WAEC - Mathematics (1999 - No. 23)

A pole of length L leans against a vertical wall so that it makes an angle of 60^o with the horizontal ground. If the top of the pole is 8m above the ground, calculate L.

\(16\sqrt{3}m\)

\(4\sqrt{3}m\)

\(\frac{\sqrt{3}}{16}\)

\(\frac{16\sqrt{3}}{3}\)

Explanation

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

WAEC - Mathematics (1999 - No. 23)

Explanation

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