Question

Help out me with this question please?

Answer 1

HEY Buddy.....!! here is ur answer..

Answer 2

In ∆ ABD,

$\tan(45) = \frac{300}{x} \\ 1 = \frac{300}{x} \\ = > x = 300m$
and,

In ∆ BCD,
$\tan(60) = \frac{300}{y} \\ \sqrt{3} = \frac{300}{y} \\ = > y = \frac{300}{ \sqrt{3} }$
Now

Width of the river = AC = x+y

so,
$AC\: = 300 + \frac{300}{ \sqrt{3} } \\AC = 300(1 + \frac{1}{ \sqrt{3} } ) \\ AC= 300(1 + \frac{1}{1.732} ) \\ AC= 300(1.58) \\AC = 474m$

Therefore, width of river = AC = 474m