The Shadow of a tower Standing on a Level ground is Found to be 40 m Longer when Suns Altitude is 30 than When it was 60. Find the Height of the tower | Bzziii

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The shadow of a tower standing on a level ground is found to be 40 m longer when Suns altitude is 30 than when it was 60. Find the height of the tower. (Take = `\sqrt{3}` = 1.732)







Given tower be AB

When Sun's altitude is `60^0`

`\angle` ACB = `60^0`

& Length of shadow = BC

When Sun's altitude is `30^0`

`\angle` ADB = `30^0`

& Length of shadow = DB


Shadow is 40 m when angle changes from 60° to 30° CD = 40 m

We need to find height of tower i.e. AB Since tower is vertical to ground ∴ ∠ ABC = 90°

From (1) and (2)

`\frac{"AB"}{\sqrt{3}}` = `\sqrt{3}`AB - 40

AB = `\sqrt{3}` (`\sqrt{3}`AB) - 40 `\sqrt{3}`

AB = 3AB - 40 `\sqrt{3}`

40 `\sqrt{3}` = 3AB - AB

40 `\sqrt{3}` = 2AB

AB = `\frac{ 40 \sqrt{3}}{2}`

AB = 20`\sqrt{3}`


Hence, Height of the tower = AB = 20`\sqrt{3}`metre



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