Question

an observer of height 1.8m is 13.2m away from a plam tree. the angle of elevation of the top of tree from his eyes is 45°. what is the height of the palm tree

Answer 1

hope it's helpful to ypu

Answer 2

We need to recall the following trigonometric ratios.

• $tan(A)=\frac{Opposite}{Adjacent}$
• $tan45{\textdegree}=1$

This problem is about the trigonometric ratio.

Given:

Height of observer $=1.8$ m

Distance between an observer and a palm tree $=13.2$ m

The angle of elevation of the top of the tree from his eyes  $=45{\textdegree}$

Let $h$ m be the height of the palm tree above the height of the observer.

In Δ $BCD$,

∠ $B=45{\textdegree}$ , $CD=h$ m and $BD=13.2$ m

Using trigonometric ratio, we get

$tan(B)=\frac{CD}{BD}$

$tan45{\textdegree}=\frac{h}{13.2}$

$1=\frac{h}{13.2}$

$h=13.2$ m

Hence, the height of the palm tree is,

$=1.8+h$

$=1.8+13.2$

$=14$ m