Question

answer to this question.

Answer 1

HELLO DEAR,

in fig. AB is the tower and CD =2k

im ∆ABC

AB/BC= tan45°

AB/BC=1

=>AB=BC---------(1)

IN ∆ ABD

AB/BD=tan30°

=>AB/BD=1/√3

=>√3 AB=BD---------(2)


.
given that:-

CD=2K

=> BD-BC =2K

from--(1) and----(2)

we get,

=> √3 AB-AB=2k

=>AB(√3-1)=2K

.

$ab = \frac{2k}{( \sqrt{3} - 1)} \\ = > \frac{2k( \sqrt{3} + 1) }{(1 + \sqrt{3} )( \sqrt{3} - 1) } \\ = > \frac{2k( 1 + \sqrt{3} ) }{(3 - 1)} \\ = > ab = \frac{2k(1 + \sqrt{3}) }{2} \\ = > ab = k(1 + \sqrt{3} )$

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Answer:

in fig. AB is the tower and CD =2k

im ∆ABC

AB/BC= tan45°

AB/BC=1

=>AB=BC---------(1)

IN ∆ ABD

AB/BD=tan30°

=>AB/BD=1/√3

=>√3 AB=BD---------(2)

.

given that:-

CD=2K

=> BD-BC =2K

from--(1) and----(2)

we get,

=> √3 AB-AB=2k

=>AB(√3-1)=2K

.

ab=2k(3−1)=>2k(3+1)(1+3)(3−1)=>2k(1+3)(3−1)=>ab=2k(1+3)2=>ab=k(1+3)\begin{lgathered}ab = \frac{2k}{( \sqrt{3} - 1)} \\ = > \frac{2k( \sqrt{3} + 1) }{(1 + \sqrt{3} )( \sqrt{3} - 1) } \\ = > \frac{2k( 1 + \sqrt{3} ) }{(3 - 1)} \\ = > ab = \frac{2k(1 + \sqrt{3}) }{2} \\ = > ab = k(1 + \sqrt{3} )\end{lgathered}

ab=

(

3

−1)

2k

=>

(1+

3

)(

3

−1)

2k(

3

+1)

=>

(3−1)

2k(1+

3

)

=>ab=

2

2k(1+

3

)

=>ab=k(1+

3

)

hope it helps u