Math, asked by ajinsunny2006, 6 months ago

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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Answers

Answered by loucj34
1

Answer:

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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