In a right angle triangle one acute angle is double the other. Prove that the hypotenuse is double the smallest side A ?
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Let ∠BAC=p
then ∠ACB=2p
Now in △ABC,
ABC+BCA+ACB=180∘
90∘ +2p+p=180∘
p=30∘ =∠ACB
∠ACB=2(30°)=60∘
Side opposite to the smallest angle is smallest
Hence, BC is smallest
Now
cos2p=
BC
AC
cos60∘ =
AC
BC
2
—
1 =
AC
BC
AC=2BC. Hence Proved the result.
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