This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
ScanSolve AI Answer
Hey I — ready when you are. Here is the solution to question 47, continuing from where we left off. 47. Town B is 150 km on a bearing of 050^ from town A. Another town C is on a bearing of 110^ from town A and on a bearing of 150^ from town B. A fourth town D is 240 km on a bearing of 320^ from town A. Without using a scale drawing, calculate to the nearest kilometre: From the previous calculations: AB = 150 km AD = 240 km BAC = 60^ ABC = 100^ BCA = 20^ CAD = 150^ AC ≈ 431.790 km a) The distance AC We found this in the previous step: AC ≈ 432 km The distance AC is 432 km. b) The distance CD Step 1: Use the Cosine Rule in ACD. We have AD = 240 km, AC ≈ 431.790 km, and CAD = 150^. CD^2 = AD^2 + AC^2 - 2(AD)(AC) ( CAD) CD^2 = (240)^2 + (431.790)^2 - 2(240)(431.790) (150^) Step 2: Calculate the values. CD^2 = 57600 + 186442.664 - 207259.2 × (-0.8660254) CD^2 = 244042.664 + 179505.04 CD^2 = 423547.704 Step 3: Find CD. CD = sqrt(423547.704) CD ≈ 650.805 km Rounding to the nearest kilometre: CD ≈ 651 km The distance CD is 651 km. c) The distance BC Step 1: Use the Sine Rule in ABC. We have AB = 150 km, BAC = 60^, and BCA = 20^. (BC)/(( BAC)) = (AB)/(( BCA)) (BC)/((60^)) = (150)/((20^)) Step 2: Solve for BC. BC = (150 × (60^))/((20^)) BC = (150 × 0.866025404)/(0.342020143) BC = (129.9038106)/(0.342020143) BC ≈ 379.814 km Rounding to the nearest kilometre: BC ≈ 380 km The distance BC is 380 km. Drop the next question.