Question

If in a `triangle"ABC",` a²cos² A - b² - c² = 0, then ______.

Options

• `(A+B)/2`

• `pi/2<A<pi`

• `A=pi/2`

• `A<(1/2ac sinB)/(1/2(a+b+c))`

Answer 1

If in a `triangle"ABC",` a²cos² A - b² - c² = 0, then `underline(pi/2<A<pi)`.

Explanation:

a²cos² A - b² - c² = 0

`=>cos^2A=(b^2+c^2)/a^2`

Since cos²A ≤ 1 i.e., cos²A < 1

`therefore(b^2+c^2)/a^2< 1 =>b^2+c^2-a^2<0`

`(a+c-b)/(a+c-b)`   ....[∵ 2bc > 0]

∴ cosA < 0 ⇒ A ∈ `(pi/2,pi)`