Given: ∆ABC, AB = 12, AC = 17 Area ∆ABC = 65 Find: BC, m∠A, m∠B, m∠C

Answer:   BC = 10.89, ∠A = 39.59°, ∠B = 95.80°, ∠C = 44.61°   BC = 27.34, ∠A = 140.41°, ∠B = 23.35°, ∠C = 16.24°Step-by-step explanation:Using the area formula to find angle A, we get ...   Area = (1/2)bc·sin(A)   65 = (1/2)(12)(17)sin(A)   sin(A) = 65/102 . . . . divide by the coefficient of the sine term   A = arcsin(65/102) = 39.59°   or   180°-39.59° = 140.41°Then from the law of cosines*, ...   BC² = AB² +AC² -2·AB·AC·cos(A) = 12² +17² ±2·12·17·cos(39.59°)   BC² = 433 ± 314.426   BC = √118.574   or   √747.426   BC = 10.89 or 27.34___For A = 39.59° and BC = 10.89, the remaining angles are ...   sin(C)/AB = sin(A)/BC   C = arcsin(12·(65/102)/10.89) = arcsin(.7023) = 44.61°   B = 180° -39.59° -44.61° = 95.80°__For A = 140.41° and BC = 27.34, the remaining angles are ...   sin(C) = AB·sin(A)/BC = 12(65/102)/27.34   C = arcsin(0.2797) = 16.24°   B = 180° -140.41° -16.24° = 23.35°__In summary, the solutions are ...   BC = 10.89, ∠A = 39.59°, ∠B = 95.80°, ∠C = 44.61°   BC = 27.34, ∠A = 140.41°, ∠B = 23.35°, ∠C = 16.24°_____* For a given value of 0 < (x=sin(α)) < 1, there are two possible positive angles: α = arcsin(x) and 180°-α. In the Law of Cosines formula, these different angles result in cos(α) and cos(180°-α) = -cos(α)._____The solution process is the same for the remaining sides and angles, once you recognize that the initial value of sin(A)=65/102 can have two different angles as its solution.