Find the sides and angles of triangle ABC if angle B = 45, angle C = 60, BC = √3.

1. Since the sum of the angles of the triangle is 180 °, then we find the angle A:
180 ° – 45 ° – 60 ° = 75 °;
2. By the sine theorem, we find the side AB:
BC / sinA = AB / sinC;
AB = BC * sinC / sinA;
AB = √3 * sin60 / sin75;
AB = √3 * √3 / 2 / (√6 + √2) / 4 = 3 (√6 + √2) / 8;
3. Using the same theorem of sines, we find the side of the AC:
AB / sinC = AC / sinB;
AC = AB * sinB / sinC;
AC = 3 (√6 + √2) / 8 * √2 / 2 / √3 / 2 = 3√2 (√6 + √2) / 8√3 = √6 (√6 + √2) / 8 = (6 + 2√3) / 8 = (3 + √3) / 8.
Answer: angle A = 75 °, side AB = 3 (√6 + √2) / 8, AC = (3 + √3) / 8.