Question

In a right angled triangle, one acute angle is double the other. Prove that the hypotenuse is double the smallest side.

Answer 1

Angle= A
double angle= 2A
so A+2A+90=180 
3A= 180-90
3A=90
A=90/3
A=30

as, cos theta= x/H

cos (30)= x/H
1/2=x/H
or H=2x 

Answer 2

Solution:-
Let there be a right angled triangle ABC and angle B is the right angle.
Let ∠ BAC = x
Then, ∠ ACB = 2x
Now, in triangle ABC,
x+2x+90 = 180°
3x = 90°
x = 30°
So, ∠ BAC = 30° and ∠ ACB = 2x = 60° 
Side opposite to the smallest angle is the smallest side,
Hence, side BC is the smallest side.
Now, cos 2x = BC/AC
⇒ cos 60° = BC/AC
⇒ 1/2 = BC/AC
⇒ AC = 2BC
Hence proved