Take a Test JARRET MILLER Mozilla Firefox mStudentPlayer Tes

Take a Test JARRET MILLER Mozilla Firefox m/Student/Player Test aspx?test 5062 yes https://www. mathxl.col UNO- Math 1116- College Trigonometry Stephen Andrus, Jr. Test: Test 4 (copy) This Question: 1 pt 5 of 10 (3 complete Solve the triangle ABC he triangle exists. A 41.5% a 8.9 m b 10.3 m select ne correct cnolce low ana nil in tne answer Doxes wnnin tne cnoce. O A. There are 2 possible solutions for the triangle. The measurements for the solution with the longer side c are as follows. mLB 54.3 mLC 84.7 The length of side C 12.6 Round to the nearest Round to the nearest Round to the nearest tenth as tenth as needed.) tenth as needed.) needed The measurements for the Solution with the shorter side are as follows. mLB 125.7 m.ZC 12.8 The length of side c 2.8 Round to the nearest Round to the nearest Round to the nearest tenth as tenth as needed.) tenth as needed.) needed O B. There is only 1 possible solution for the triangle. The measurements for the remaining angles B and C and side care as follows. mLB The length of side c m.ZC Round to the nearest Round to the nearest Round to the nearest tenth as tenth as needed.) tenth as needed.) needed O c. There are no possible solutions for this triangle. Click to select and enter your answer(s). Ask me anything JARRET MILLER 2/1/16 7:14 PM Submit Test This Test 10 pts possible 12/1/2016

Solution

A = 41.5

a = 8.9 m

b = 10.3 m

a^2 = b^2 + c^2 - 2bc*cosA

8.9^2 = 10.3^2 + c^2 - 2*10.3*c*cos41.5

c^2 - 15.43*c + 26.88 = 0

c = 13.4 m

Now c^2 = a^2 + b^2 - 2ab*cosC

13.4^2 = 8.9^2 + 10.3^2 - 2*8.9*10.3*cosC

C = 88.2 degree

B = 180 - 41.5 - 88.2 = 50.3 degree