Question

Two stations due south of a leaning tower which leans towards the north are at distances a and b from its foot. If $\alpha$, $\beta$ be the elevations of the top of the tower from these stations,
Prove that its inclination $\theta$ to the horizontal is given by,
cot $\theta$ = (b cot $\alpha$ - a cot $\beta$)/b - a​

Answer 1

$\underline{\sf{According\:to\:the\:Question}}$

It is given that, two stations due south of a leaning tower which leans towards the north are at distances a and b from its foot. If $\alpha$, $\beta$ be the elevations of the top of the tower from these stations. We need to prove that,

• $\bf cot\:\theta = \dfrac{b\:cot\:\alpha - a\:cot\:\beta}{b - a}$

$\underline{\sf{Proof}}$

Let AB be the leaning tower and let C and D be two given stations at distances a and b respectively from the foot A of the tower.

Let AE = x and BE = h

In ∆ AEB we have,

• $\sf{\pink{tan\:\theta = \dfrac{BE}{AE}}}$

$\\ :\implies\:\sf tan\:\theta = \dfrac{h}{x}$

$\\ :\implies\:\sf x = h\:\dfrac{1}{tan\:\theta}$

$\\ :\implies\:\bf x = h\:cot\:\theta\qquad...(1)$

In ∆ CEB we have,

• $\sf{\blue{tan\:\alpha = \dfrac{BE}{CE}}}$

$\\ :\implies\:\sf tan\:\alpha = \dfrac{h}{a + x}$

$\\ :\implies\:\sf a + x = h\:\dfrac{1}{tan\:\alpha}$

$\\ :\implies\:\sf a + x = h\:cot\:\alpha$

$\\ :\implies\:\bf x = h\:cot\:\alpha - a\qquad...(2)$

In ∆ DEB we have,

• $\sf{\gray{tan\:\beta = \dfrac{BE}{DE}}}$

$\\ :\implies\:\sf tan\:\beta = \dfrac{h}{b + x}$

$\\ :\implies\:\sf b + x = h\:\dfrac{1}{tan\:\beta}$

$\\ :\implies\:\sf b + x = h\:cot\:\beta$

$\\ :\implies\:\bf x = h\:cot\:\beta - b\qquad...(3)$

On equating the values of x obtained from (1) and (2) we have,

• $\sf{\orange{h\:cot\:\theta = h\:cot\:\alpha - a}}$

$\\ :\implies\:\sf h\:(cot\:\alpha - cot\:\theta) = a$

$\\ :\implies\:\bf h = \dfrac{a}{cot\:\alpha - cot\:\theta}\qquad...(4)$

On equating the values of x obtained from (1) and (3) we have,

• $\sf{\purple{h\:cot\:\theta = h\:cot\:\beta - b}}$

$\\ :\implies\:\sf h\:(cot\:\beta - cot\:\theta) = b$

$\\ :\implies\:\bf h = \dfrac{b}{cot\:\beta - cot\:\theta}\qquad...(5)$

Equating the values of h from (4) and (5) we get,

• $\sf{\green{\dfrac{a}{cot\:\alpha - cot\:\theta} = \dfrac{b}{cot\:\beta - cot\:\theta}}}$

$\\ :\implies\:\sf a\:(cot\:\beta - a\:cot\:\theta) = b\:(cot\:\alpha - cot\:\theta)$

$\\ :\implies\:\sf (b - a)\:cot\:\theta = b\:cot\:\alpha - a\:cot\:\beta$

$\\ :\implies\:{\underline{\boxed{\bf{\red{cot\:\theta = \dfrac{b\:cot\:\alpha - a\:cot\:\beta}{b - a}}}}}}$

$\underline{\sf{Hence,\:Proved}}$

Answer 2

Given ✓

• Two stations due south of a leaning tower which leans towards the north are at distances a and b from its foot.

• If $\alpha$, $\beta$ be the elevations of the top of the tower from these stations,

To Proof :-

• Prove that its inclination $\theta$ to the horizontal is given by,

cot $\theta$ = (b cot $\alpha$ - a cot $\beta$)/b - a

Solution :-

• I solve this question and write in notebook refer the attachment

Thanks