(b)
Consider the Δ ABC in which ∠B = 90°
Since, tan A = \(\frac{1}{\sqrt3}\)
∴ \(\frac{1}{\sqrt3}\) = \(\frac{BC}{AB}\)
⇒ \(\frac{AB}{\sqrt3}\) = \(\frac{BC}{1}\) = k
⇒ AB = √3k and BC = k
By Pythagoras’ Theorem,
AC2 = AB2 + BC2
= (√3 k)2 + k2 + 4k2
⇒ AC = 2k
Now for ∠A, Base = AB, Perp. = BC, Hyp.= AC