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$Tan \ 30^o = \frac{1}{\sqrt 3} = \frac{x}{d}$
$= x \sqrt 3$
$Tan \ 60^o = \frac{x + 10}{d}$
$\sqrt 3 = \frac{x + 10}{d}$
$d = \frac{x+ 10}{\sqrt 3}$
$x \sqrt 3 = \frac{x + 10}{\sqrt 3}$
3x = x + 10
2x = 10
x = 5
$\therefore AB,$ the height of building = 10 + 5 = 15 metres
raphael answered the question on January 20, 2017 at 06:18
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