Question

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

Answer 1

Let BAC = Ф
then AcB = 2 Ф

Now ABC,
ABC + BCA + ACB = 180 
90 + Ф + 2 = 180
Ф = 30 
ACB = 2 ( 30 ) = 60 
Side opposite to the smallest angle is smallest
Hence BC is smallest 
Now 
Cos 2Ф = BC / Ac , Cos 60 = BC / AC , 1 / 2 = BC / AC , AC = 2BC

Answer 2

Consider a ∆ABC in which ∠B is a right angle and ∠C is double of ∠A

To Prove: AC = 2BC

Construction: produce CB upto D such that BC = BD & join AD

Proof:

In ∆ ABD & ∆ ABC

BD= BC. (By construction)

∠ABD = ∠ABC ( 90°)

AB = AB (Common)

∆ABD ∆ABC. ( BY SAS)

AD= AC. (CPCT)........(1)

∠DAB = ∠CAB = P ( Let) (By CPCT).......(2)

∠DAC = ∠DAB + ∠CAB

∠DAC= P + P = 2P. ( From eq 2)

∠DAC = ∠ACB ( ∠ACB= 2P)

DC = AD

(Since, sides opposite to equal angles of a triangle are equal)

2BC = AD= AC ( BC= DB from eq 1)

Hence, AC= 2BC

==================================================================

Hope this will help you...