Question

A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height 'h'as shown in the figure. At a point on a plane, the angle of elevation of bottom of the flagstaff is $\alpha$ and that of the top of flagstaff is $\beta$. Prove that the height of the tower is $\frac{h.tan\alpha}{tan\beta-tan\alpha}$.

Answer 1

$\bf{ \underline { \red{step \: 1 - \:}take \: out \: tan \: \alpha }}$


$\bf{\tan( \alpha ) = \frac{bc}{ab} }$


$\bf{ \tan( \alpha ) = \frac{x}{ab} }$


$\bf{ab = \frac{x}{tan \: \alpha } - - (1) }$

$\bf{ \underline{ \green{step \: 2 - }take \: out \: tan \: \beta }}$


$\bf{ \tan( \beta ) = \frac{db}{ab} }$


$\bf{ \tan( \beta ) = \frac{h + x}{ab} }$


$\bf{tan \: \beta = \frac{h + x}{ \frac{x}{tan \: \alpha } } \: \: (from \: equation \: 1) }$


$\bf{tan \: \beta = \frac{(h + x)tan \: \alpha }{x} }$




$\bf{ \: x \: tan \beta = h \: tan \: \alpha + x \: \tan( \alpha ) }$



$\bf{x \: tan \beta - x \: tan \: \alpha = h \: tan \alpha }$

$\bf{ \underline{ \blue{step \: 3 - }take \: x \: common}}$


$\bf{x( tan \: \beta - tan \alpha) = h \: tan \: \alpha }$


$\bf{ \underline{ \pink{step \: 4 - }keep \: x \: on \: a \: side }}$


$\bf{x = \frac{h \: tan \: \alpha }{tan \: \beta - tan \: \alpha } }$


Since, x= height of tower, so

$\bf{height \: of \: tower \: = \frac{h \: tan \: \alpha }{tan \: \beta - tan \: \alpha } }$


Hence Proved

Answer 2

$\sf{\bigstar\;\;We\;know\;that : \boxed{\sf{Tangent\;of\;Triangle = \dfrac{Opposite\;Side}{Adjacent\;Side}}}}$

$\textsf{Consider Triangle ACB in the given Figure :}$

✿  $\textsf{In Triangle ACB : The Opposite Side is BC}$

✿  $\textsf{In Triangle ACB : The Adjacent Side is AB}$

$\implies \sf{Tangent\;of\;Triangle\;ACB = \dfrac{BC}{AB}}$

$\textsf{Given : Angle of Elevation of bottom of the Flagstaff\;(Top of Tower) = \alpha }$

$\textsf{Given : The Height of the Tower (BC) = x}$

$\sf{\implies Tan\alpha = \dfrac{x}{AB}}$

$\sf{\implies AB = \dfrac{x}{Tan\alpha}}$

$\textsf{Now, Consider Triangle ADB in the given Figure :}$

✿  $\textsf{In Triangle ADB : The Opposite Side is DB}$

✿  $\textsf{In Triangle ADB : The Adjacent Side is AB}$

$\implies \sf{Tangent\;of\;Triangle\;ADB = \dfrac{DB}{AB}}$

$\textsf{Given : Angle of Elevation of Top of the Flagstaff = \beta }$

$\textsf{From the Figure, We can Notice that : DB = DC + CB}$

$\textsf{Given : The Height of the Flagstaff (DC) = h}$

$\sf{\implies DB = h + x}$

$\sf{\implies Tan\beta = \dfrac{h + x}{AB}}$

$\sf{\implies AB = \dfrac{h + x}{Tan\beta}}$

$\textsf{As, In both Triangles ADB and ACB, Length of Side AB is Same and Equal : }$

$\implies \textsf{We can equate both the Lengths of Side AB}$

$\sf{\implies \dfrac{x}{Tan\alpha} = \dfrac{h + x}{Tan\beta}}$

$\sf{\implies \dfrac{Tan\beta}{Tan\alpha} = \dfrac{h + x}{x}}$

$\sf{\implies \dfrac{Tan\beta}{Tan\alpha} = \dfrac{h}{x} + 1}$

$\sf{\implies \dfrac{h}{x} = \bigg(\dfrac{Tan\beta}{Tan\alpha} - 1\bigg)}$

$\sf{\implies \dfrac{h}{x} = \bigg(\dfrac{Tan\beta - Tan\alpha}{Tan\alpha}\bigg)}$

$\sf{\implies \dfrac{x}{h} = \bigg(\dfrac{Tan\alpha}{Tan\beta - Tan\alpha}\bigg)}$

$\sf{\implies {x} = \bigg(\dfrac{h.Tan\alpha}{Tan\beta - Tan\alpha}\bigg)}$

$\sf{\implies Height\;of\;the\;Tower = \bigg(\dfrac{h.Tan\alpha}{Tan\beta - Tan\alpha}\bigg)}$