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 ${60}^0$ to ${30}^0$.

Answer 1

Let's assume AB to be the height of the tree, h m.
Let the two points be C and D be such that $CD=20m.<ADB=$${30}^0$ and < ACB =${60}^0$
In $△ABC$,
$AB/BC=tan{60}^0=\sqrt{3}$
$BC=AB/\sqrt{3}=h/\sqrt{3}$...... (i)
In $△ABD$,
$AB/BD=tan{30}^0$
$h/(20+BC)=1/\sqrt{3}$
$\sqrt{3}h=20+BC$
$\sqrt{3}h=20+h/\sqrt{3}$ ......[From (i) ]
$h(\sqrt{3}-1/\sqrt{3})=20$
$h=20/(\sqrt{3}-1/\sqrt{3})=20/1.154=17.32m$
Therefore. the height of the wee is $17.32m.$