Let the ∆ABC be right angled at C.
Then, AB = c, BC = a, AC = b
Given, the sides of the right angled ∆ are in A.P. ⇒ a, b, c are in A.P.
Now, let a = x – d, b = x, c = x + d
(d being a +ve quantity as c being the hypotenuse is the greatest side)
∴ c2 = a2 + b2 (Pythagoras’ Theorem)
⇒ (x + d)2 = (x – d)2 + x2
⇒ x2 + 2xd + d2 = x2 – 2xd + d2 + x2
⇒ 4xd = x2 ⇒ . d = \(\frac{x}{4}\)
∴ a = x – d = x - \(\frac{x}{4}\) = \(\frac{3x}{4}\), b = x, c = x + d = x + \(\frac{x}{4}\) = \(\frac{5x}{4}\)
C being the right angle, A and B are the acute angles.
∴ Sin A = \(\frac{a}{c}\) = \(\frac{3x/4}{5x/4}\)= \(\frac{3}{5}\)
and Sin B = \(\frac{b}{c}\) = \(\frac{x/5}{x/4}\) = \(\frac{x}{\frac{5x}{4}}\) = \(\frac{4}{5}\)
∴ The sines of the acute angles are \(\frac{3}{5}\), \(\frac{4}{5}\)