Question

in a right angle triangle one acute angle is double of the other prove that the hypotenuse is double of the small side

Answer 1

Let ∠BAC = θ

Then, ∠ACB = 2θ

Now, In ΔABC,

∠ABC + ∠BAC + ∠ACB = 180°

⇒ 90° + θ + 2θ = 180°

⇒ θ = 30°

∴ ∠ACB = 2 (30°) = 60°

Side opposite to the smallest angle is smallest.

Hence, side BC is the smallest.