Question

There exists a triangle ABC satisfying the conditions

(A) bsinA = a, A  < π/2

(B) bsinA > a, A > π/2

(C) bsinA > a, A < π/2 

(D) bsinA < a, A < π/2, b < a

Answer 1

Correct option (A, D)

Explanation:

The sine formula is

a/sinA = b/sinB

⇒ a sin B = b sin A

(a) b sin A = a ⇒ a sin B = a ⇒ B = π/2

Since, ∠A  < π/2 therefore, the triangle is possible

(b) and (c) : b sin A > a ⇒ a sin B > a ⇒ sin B > 1

∴ ∆ ABC is not possible

(d): b sin A < a ⇒ a sin B < a

⇒ sinB < 1 ⇒ ∠B exists

Now, b > a ⇒ B > A since A < π/2

∴ The triangle is possible.