Question

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

Answer 1

Here you go...... .

Answer 2

Answer:

Answer:

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

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Step-by-step explanation:

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Step-by-step explanation: