Question

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

Answer 2

Answer:

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