Question

In ΔАВС, ∠ACB < 90°. The correct relation is ______.

पर्याय

• AB² = BC² + AC²

• BC² < AC² + AB²

• AB² < BC² + AC²

• AB² > BC² + AC²

Answer 1

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.