1.

The shadow of a tower standing on a level ground is found to be 40 m longer when the Suns altitude is `30o`than when it is `60o`. Find the height of the tower.

Answer» let h be the height of th tower
D is the point where it is forming angle `30^@`
& C is the point where it is forming angle`60^@`
now `tan 60^@ = h/(BC)`
`BC= h/tan 60^@ = h/sqrt3`
now, in `/_ ABD`
`In 30^@ = (AB)/(BD)= h/(BD)`
`BD= h/(tan 30^@) = h/(1/sqrt3)= hsqrt3`
as we can see, `BD-BC=CD`
`(sqrt3h) -(1/sqrt3)h = 40`
`h(sqrt3 - 1/sqrt3) = 40`
`h(3-1)/sqrt3 = 40`
`h(2/sqrt3) = 40`
`h=40*sqrt3/2 = 20sqrt3`m
answer


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