In a right angled triangle if one acute angle is doubled prove that the hypotenuse is equal to the double of the smallest side.
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Solution:-
Let ∠ BAC = θ
Then ∠ ACB = 2θ
Now, in Δ ABC,
∠ 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 side of the triangle.
Now,
cos 2θ = BC/AC. ⇒ cos 60° = BC/AC ⇒ 1/2 = BC/AC
Hence AC = 2BC
Proved.
Let ∠ BAC = θ
Then ∠ ACB = 2θ
Now, in Δ ABC,
∠ 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 side of the triangle.
Now,
cos 2θ = BC/AC. ⇒ cos 60° = BC/AC ⇒ 1/2 = BC/AC
Hence AC = 2BC
Proved.
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