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Question
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.
Solution
Let AB is the tower and BD is the length of the shadow when the sun's altitude is 60°.
Let, AB be x m and BD be y m.
So, CB = (40 + y) m
Now, we have two right-angled triangles ΔABD and ΔABC
In ABD
`tan 60^@ = "AB"/"BD"`
`sqrt3 = "x"/"y"`
In ABC
`tan 30^@ = "AB"/"BD"`
`1/sqrt3 = "x"/("y"+40)` ...(1)
We have x = y `sqrt3`
substituting the value in (1)
`("y"sqrt3)sqrt3 = "y" + 40`
⇒ 3y = y + 40
⇒ y = 20
⇒ `"x" = 20 sqrt3"m"`
If we take the value of `sqrt3` = 1.73
⇒ x = 20 × 1.73
= 34.64 m
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