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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. |
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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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