Question

$\Large\frak{\underline{\red{Question:}}}$
(In the picture attached)
________________
$\Large\frak{\underline{\red{Information\:'bout\:Question:}}}$
Chapter name - Heights & Distances
Class - 10th
Source - NCERT Exemplar/ R.D Sharma book{CBSE}
____________________

Diagram needed!
It would be okay if you don't skip any step while solving.

All the best :)​

Answer 1

Answer:

see the attachments

actually, this question appeared in my pre boards and I got full marks, that's why I am suggesting you this answer

Step-by-step explanation:

please mark it as brainliest and follow

Answer 2

⠀⠀⠀⠀⠀⠀⠀⠀⠀Diagram in the attachment.

$\huge\bf\underline\mathfrak\red{Given :}$

• $\text{Height of the vertical flag-staff,AD = h.}$
• $\text{Angles of elevation of bottom and top = α and β.}$

$\huge\bf\underline\mathfrak\red{To \: Prove :}$

• $\text{Height of the tower =}$$\sf \dfrac{h \: tan \: \alpha }{h \: tan \: \beta }$

$\huge\bf\underline\mathfrak\red{Answer :}$

It is given that, height of the flag staff, AD = h.

Let height of the tower, AB be denoted by (H).

• It is also provided that, Angle of elevation of top and bottom of the flag-staff of the plane is ∠DCB = β and ∠ACB = α, respectively.

⠀⠀⠀⠀⠀⠀⠀⠀⠀Let CB be x

⠀⠀ $\underline\text{Now In ∆ACB,}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀$\sf tan \: \alpha = \dfrac{AB}{CB \:} = \dfrac{H}{x}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀[ $\text{Since, tanΘ}$ = $\sf \dfrac{opposite \: side \: }{base}$ ]

⠀⠀⠀⠀⠀⠀⠀⠀⠀$\implies$ $\sf x \: = \: \dfrac{h}{tan \: \alpha }$ $\text{[Equation (i)]}$

⠀⠀$\underline\text{In ∆DCB,}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀$\text{tanβ =}$ $\sf \dfrac{DB}{CB} = \: \dfrac{DA \: + \: AB}{x} = \dfrac{h \: + \: H}{x}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀$\implies$ $\sf x = \dfrac{h \: + \: H}{tan \: \beta }$ $\text{[Equation (ii)]}$

⠀ $\underline\text{From Equation 1 and 2,}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀$\sf \dfrac{H}{tan \: \alpha } = \dfrac{H \: + \: h}{tan \: \beta }$

⠀$\underline\text{On cross multiplying :-}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀$\sf H \: tan \: \beta = (H \: + h)tan \: \alpha$

⠀⠀⠀⠀⠀⠀⠀⠀⠀$\sf H \: tan \: \beta = \: H \: tan \: \alpha \: + h \: tan \: \alpha$

⠀⠀⠀⠀⠀⠀⠀⠀⠀$\sf H \: tan \: \beta \: - H \: tan \: \alpha = h \: tan \: \alpha$

⠀$\underline\text{Taking H as common :-}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀$\sf H \: ( \: tan \: \beta \: - tan \: \alpha \: ) \: = \: h \: tan \: \alpha$

⠀$\underline\text{On re-arranging :-}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀$\sf\purple{H \: = \: \dfrac{h \: tan \: \alpha }{tan \: \beta \: - \: tan \: \alpha }}$

$\bf\underline\mathfrak{Hence, \: Proved }$

⠀⠀⠀⠀⠀⠀⠀⠀⠀