Can someone PLEASE help me answer these questions down below on the bottom. I am not sure how to answer these. I would appreciate if someone can help me from the data I have produced please.

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How well did the sum of the interior angles compare with the expected value of 180 degrees? If you had an error, what do you think is the source ofthis error? How well were you able to verify the Pythagorean theorem? Ifthere was a difference computed, what do you think is the source of the error? Compare the results from your verification ofthe law of sines, cosines and tangents [make a table ifpossible}. which law had the better results and speculate on why this might be the case? Solve the following problem using the trigonometric relationships described in this lab. Two very small spherical conductors are charged with equal charge and suspended by a non-conducting string [each of equal length L] attached at a common point. {See the included figure}. The conductors repel each other and are forced apart at the angle 19 given in the figure. If each spherical conductor has a mass m, calculate the charge on the spherical conductors. Use the following values for your calculation L = 1 meter, and 19=30 deg, and m = 10 gm [.01 kg}. Here are the steps to solve this problem. a. First use the Law of cosines to calculate the distance 'r' between the spherical conductors {assume the radius ofthe sphere is very small and can be neglected assume a point charge]. (Please show your work.',| b. Use this distance 'r' in Coulomb's Equation for the force. [Please show your work.) c. Write an equation for the horizontal and vertical components ofthe tension [T] in the string. {Please show your work.) d. Set the vertical component of the tension equal to the gravitational force and solve for T. [Please show your work.) e. Set the horizontal component ofthe tension equal to the Coulombic repulsion . i Charged / Spherical Conductors [step b} and solve for the charge. (Please show your work.) Table 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 2.13 0.85 2.0 60 41 88 For at least the first row above, show all of your work in typed format. Data Analysis Procedure: You may use a spreadsheet to expedite the analysis of the data. The % percent error = (difference/average) * 100. BUT show your full typed work for at minimum the first row. 1. Compute the Interior Angles and Error From the Data collected above, fill out the following Table. Triangle A+B+C Accuracy % Error 181 0.5 189 5 Calculate the % error = ((A+B+C) -180) *100/180 (The theoretical sum should be 180 degrees). error = (181)-180 *100/180 = 0.5 % % error = (189)-180 *100/180 = 5.0% 2. Pythagorean Error Use the data from Triangle 2 to complete the entries to this table. Triangle a-+ b2 Error Difference 5.26 4.00 27.2 The % percent error = (difference/average) * 100 Difference = | c2 - a2 - b2| = | (2.0m)2 - (2.13m)2 - (0.85m)2 | = 1.26 m2 Average = (c2 + a2 + b2)/2 = 9.26 / 2 = 4.63 m2 The % percent error = (1.26 m2/ 4.63 m2) * 100 = 27.2% 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 2.12 2 (0.22 , 0.42) 2 2.46 1.30 2.00 1.9 (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) s sin(98) 0.52 0.51 _0.47 0.42 = 0.39 = 0.22 1.25 1.32 2.10 1.25 1.32 2.10The 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) (atb-c) % Difference 2ab Error -0.1391 -0.335 -82.658% 0.03 0.34 163.54% 48 For at least the first row above, show all of your work in typed format. difference * 100 %difference error = average Difference = cos(C) - + for triangle 1 diff = |-0.1391 - (-0.335)| = 0.1959 for triangle21 diff = 10.0348 - (0.347)| = 0.3122 Average = [cos(C) + + +21/2 for triangle 1 avg= (-0.1391 -0.335)/2 = -0.237 for triangle 2 avg = (0.0348 + 0.347)/2 = 01909 % difference error: difference # 100 %difference error = average for triangle 1 = ((0.1959)/(-0.237))*100 = -82.658% for triangle 2 = ((0.3122)/(0.1909)) *100 = 163.54% The % percent error = (difference/average) * 100. Difference = |cos(C) - [a-+bi-c?) 2ab Average = (cos(C)+) [a-+b--cz). 2 ab -)/2 In principle, one could apply the law of Cosines to calculate cos(A) and cos(B). 5. Verify the Law of Tangents Use the Data collected in Table 0 above to complete the Table below Triangle (a - b) Tan(- - B Difference ( atb ) Error Tan( -0.027 -0.027 -0.858 0.429 0.429 0.105 For at least the first row above, show all of your work in typed format. Triangle 1: Difference = tan -0.07 2 -0.027 - - 2.32790597 x 10-4 tan 2.57 2 Average = tan -0.07 2 -0.027 + tan 2.57 = -0.0271163953 2 Percent Error 2.32790597 x 10-4 - 100 = -0.858486515 -0.0271163953 Triangle 2: Difference = tan 1.28 2 0.429 tan 2.98 = 4.51230463 x 10-4 2 Average = tan 1.28 2 0.429 + tan 2.98 - 0.4292256152 2