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Question
In ΔАВС, ∠ACB < 90°. The correct relation is ______.
Options
AB2 = BC2 + AC2
BC2 < AC2 + AB2
AB2 < BC2 + AC2
AB2 > BC2 + AC2
MCQ
Fill in the Blanks
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Solution
In ΔАВС, ∠ACB < 90°. The correct relation is AB2 < BC2 + AC2.
Explanation:
By the cosine rule,
AB2 = BC2 + AC2 – 2·(BC)(AC)·cos(∠ACB).
If ∠ACB < 90°.
Then cos(∠ACB) > 0.
So, the last term is positive and AB2 is less than BC2 + AC2.
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