Question

In a right angled triangle
,one acute angle
Is double the other prove that the hypotenuse is double
the smallest side

Answer 1

Answer:

The answer is:

Step-by-step explanation:

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=  AC /BC

cos60∘  =  AC /BC

$\frac{1}{2} = \frac{BC}{AB}$

AC=2BC.

Hence Proved the result.

Please mark me as brainliest.