Question

in a right angle triangle if one acute angle is twice to the another angle then prove that the hypotenuse is twice to the smallest 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.

Now,