Question

please answer that question fast..​

Answer 1

Answer:

Let AB be the tower then,

$\tan30 = \frac{AB}{BC } \\ \frac{1}{ \sqrt{3} } = \frac{x}{30} \\ 1 \times 30 = \sqrt{3} x \\ \frac{30}{ \sqrt{3} } = x \\ \frac{30 \sqrt{3} }{3} = x \\ x = 10 \sqrt{3} \: m \: or \: 17.32 \: m$

So, the tower is approximately 17.32m...

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