In aright angled triangle one acute angle is double the other. show that the hypotenuse is double the smallest side
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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,
cos2theta= BC/AC
cos60=BC/AC
1/2= BC/AC
AC=2 BC
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,
cos2theta= BC/AC
cos60=BC/AC
1/2= BC/AC
AC=2 BC
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