Question

in a right angle triangle if one of the acute angle is doubke the other then prove that the hypotenuse is double the shorter side

Answer 1

in triangle ABC
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,

cos 2theta = BC/AC
cos60=BC/AC
1/2= BC/AC

AC=2 BC