The shadow of a tower standing on a level ground is found to be 40m longer when the sun altitude

The shadow of a tower standing on a level ground is found to be 40 m longer when Sun’s altitude is 30° than when it was 60°. Find the height of the tower

Let AB be the tower of height h.given the shadow of tower DC = 40 m. the attitude of the sun is ∠D = 30° and ∠C = 60°. Here we have to find the height of the tower. Let BC = x and DC= 40.

So we have trigonometric ratios in ΔACB

`=> tan C = (AB)/(BC)`

`=> tan 60^@ = h/x`

`=> sqrt3 = h/x`

`=> x = h/sqrt3`

Again in ΔADB

`=> tan D = (AB )/(DB)`

`=> tan 606^@ = h/x`

`=> sqrt3 = h/x`

`=> x = h/sqrt3`

Again in ΔADB

`=> tan D = (AB)/(DB)`

`=> tan 30^@ = h/(40+ x)`

`=> 1/sqrt3 h/(40 + x)`

`=> 40 + x = sqrt3h`

`Put x = h/sqrt3`

`=> 40 + h/sqrt3 \ sqrt3h`

`=> 40 = sqr3h = h/sqrt3`

`=> 40 = (2h)/sqrt3`

`=> h = 20sqrt3`

Hence height of tower is `20sqrt3` m

Concept: Heights and Distances

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Text Solution

Solution : 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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