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Math Help - height and distance problem

  1. #1
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    height and distance problem

    1. A fire in a building B is reported on telephone to two fire stations P and Q , 20 km apart from each other on a straight road. P observes that the fire is at an angle of 60 degrees to the road and Q observes that it is at an angle of 45 degrees to the road. Which station should send its team and how much will its team have to travel?

    2. An aeroplane flying horizontally 1 km above the ground is observed at an elevation of 60 degrees. After 10 seconds , its elevation is observed to be 30 degees. Find the speed of the aeroplane in km/hr.
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  2. #2
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    Quote Originally Posted by aman View Post
    1. A fire in a building B is reported on telephone to two fire stations P and Q , 20 km apart from each other on a straight road. P observes that the fire is at an angle of 60 degrees to the road and Q observes that it is at an angle of 45 degrees to the road. Which station should send its team and how much will its team have to travel?

    2. An aeroplane flying horizontally 1 km above the ground is observed at an elevation of 60 degrees. After 10 seconds , its elevation is observed to be 30 degees. Find the speed of the aeroplane in km/hr.
    To #1:
    You are dealing with the triangle PQB where you know
    PQ = 20\ km
    \angle(QPB)=60^\circ
    \angle(BQP)=45^\circ ......and consequently
    \angle(PBQ)=75^\circ

    Use Sine rule to calculate the length of the other two sides of the triangle (14.64 | 17.93)

    To #2:
    From the height of the plane and the angle of elevation you can calculate the ditance of the observer to the footpoint of the airplane:

    \tan(60^\circ)=\dfrac{1000}x~\implies~x\approx 577.35\ m

    \tan(30^\circ)=\dfrac{1000}y~\implies~y\approx 1732.05\ m

    Depending on the position of the observer with respect to the course the airplane is flying you'll get a minimum or maximum speed:

    <br />
v_{min}=\dfrac{1732.05 - 577.35\ m}{10\ s}= 115.47\ \frac ms = 415.7\ \frac{km}h
    or
    <br />
v_{min}=\dfrac{1732.05 + 577.35\ m}{10\ s}= 230.94\ \frac ms = 831.4\ \frac{km}h
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