Question

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


give suitable explanation too thanks!

Answer 1

Answer:

Let /BAC = p

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then ZACB = 2p

Now in ΔΑΒC,

ABC + BCA + ACB = 180°

90° +2p+ p = 180°

p= 30° ZACB

ZACB = 2(30°) = 60°Side opposite to the smallest angle is smallest

Hence, BC is smallest

Now

BC

cos 2p - AC

BC

AC

cos 60°

=

1 2 BC AC

AC =

2BC. Hence Proved the result.

Answer 2

i am your friend dvenkat27

answer: Use the fact that sin30∘=12 and hence prove that hypotenuse is double the smallest side. BC = 2AB. Hence the hypotenuse is double the smallest side. Hence proved.

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