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.
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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.....
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