If in a triangle A B C , A B = 5 units, ∠ B = cos - 1 3 5 and radius of circumcircle of ∆ A B C is 5 units, then the area (in sq. units) of Δ A B C is:

Step-by-step solution

We have, As, cos B = 3 5 ⇒ sin B = 4 5 Given: R = 5 By sine rule, we have c sin C = 2 R ⇒ 5 10 = sin C ⇒ C = 30 ° Now, b sin B = 2 R ⇒ b = 4 5 × 10 = 8 Now, by cosine formula cos B = a 2 + c 2 - b 2 2 a c ⇒ 3 5 = a 2 + 25 - 64 2 5 a ⇒ a 2 - 6 a - 39 = 0 ⇒ a = 6 ± 192 2 = 6 ± 8 3 2 ⇒ a = 3 + 4 3 (Reject a = 3 - 4 3 < 0 ) Now, Δ = a b c 4 R = 3 + 4 3 · 8 · 5 4 · 5 = 2 3 + 4 3 ⇒ Δ = 6 + 8 3 sq. units.