Where do I start?
A triangle with sides k + 1 , k + 1 and k + 2 , where k is a positive integer, must be
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1 solution
How?please explain
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Pythagoras Theorem extended.
If Δ A B C is acute, then
a 2 + b 2 > c 2
If Δ A B C is right-angled, then
a 2 + b 2 = c 2
If Δ A B C is obtuse, then
a 2 + b 2 < c 2
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Got it ....Thnx
This could be proved by using cosine rule.
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@Anurag Pandey – Isn't it obvious from Pythagoras Theorem? Also, cosine rule is proven by Pythagoras Theorem.
Maybe it is not like the way you think..............
Since ( k + 1 ) 2 + k + 1 2 = k 2 + 3 k + 2 < ( k + 2 ) 2 , this triangle is obtuse.