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