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 \(\frac{16\sqrt{3}}{3}\)
waec 1999mathematics

The correct answer is D. \(\frac{16\sqrt{3}}{3}\)

\(sin 60 = \frac{8}{L} = 8 \div \frac{\sqrt{3}}{2}; L = \frac{8}{sin 60}=\frac{16}{\sqrt{3}}\)

When we rationalize gives \(\frac{16\sqrt{3}}{3}\)

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