Can someone please help me find the values of Cos(C) and difference error I have submitted the charts for it

3. Verify the Law of Sines: Use the data collected above to fill out the Table below Triangle a/sin(A) b/sin(B) c/sin(C) Average Range (min, max) 1 1.91 1.97 2.1 2 (0.22 , 0.42 2 2.46 1.30 2.00 1.92 (0.47, 0.70) Triangle 1: a / sin(A) = 1.25 / sin(41) = 1.91 b / sin(B) = 1.32 / sin(42) = 1.97 c / sin(C) = 2.10 / sin(98) = 2.12 Average: (1.91 + 1.97 + 2.12) / 3 =6 /3 =2 Triangle 1. Sin(41) _ sin(42) sin(98) 0.52 0.51 0.47 1.25 1.32 2.10 1.25 1.32 2.10 0.42 = 0.39 = 0.22 Triangle 2: a / sin(A) = 2.13 / sin(60) = 2.46 b / sin(B) = 0.85 / sin(41) = 1.30 c / sin(C) = 2.00 / sin(88) = 2.00 Average: (2.46 + 1.30 + 2.00) / 3 = 5.76 /3 = 1.92 Triangle 2: . sin (60) sin(41) sin (88) 0.87 0.66 0.99 2.13 0.85 2.00 1.25 1.32 2.10 0.70 = 0.50 = 0.47 The average in the table is the average of each of ratio computed in the first three columns. Instead of computing a % error, we are going to use another metric - the range to quantify the precision of our results. Range = [Minimum of Sine Law ratio, Maximum of Sine Law Ratio] 4. Verify the Law of Cosines Use the data collected to compute the table below. Triangle cos(C) (az + b2 - c2) % Difference 2 ab Error 1 -0.335 2 0.347 For at least the first row above, show all of your work in typed format. The % percent error = (difference/average) * 100. Difference = |cos(C) - (at+b2-c2) 2 ab Average = (cos(C)+) (a2+62-c2) 2 ab 2)/2Table 0: Triangle Data Collection Triangle a (m) b (m) c (m) Angle A Angle B Angle C 1 1.25 1.32 2.10 41 42 98 2 2.13 0.85 2.0 60 41 88 For at least the first row above, show all of your work in typed format