How do you solve the triangle given #B=15^circ30', a=4.5, b=6.8#? Trigonometry Triangles and Vectors The Law of Sines 1 Answer sankarankalyanam Mar 28, 2018 #color(blue)("Area of Triangle " A_t = 6.63" sq units"# Explanation: #hat B = 15.5^@, a = 4.5, b = 6.8# Applying the law of sines, #sin A = (sin B * a) / b = (sin (15.5) * 4.5) / 6.8 = 0.1768# #hat A = sin ^_1 0.1768 = 10.18^@# #hat C = 180 - hat A - hat B = 180 - 10.18 - 15.5 = 154.32^@# #"Area of Triangle " A_t = (1/2) a b sin C = (1/2) * 4.5 * 6.8 * sin 154.32 = 6.63" sq units"# Answer link Related questions What is the Law of Sines? Does the law of sines apply to all triangles? What is the ambiguous case of the law of sines? When can the law of sines be used? How do you find #angleB# if in triangle ABC, #a = 15#, #b = 20# , and #angle A=30^@#? How do you prove #\frac{a-c}{c} = \frac{\sin A - \sin C}{\sin C}# using the law of sines? How do you find all possible measures of B if #A = 30^\circ#, #a = 13#, #b = 15# for triangle ABC? How do you use Law of sines, given A=102, b=13, c=10? In triangle ABC, sin A=1/3, sin B=1/5, and b=6, how do you find side a? The length of the base of an isosceles triangle is 30 m. The angle opposite the base measures 32... See all questions in The Law of Sines Impact of this question 2633 views around the world You can reuse this answer Creative Commons License