Question

The shadow of the tower standing on a level plane is found to be 100 m longer when the sun's is angle of elevation is 30, than when it is 45. The height of the tower is

Answer 1

Answer:

Height of tower (AB) = 136.6 m

$\rule{200}2$

Step-by-step explanation:

In ∆ABC

$\implies\:\sf{\dfrac{AB}{BC}\:=\:tan45^{\circ}}$

$\implies\:\sf{\dfrac{AB}{BC}\:=\:1}$

$\implies\sf{AB\:=\:BC}$ ...(1)

Similarly, In ∆ABD

$\implies\:\sf{\dfrac{AB}{BD}\:=\:tan30^{\circ}}$

$\implies\:\sf{\dfrac{AB}{BC\:+\:CD}\:=\:\dfrac{1}{\sqrt{3}}}$

$\implies\:\sf{\dfrac{AB}{BC\:+\:100}\:=\:\dfrac{1}{\sqrt{3}}}$

$\implies\:\sf{AB\sqrt{3}\:=\:BC\:+\:100}$

$\implies\:\sf{AB\sqrt{3}\:=\:AB\:+\:100}$ [From (1)]

$\implies\:\sf{AB\sqrt{3}\:-\:AB\:=\:100}$

$\implies\:\sf{AB(\sqrt{3}\:-\:1)\:=\:100}$

$\implies\:\sf{AB\:=\:\dfrac{100}{\sqrt{3}\:-\:1}}$

Rationalise it

$\implies\:\sf{AB\:=\:\dfrac{100}{\sqrt{3}\:-\:1}\:\times\:\dfrac{\sqrt{3}\:+\:1}{\sqrt{3}\:+\:1}}$

$\implies\:\sf{AB\:=\:\dfrac{100(\sqrt{3}\:+\:1)}{3\:-\:1}}$

$\implies\:\sf{AB\:=\:\dfrac{100(\sqrt{3}\:+\:1)}{2}}$

$\implies\:\sf{AB\:=\:50(\sqrt{3}\:+\:1)}$

Substitute value of AB in equation (1)

$\implies\:\sf{BC\:=\:50(\sqrt{3}+\:1)}$

$\rightarrow\:\sf{BC\:=\:50(1.732\:+\:1)}$

$\rightarrow\:\sf{BC\:=\:50(2.732)}$

$\rightarrow\:\sf{BC\:=\:136.60\:m}$

As, AB = BC

So,

$\rightarrow\:\sf{AB\:=\:136.60\:m}$

Answer 2

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

• shadow of tower increase 100m longer , when angle of elevation changes from 45° to 30° .

$\Large\underline\mathfrak{Question}$

• Find the Height of Tower ...

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$\Large\bold\star\underline{\underline\textbf{Solution}}$

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

From image we can see that :---

• AB = h = Height of tower .
• BC = let x m
• CD = 100m
• Angle ACB = 45°
• Angle ADC = 30°

Formula used :---

• Tan@ = Perpendicular/Base
• Tan45° = 1
• Tan30° = 1/√3

In Rt ∆ ABC , we have

$\tan(45) = \frac{AB}{BC} = \frac{h}{x} \\ \\ \red \leadsto \: 1 = \frac{h}{x} \\ \\ \leadsto \: \pink{\large\boxed{\bold{h = x}}}$

Now, in Rt ∆ ABD , we have ,,

$tan(30) = \frac{AB}{BD} = \frac{h}{(100 + x)} \\ \\ \red \leadsto \: \frac{1}{ \sqrt{3} } = \frac{h}{(100 + x)} \\ \\ \red \leadsto \: \red{\large\boxed{\bold{h \sqrt{3} = x + 100}}}$

Now, putting value of x in This Equation we get,,,,

$h \sqrt{3} = h + 100 \\ \\ \red\leadsto \: h( \sqrt{3} - 1) = 100 \\ \\\red\leadsto \: h = \frac{100}{( \sqrt{3} - 1) } \\ \\ \textbf{Rationalize Denominator} \\ \\ \red\leadsto h = \frac{100( \sqrt{3} + 1)}{ (\sqrt{3}) ^{2} - 1 } \\ \\ \red\leadsto h = \frac{100( \sqrt{3} + 1) }{2} \\ \\ \large\boxed{\bold{h = 50( \sqrt{3} + 1) }}$

Hence , Height of Tower will be 50(√3+1) m....

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