,\frac{10}{x} \quad\Rightarrow\quad x \,=\,\frac{10}{\tan70^o}\quadf[1]\)

In right triangle BQR ⁣:    tan33o=10d+xd=10tan33ox[2]\displaystyle \text{In right triangle }BQR\!:\;\;\tan33^o \,=\,\frac{10}{d+x} \quad\Rightarrow\quad d \:=\:\frac{10}{\tan33^o} - x\quad[2]

Substitute [1] into [2]:   d  =  10tan33o10tan70o  =  11.7589473\displaystyle \text{Substitute [1] into [2]: }\;d \;=\;\frac{10}{\tan33^o} - \frac{10}{\tan70^o} \;=\;11.7589473

The plane flew about 11.8 km in one minute,\displaystyle \text{The plane flew about 11.8 km in one minute,}


Therefore, its speed is: 60×11.8  =  705.6    706 km/hr.\displaystyle \text{Therefore, its speed is: }\:60 \times 11.8 \;=\;705.6 \;\approx\;706\text{ km/hr.}

 
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