Problem 3 - Entrance Test

Since the sum of the angles in a triangle is 180°, we can use the Law of Sines to relate the lengths of the sides of the triangle to the sines of the angles. The Law of Sines states that for any triangle with sides a, b, and c, and opposite angles A, B, and C, respectively, we have a/sin(A) = b/sin(B) = c/sin(C). In this case, we have ∠A = 60°, ∠B = 80°, and ∠C = 40°, so we can write AB/sin(40°) = AC/sin(80°). Therefore, the ratio of the length of side AB to the length of side AC is AB/AC = sin(80°)/sin(40°). Using the double-angle formula for sine, which is sin(2x) = 2sin(x)cos(x), we can rewrite sin(80°) as sin(2*40°) = 2sin(40°)cos(40°), so AB/AC = 2sin(40°)cos(40°)/sin(40°) = 2cos(40°). Using the fact that cos(40°) = cos(60° - 20°) and the angle subtraction formula for cosine, which is cos(a - b) = cos(a)cos(b) + sin(a)sin(b), we can rewrite cos(40°) as cos(60° - 20°) = cos(60°)cos(20°) + sin(60°)sin(20°) = (1/2)cos(20°) + (√3/2)sin(20°). However, to solve the problem we use the law of sines: AB/AC = sin(40°)/sin(80°) = sin(40°)/sin(2*40°) = sin(40°)/(2*sin(40°)*cos(40°)) = 1/(2cos(40°)). Since cos(40°) = cos(60° - 20°) we need to calculate this value or find an alternative approach to get the value of the ratio.