Question

If the sides of a right-angled triangle form an A.P., then find the sines of the acute angles.

Answer 1

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 \(\frac{3}{5}\), \(\frac{4}{5}\)