Question

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

Answer 1

Answer:

Step-by-step explanation:

Let ABC is a right angle triangle, right angle at B as shown in the above figure.

Given, in a right-angled triangle, one acute angle is double the other.

Let ∠BAC = x, then ∠ACB = 2x

Now, in triangle ABC

=> ∠ABC + ∠BAC + ∠ACB = 180   {Since sum of all the angles in a trinagle is equal to 180 degree}

=> 90 + x + 2x = 180    {Since triangle is right angle at B}

=> 90 + 3x = 180

=> 3x = 180 - 90

=> 3x = 90

=> x = 90/3

=> x = 30

So, ∠BAC = x = 30 degree

and ∠ACB = 2x = 2 * 30 = 60 degree

Now, In triangle ABC,

     cos 2x = BC/AC

=> cos 60 = BC/AC

=> 1/2 = BC/AC

=> AC = 2*BC

Hence, if in a right angled triangle one acute angle is double the other then the hypotenuse is

double the smallest side.