Question

The height of a tower is 60 m. Watchingfrom a point on the ground, the angle ofelevation of the top of the tower is found
to be 60°. Find the distance of that pointfrom the base of the tower.​

Answer 1

Answer:

hiii

Step-by-step explanation:

Given ....

H = 60m

Angle of elevation = 60degree.

we know....

Tan thita = P/B.

NOE SOLVE.....

\begin{gathered} \tan(60) = \: \frac{60}{b} = \frac{p}{b} \\ \sqrt{3} = \: \frac{60}{b} \\ \sqrt{3}b = 60 \\ b = \frac{60}{ \sqrt{3} } \\ \\ b = \frac{60 \times \sqrt{3} }{ \sqrt{3 \times \sqrt{3} } } \\ b = \frac{60 \sqrt{3} }{3} \\ b = 30 \sqrt{3} \end{gathered}

tan(60)=

b

60

=

b

p

3

=

b

60

3

b=60

b=

3

60

b=

3×

3

60×

3

b=

3

60

3

b=30

3

I hope this helps you out.....