90
From a point on the ground, the angle of elevation of the top of a tree observed by a person is $60^\circ$. When moved back by $28$ m, in the same line, the angle of elevation from another point on ground becomes $30^\circ$. Find the height of the tree and its distance from the initial point. (Use $\sqrt{3} = 1.73$)
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Solution:
Let the height $AD$ of the tree be $h$ m and its distance from the initial point $B$ be $x$ m
In $$\begin{aligned}& \Delta CAD, \tan 30^\circ = \frac{h}{x + 28} \Rightarrow x + 28 = h\sqrt{3} \dots (i) \\ & In\end{aligned}$$
Delta BAD,
tan 60^
circ =
frac{h}{x}
Rightarrow h = x
sqrt{3}
dots (ii)
Solving to get, $x = 14, h = 14 \times 1.73 = 24.22$
Height of the tree $= 24.22$ m and distance from the initial point $= 14$ m.
Let the height $AD$ of the tree be $h$ m and its distance from the initial point $B$ be $x$ m
In $$\begin{aligned}& \Delta CAD, \tan 30^\circ = \frac{h}{x + 28} \Rightarrow x + 28 = h\sqrt{3} \dots (i) \\ & In\end{aligned}$$
Delta BAD,
tan 60^
circ =
frac{h}{x}
Rightarrow h = x
sqrt{3}
dots (ii)
Solving to get, $x = 14, h = 14 \times 1.73 = 24.22$
Height of the tree $= 24.22$ m and distance from the initial point $= 14$ m.