A pole of length L leans against a vertical wall so that it makes an angle of 60o with the horizontal ground. If the top of the pole is 8m above the ground, calculate L.
A.
\(16\sqrt{3}m\)
B.
\(4\sqrt{3}m\)
C.
\(\frac{\sqrt{3}}{16}\)
D.
\(\frac{16\sqrt{3}}{3}\)
Correct answer is D
\(sin 60 = \frac{8}{L} = 8 \div \frac{\sqrt{3}}{2}; L = \frac{8}{sin 60}=\frac{16}{\sqrt{3}}\)
When we rationalize gives \(\frac{16\sqrt{3}}{3}\)