First, use the law of cosines to find the length of BC. BC² = AB²+AC²-2AB·AC·cosA ⇒ BC = √(BC²) = √(AB²+AC²-2AB·AC·cosA) ··· Eq.1
Then, use the law of sines to find the angle of B. BC/sinA = AC/sinB ⇒ sinB = (AC·sinA)/BC ⇒ ∠B = arcsin{(AC·sinA)/BC)} = arcsin{(AC·sinA)/√(AB²+AC²-2AB·AC·cosA)} ··· Eq.2
Plug ∠A = 31°, AB = 35 and AC = 36 into both Eq.1 and Eq.2. BC = √(35²+36²-2x35x36xcos31°) = 18.99837894…, ∠B = arcsin{(36xsin31°)/√(35²+36²-2x35x36xcos31°) = 77.40738802…
Therefore, BC ≈ 19.00 and ∠B ≈ 77.41°