Question

ans this! as soon as

Answer 1

tan30°=AB/BC
tan30°=1/√3
1/√3=AB/48
AB = 16√3

Answer 2

${\boxed {\boxed {\underline{ \sf{Here's \: the \: answer}}}}}$

In the given figure ,

AB represents the height of building

C represents the position of boy at a distance of 48 m from the building

Angle ACB is the angle of elevation.

$\sf{ \therefore \angle \: ACB = 30 \degree}$

Now,

In ∆ ABC ,
Angle ABC = 90°

$\sf{ \therefore tan \angle \: ACB = \frac{AB}{BC}}$

$\sf{\therefore \: tan \: 30 \degree = \frac{AB}{48}}$

$\sf {\frac{1}{ \sqrt{3} } = \frac{AB}{48} }$

$\sf{AB = \frac{48}{ \sqrt{3}} }$

$\sf{AB = \frac{48 \times \sqrt{3} }{ \sqrt{3} \times \sqrt{3} }}$

$\sf{AB = \frac{48 \sqrt{3} }{ 3}}$

$\sf{AB = 16 \sqrt{3}}$

We know that,

√3 = 1.73 ( Approx. )

$\sf{ \therefore \: AB = 16 \times 1.73}$

$\sf{AB = 27.68 \: m}$

$\sf { \underline{Final \: answer =>}}$

Height of the building is 27.68 m.

______________________________

Thanks!!