Question

From the top of a building 60m high, the angle of elevation and depression of the top

and the foot of another building are α and β respectively. Find the height of the second

building.

a) 60(1+ tan α tanβ)

b) 60(1+ cot α tanβ)

c) 60(1+ tan α cotβ)

d) 60(1- tan αcotβ)​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that AB be the building which is 60 m high.

Let assume that height of the other building CD is 'h' m.

Now, it is given that,

The angle of elevation and depression of the top D and the foot C of another building are α and β respectively from the top B of 60 m high building.

Let BE the perpendicular on CD.

So,

AB = CE = 60 m

DE = CD - CE = h - 60 m

Let assume that BE = AC = x m.

Now,

$\red{\bf :\longmapsto\:In \: \triangle \: BEC}$

$\rm :\longmapsto\:tan \beta = \dfrac{EC}{BE}$

$\rm :\longmapsto\:tan \beta = \dfrac{60}{x}$

$\rm :\longmapsto\: \: x = \dfrac{60}{tan \beta }$

$\bf\implies \:x = 60 \: cot \beta - - - (1)$

Now,

$\red{\bf :\longmapsto\:In \: \triangle \: BED}$

$\rm :\longmapsto\:tan \alpha = \dfrac{ED}{BE}$

$\rm :\longmapsto\:tan \alpha = \dfrac{h - 60}{x}$

$\rm :\longmapsto\:tan \alpha = \dfrac{ h - 60}{60 \: cot \beta }$

$\rm :\longmapsto\:60 \: tan \alpha \: cot \beta = h - 60$

$\rm :\longmapsto\:h \: = \: 60 + 60 \: tan \alpha \: cot \beta$

$\bf :\longmapsto\:h \: = \: 60(1 \: + \: tan \alpha \: cot \beta )$

So, Height of second building is

$\boxed{ \bf{ \: \bf :\longmapsto\:h \: = \: 60(1 \: + \: tan \alpha \: cot \beta ) \: \: \: \: \: \: }}$

Answer 2

Answer:

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Step-by-step explanation:

Answer is the image... there are 2 ways to express H. According to question, we can use the second way ✌️