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

  • Sunil Kumar Sunil
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  • (c) 173 mbr /Angle of elevation is equal to thirty degrees;br /let angle of elevetion be =a;br /a=30 degreesbr /tan (30 degrees)=(prependicular/base);br /prependicular= height of tower;br /base = distance of point p from foot of tower;br /tan(30 degrees)=1/root(3);br /(1/root(3))=100/base          eqn(1)br /from eqn(1) we get base equal to 173mbr / 

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  • from the above data, we should form a triangle of type APB. br /such that PB is the hypotenuse, AB is the tower and AP is the distance between foot of the tower and P.br /angle of elevation= <APB = 30 degrees and AB = 100 mbr /AB/AP = tan 30 = 1/1.7320 br /            AP =  (ABX1.7320) mbr /                  = 100X1.7320br /                  = 173 m

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