Question

A man standing on the bank of a river observes that the angle of elevation of a tree on the opposite bank is 60° . when the moves 50m.away from the bank, he finds the angle of elevation to be 30° . calculate the height of the tree?

Answer 1

Let a $\bold{\triangle}$ABD having AB = height of tree and BC = x.

A man was standing on the bank of river and observes that angle of elevation of tree which is on the opposite bank of river is 60°.

When the man moves 50 m away from the bank i.e. CD the angle of elevation becomes 30° as shown in figure.

We have to calculate the height of the tree.

In $\triangle$ABC

$\implies\:\dfrac{AB}{BC}\:=\:\tan\:\theta$

$\implies\:\dfrac{h}{x}\:=\:\tan60^{\degree}$

$\implies\:\dfrac{h}{x}\:=\:\sqrt{3}$

$\implies\:h\:=\:x\sqrt{3}$ ____ (eq 1)

Similarly,

In $\triangle$ABD

$\implies\:\dfrac{AB}{BD}\:=\:\tan\:\theta$

$\implies\:\dfrac{AB}{BC\:+\:CD}\:=\:\tan\:\theta$

$\implies\:\dfrac{h}{x\:+\:50}\:=\:\tan30^{\degree}$

$\implies\:\dfrac{h}{x\:+\:50}\:=\:\dfrac{1}{\sqrt{3}}$

$\implies\:h\sqrt{3}\:=\:x\:+\:50$

$\implies\:h\:=\:\dfrac{x\:+\:50}{\sqrt{3}}$ ____ (eq 2)

From (eq 1) and (eq 2) we get,

$\Rightarrow\:x\sqrt{3}\:=\:\dfrac{x\:+\:50}{\sqrt{3}}$

$\Rightarrow\:3x\:=\:x\:+\:50$

As, √3 × √3 = 3

$\Rightarrow\:2x\:=\:50$

$\Rightarrow\:x\:=\:25\:m$

Put value of x in (eq 1)

$\implies\:h\:=\:25\sqrt{3}\:m$

∴ Height of tree is 25√3 m.

Answer 2

1. [Figure]

Answer:

$\huge{\pink{\green{\sf{\therefore HEIGHT=25\sqrt{3}\:m}}}}$

step-by-step explanation:

$\huge{\pink{\green{\underline{\red{\sf{SOLUTION-}}}}}}$

• In the given question information given about a man standing on the bank of a river and he observes the angle of elevation is 60°.

• After travelling 50 m angle of elevation changes to 30°.

• We have to find the height of tree.

$\underline \bold{Given : } \\ \implies Angle \: of \: elevation \: of \: tree ( \angle ACB) = 60 \degree \\ \implies distance(DC) = 50m \\ \implies let \: distance (BC) = x \: m \\ \\ \underline \bold{To \:Find : } \\ \implies height \:of \: tree( AB) = ?$

• According to given question :

$\bold{In \triangle ABC : } \\ \implies tan \: \theta= \frac{p}{b} \\ \implies tan \: 60 \degree = \frac{h}{x} \\ \implies \sqrt{3} = \frac{h}{x} \\ \bold{\implies x = \frac{h}{ \sqrt{3} } m} \\ \\ \bold {In \triangle ABD : } \\ \implies tan \theta = \frac{p}{b} \\ \implies tan \: 30 \degree = \frac{AB}{BD} \\ \implies \frac{1}{ \sqrt{3} } = \frac{h}{50 + x} \\ \implies 50 + x = \sqrt{3} h \\ \bold {putting \: value \: of \: x }\\ \implies 50 + \frac{h}{ \sqrt{3} } = \sqrt{3} h \\ \implies \frac{50 \sqrt{3} + h }{ \sqrt{3} } = \sqrt{3} h \\ \implies 50 \sqrt{3} + h = 3h \\ \implies 50 \sqrt{3} = 3h - h \\ \implies h = \frac{50 \sqrt{3} }{2} \\ \bold{ \therefore h = 25 \sqrt{3} m}$