Question

a person standing on the bank of a
river observes that the angle of elevation
of the top of a tree standing on the
opposite bank is 60° .When he moves 40m
away from the bank he finds the angle
of elevation to be 30°. Find the
height of the tree and the width of
the river.​

Answer 1

$\bold\green{Height\:of\:tree}$= 34.64 metres.

$\bold\green{Width\:of\:Canal}$= 20 meters.

EXPLANATION:

Let,

• CD be the tree of height h meter.
• B be the position of a man standing on the opposite bank of the river.

After moving 40 M away from point B.

so, let new position of man be A

hence, AB= 40 meter.

• The angles of elevation of the top of the tree from point A and B are 30° and 60° respectively. i.e,, ∠CAD = 30° and ∠CBD = 60°
• Let BC = x meter.

$\rule{300}{2}$

In right triangle BCD , we have

$\tan60= \frac{CD}{BC} \\ ⇛ \sqrt{3} = \frac{h}{x} \\ ⇛x = \frac{h}{ \sqrt{3} } ..............(i)$

In right triangle ACD, we have

$\tan30 = \frac{CD}{AC} \\ ⇛ \frac{1}{ \sqrt{3} } = \frac{h}{x + 40} \\ ⇛x + 40 = \sqrt{3} h \\ x = \sqrt{3} h - 40...........(ii)$

Comparing (¡) and (ii), we get

$\frac{h}{ \sqrt{3} } = \sqrt{3} h - 40 \\ h = 3h - 40 \sqrt{3} \\ - 2h = - 40 \sqrt{3} \\ h = 20 \sqrt{3} \\ ⇛20 \times 1.732 \\ ⇛34.64 \: meter$

hence, the height of the tree is 36.64 metres.

Now substituting the value of h in (i), we get

$x = \frac{h}{ \sqrt{3} } \\ x = \frac{20 \sqrt{3} }{ \sqrt{3} } \\ ⇛x = 20 \: meter.$

hence, width of River is 20 meters.

$\rule{300}{2}$

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Height\:of\:tree=20\sqrt{3}\:m}}}$

$\green{\tt{\therefore{Width\:of\:river=20\:m}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given: }} \\ \tt: \implies length \: DC = 40 \: m \\ \\ \tt: \implies First \: angle \: of \: elevation = 60 \degree \\ \\ \tt: \implies Second\: angle \: of \: elevation = 30 \degree \\ \\ \red{\underline \bold{To \: Find: }} \\ \tt: \implies Height \: of \: tree = ? \\ \\ \tt: \implies Width \: of \: river =?$

• According to given question :

$\bold{In \:right \: angled \: \triangle \: ABD} \\ \tt: \implies tan \: \theta = \frac{p}{b} \\ \\ \tt: \implies tan \:30 \degree = \frac{AB}{BD} \\ \\ \tt: \implies \frac{1}{ \sqrt{3} } = \frac{AB}{40 + x} \\ \\ \tt: \implies 40 + x = AB \sqrt{3} \\ \\ \tt: \implies x = AB \sqrt{3} - 40 - - - - - (1) \\ \\ \bold{In \: right \: angled \: \triangle \: ABC} \\ \tt: \implies tan \: \theta = \frac{p}{b} \\ \\ \tt: \implies tan \: 60 \degree = \frac{AB}{BC} \\ \\ \tt: \implies \sqrt{3} = \frac{AB}{x} \\ \\ \tt: \implies x = \frac{AB}{ \sqrt{3} } - - - - - (2)$

$\text{From \: (1) \: and \: (2)} \\ \tt: \implies \frac{AB}{ \sqrt{3} } = AB \sqrt{3} - 40 \\ \\ \tt: \implies AB= 3AB - 40 \sqrt{3} \\ \\ \tt: \implies 40 \sqrt{3} = 2AB \\ \\ \tt: \implies AB = \frac{40 \sqrt{3} }{2} \\ \\ \green{\tt: \implies AB = 20 \sqrt{3} \: m } \\ \\ \text{Putting \: value \: of \: ab \: in \: (2)} \\ \tt: \implies x = \frac{20 \sqrt{3} }{ \sqrt{3} } \\ \\ \green{\tt: \implies x = 20 \: m} \\ \\ \green{\tt\therefore Height \: of \: tree \: is \: 20 \sqrt{3 } \: m} \\ \\ \green{\tt\therefore Width \: of \: river \: is \: 20 \: m}$