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
Here is the solution to question 47. 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: First, let's determine the angles within the triangles formed by towns A, B, C, and D. Let N be the North direction from town A. Bearing of B from A is 050^. So, NAB = 50^. Bearing of C from A is 110^. So, NAC = 110^. Bearing of D from A is 320^. So, NAD = 320^. From these, we can find the angles at A: BAC = NAC - NAB = 110^ - 50^ = 60^ CAD = 360^ - ( NAD - NAC) = 360^ - (320^ - 110^) = 360^ - 210^ = 150^ Alternatively, CAD = NAD - NAC = 320^ - 110^ = 210^. The interior angle is 360^ - 210^ = 150^. Now, let's find the angle ABC. Draw a North line at B, parallel to the North line at A. The alternate interior angle to NAB is the angle between the North line at B and the line BA (pointing towards A). So, NBA = 50^. The bearing of C from B is 150^. This means the angle from the North line at B, clockwise to BC, is 150^. So, NBC = 150^. ABC = NBC - NBA = 150^ - 50^ = 100^ In ABC, we have: AB = 150 km BAC = 60^ ABC = 100^ The third angle BCA = 180^ - (60^ + 100^) = 180^ - 160^ = 20^. a) The distance AC Step 1: Use the Sine Rule in ABC to find AC. (AC)/(( ABC)) = (AB)/(( BCA)) (AC)/((100^)) = (150)/((20^)) Step 2: Solve for AC. AC = (150 × (100^))/((20^)) AC = (150 × 0.984807753)/(0.342020143) AC ≈ 431.790 km Rounding to the nearest kilometre: AC ≈ 432 km The distance AC is 432 km. b) The distance CD Step 1: Consider $