Question

Find the height of a tree when it is found that on walking away from it 20 m, in a horizontal line through its base, the elevation of its top changes from 60° to 30°.

Answer 1

Let AB be the tree and its height be x

DC = 20 m.

Now in right ΔADB,

`tan theta = (AB)/(DB)`

`\implies tan 30^circ = x/(DB)`

`\implies 1/sqrt(3) = x/(DB)`

`\implies DB = sqrt(3)x`.   ...(i)

In ΔACB, we have

`tan 60^circ = x/(CB)`

`\implies sqrt(3)/1 = x/(CB)`

∴ `CB = x/sqrt(3) = (sqrt(3)x)/3`   ...(ii)

But DB – CB = DC

`\implies sqrt(3)x - (sqrt(3)x)/3 = 20`

`\implies (3sqrt(3)x - sqrt(3)x)/3 = 20`

`\implies (2sqrt(3)x)/3 = 20`

`\implies x = (20 xx 3)/(2sqrt(3))`

= `(10 xx 3 xx sqrt(3))/(sqrt(3) xx sqrt(3))`

= `(30sqrt(3))/3`

∴ `x = 10sqrt(3)`

= 10 × (1.732)

= 17.32 m.

∴ Required height of the tree = 17.32 m