In the figure, \(\bigtriangleup\) ABC are in adjacent planes. AB = AC = 5cm, BC…

Explanation

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\(\sqrt{3}\)