from a point p on a level ground, the angle of elevation of the top of a tower is 30 degrees. if the tower is 100 m height, then the distance of point p from the foot of the tower is :

from a point p on a level ground, the angle of elevation of the top of a tower is 30 degrees. if the tower is 100 m height, then the distance of point p from the foot of the tower is : (a) 149 m (b) 156 m (c) 173 m (d) 200 m

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  • (c) 173 m
    Angle of elevation is equal to thirty degrees;
    let angle of elevetion be =a;
    a=30 degrees
    tan (30 degrees)=(prependicular/base);
    prependicular= height of tower;
    base = distance of point p from foot of tower;
    tan(30 degrees)=1/root(3);
    (1/root(3))=100/base          eqn(1)
    from eqn(1) we get base equal to 173m
     
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  • from the above data, we should form a triangle of type APB. 
    such that PB is the hypotenuse, AB is the tower and AP is the distance between foot of the tower and P.
    angle of elevation= <APB = 30 degrees and AB = 100 m
    AB/AP = tan 30 = 1/1.7320 
                AP =  (ABX1.7320) m
                      = 100X1.7320
                      = 173 m
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