Question

From the top of a 7meters high building ,the angle of elevation of the top of a cable tower 60° and the angle of despression of its foot is 45°.​

Answer 1

Answer:

In ΔABE,

tan60

o

=

AB

h

h=AB

3

....(i)

In ΔABC,

tan45

o

=

AB

7

AB=7 ...(ii)

By equation (i) and equation (ii)

h=7

3

So, height of tower is =h+7=7(

3

+1)

Answer 2

$\huge\boxed{\texttt{\fcolorbox{black}{red}{Answer}}}$

let EB be - $\sf\green{h}$

In ∆ADC - $\sf\green{∠A = ∠C= 45°}$

so, in ∆ADC ,tan45° = AD/DC

$\sf\green{1 = 7m/DC}$

$\sf\green{ DC = 7m}$

[ In figure DC = AB = 7m ]

Now, in ∆ABE

AB = 7m

so,

in ∆ABE tan60° = $\sf\green{EB/AB}$

$\sf\green{ √3 = EB/7}$

$\sf\green{ EB = 7√3m}$

$\mathtt{\sf{\red{\fbox{\underline{Hence-}}}}}$

hight of tower is $\sf\green{BC+AB}$

$\sf\green{ 7+7√3 = 7( 1 + √3 )m.}$