An observer of height $1.8 m$ is $13.2 m$ away from a palm tree. The angle of elevation of the top of the tree from his eyes is $45^{\circ}$ . What is the height of the palm tree?

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Solution

$Height of observer = A E$
$= 1.8 m$
$E D = A C$
$= 13.2 m$

In $△ A B C,$
$\begin{matrix} tan 45^{\circ} & = \frac{B C}{A C} 1 & = \frac{B C}{13.2} B C & = 13.2 m \end{matrix}$

The height of the palm tree will be equal to $B D$ . So,
$\begin{matrix} B D & = B C + C D = 13.2 + A E = 13.2 + 1.8 = 15 m \end{matrix}$

So, the height of the palm tree is equal to $15 m .$

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An observer of height 1.8 m is 13.2 m away from a palm tree. The angle of elevation of the top of the tree from his eyes is 45∘. What is the height of the palm tree?

Height of observer=AE =1.8 mED=AC =13.2 mIn △ABC,tan45∘=BCAC1=BC13.2BC=13.2 mThe height of the palm tree will be equal to BD. So,BD=BC+CD=13.2+AE=13.2+1.8=15 mSo, the height of the palm tree is equal to 15 m.

An observer of height $1.8 m$ is $13.2 m$ away from a palm tree. The angle of elevation of the top of the tree from his eyes is $45^{\circ}$ . What is the height of the palm tree?