Question

The shadow of a tower standing on level ground is found to be 40 m longer when the Sun's altitude is 30° than when it was 60°. Find the height of the tower. (Given `sqrt3` = 1.732)

Answer 1

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" = 34.64"m"`

Hence, the height of the tower is `20 sqrt3"m".`