Question on: WAEC Mathematics - 1999
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
\(\frac{16\sqrt{3}}{3}\)
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Correct Option: 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})
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