$In the figure, $\bigtriangleup$ ABC are in adjacent planes. AB = AC = 5cm, BC = 6cm and o then AE is equal to$

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 $\sqrt{3}$

$\frac{\sqrt{3}}{2}$

$\frac{2}{\sqrt{3}}$

Correct answer is B

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

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

= 30^o(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 90^o = 1

sin 60^o = $\frac{\sqrt{3}}{2}$

x = 4sin 60^o

x = 3 x $\frac{\sqrt{3}}{2}$

= 2 $\sqrt{3}$

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