Question

from the top of a building 100m high the angles of depression of the top and bottom of a tower are observed to be 45° and 60° respectively . find the height of the tower and also the distance between the foot of the building and for of the tower

Answer 1

Consider a building of height (AC) 100 m and tower (ED) be h.

So, AB = (100 - h) m

The building makes an angles of depression at the top and bottom of a tower; which are observed to be 45° and 60°.

Join B and E such that BE = CD.

Now,

In ∆ACD

$\rightarrow\:\sf{\frac{AC}{CD}\:=\:tan60^{\circ}}$

$\rightarrow\:\sf{\dfrac{100}{CD}\:=\:\sqrt{3}}$

$\rightarrow\:\sf{100\:=\:CD\sqrt{3}}$

$\rightarrow\: \sf{\dfrac{100 }{ \sqrt{3} \: } \: = \: CD}$ ...(1)

$\rightarrow\:\sf{\dfrac{100}{1.732}\:=\:CD}$

$\rightarrow\:\sf{CD\:=\:57.74\:m}$

•°• Distance between the foot of the tower and building is 57.74 m.

In ∆ABE

$\rightarrow\:\sf{\dfrac{AB}{BE}\:=\:tan45^{\circ}}$

$\rightarrow\:\sf{\dfrac{AB}{BE}\:=\:1}$

$\rightarrow\:\sf{AB\:=\:BE}$

But BE = CD

$\rightarrow\:\sf{AB\:=\:CD}$

$\rightarrow\:\sf{CD\:=\:100\:-\:h}$ ...(2)

Substitute value of CD = 100/√3 in equation (2)

$\rightarrow\:\sf{\dfrac{100}{\sqrt{3}}\:=\:10\:-\:h}$

On rationalizing we get,

$\rightarrow\:\sf{\dfrac{100\sqrt{3}}{3}\:=\:100\:-\:h}$

$\rightarrow\:\sf{h\:=\:100\:-\:\frac{100\sqrt{3}}{3}}$

$\rightarrow\:\sf{h\:=\:\dfrac{300\:-\:100\sqrt{3}}{3}}$

$\rightarrow\:\sf{h\:=\:\dfrac{100(3\:-\:\sqrt{3})}{3}}$

$\rightarrow\:\sf{h\:=\:\dfrac{100(3\:-\:1.732)}{3}}$

$\rightarrow\:\sf{h\:=\:\dfrac{100(1.268)}{3}}$

$\rightarrow\:\sf{h\:=\:100(0.4226)}$

$\rightarrow\:\sf{h\:=\:42.26\:m}$

•°• Height of tower is 42.26 m

Answer 2

$\Large\underline{\underline{\sf{Given}:}}$

• Angle of depression of The top and bottom of a tower are 45° and 60°
• Height of Building = 100 m ..

$\Large\underline\mathfrak{Question}$

• Height of tower = ?
• Distance b/w foot of building and tower ?

$\Large\underline{\underline{\sf{Solution}:}}$

$\red{\textbf{Refer To image First ...}}$

From image we can see that ,,

• AB = Height of Building = 100m
• DC = Height of tower = h m
• CB = DE = Foot distance = x m
• angle ACB = 60° ( As angle of depression = angle of elevation)
• angle ADE = 45°

___________________________

$\textbf{in right triangele ACB first}$

$tan60 = \frac{AB}{CB} = \frac{100}{x} \\ \\ \red\leadsto \: \sqrt{3} = \frac{100}{x} \\ \\ \red\leadsto \: x = \frac{100}{ \sqrt{3} } \\ \\ rationalize \\ \\ \red\leadsto \green{\large\boxed{\bold{x = \frac{100 \sqrt{3} }{3} m }}}$

$\textbf{ Hence, distance between Foot of Tower} \\ \textbf{and Building is} \: \frac{100 \sqrt{3} }{3} m$

__________________________________

Now, From diagram we can see that ,

CB = DE = 100√3/3m (Solved above)

$\textbf{in traingle AED now}$

$tan45 = \frac{AE }{ED } \\ \\ \red \leadsto \: 1 = \frac{(100 - h)}{ \frac{100 \sqrt{3} }{3} } \\ \\ \red \leadsto \frac{100 \sqrt{3} }{3} = (100 - h) \\ \\ \red \leadsto100 \sqrt{3} = 300 - 3h \\ \\ \red \leadsto3h = 300 - 100 \sqrt{3} \\ \\ \red \leadsto \: h = \frac{300 - 100 \sqrt{3} }{3} \\ \\ \red \leadsto \: h = \frac{300 - 100 \times 1.73}{3} \\ \\ \red \leadsto \: h = \: \frac{300 - 173}{3} \\ \\ \red \leadsto \: h = \frac{127}{3} \\ \\ \red \leadsto \: \large\boxed{\bold{h =42.33m }} \\ \\ \pink{\textbf{Hence, Height of tower = DC =42.33m }}$

$\large\underline\textbf{Hope it Helps You.}$