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JAMB - Mathematics (1979 - No. 42)

In the figure, \(\bigtriangleup\) ABC are in adjacent planes. AB = AC = 5cm, BC = 6cm and o then AE is equal to
3\(\sqrt{2}\)
2\(\sqrt{3}\)
\(\frac{\sqrt{3}}{2}\)
\(\frac{2}{\sqrt{3}}\)

설명

BC = 6 : DC = \(\frac{6}{2}\) = 3cm

By construction < EDE = 180o(90o + 60o) = 180o - 150o

= 30o(angle on a strt. line)

From rt < triangle ADC, AD2 = 52 - 32

= 25 - 9 = 6

AD = 4

From < AEC, let AS = x

\(\frac{x}{sin 60^o}\) - \(\frac{4}{sin 90^o}\)

sin 90o = 1

sin 60o = \(\frac{\sqrt{3}}{2}\)

x = 4sin 60o

x = 3 x \(\frac{\sqrt{3}}{2}\)

= 2\(\sqrt{3}\)

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