Question

please need the answer fast

Answer 1

Answer for 1st question is 12√3 mt

Answer 2

1) (Please refer the diagram attached first)

In the given triangle, △AGF,

$\sf \tan 60\° = \frac{AG}{FG}$

$\sf \rightarrow \sqrt{3}=\frac{28}{FG}$

$\sf \rightarrow FG = \frac{28}{\sqrt{3} } \times \frac{\sqrt{3} }{\sqrt{3} } =\frac{28}{3}\sqrt{3}$

In the triangle, △AGE,

$\sf \tan30\°=\frac{AG}{x+FG}$

$\sf \rightarrow \frac{1}{\sqrt{3} } =\frac{28}{x+\frac{28}{3}\sqrt{3} }$

$\sf \rightarrow 28\sqrt{3}=x+\frac{28}{3}\sqrt{3}$

$\sf \rightarrow x=\frac{28\sqrt{3}\times 2 }{3}$

$\sf \rightarrow x=\frac{56}{3} \sqrt{3}=18.67\sqrt{3}m$

∴ He walked for $\sf 18.67\sqrt{3}m$.

2) (Please refer the diagram attached first)

Let AB be the height of the tower.

Let BC be the height of the flag.

$\sf \tan 30\°=\frac{AB}{AP}$

$\sf AP=\sqrt{3}AB......(i)$

$\sf \tan60\°=\frac{AC}{AP}$

$\sf AP=\frac{1}{\sqrt{3} } (AB+5)......(ii)$

Now let's solve with (i) and (ii)

$\sf \frac{1}{\sqrt{3} } (AB+5)=\sqrt{3}AB$

$\sf 3AB=AB+5$

$\sf 2AB=5$

$\sf AB=2.5m$

∴The height of the tower would be $\sf 2.5 m$.