Determine the interior angle ABC.

Here's how to solve the problem: a) Calculate the length of AC: Step 1: Determine the interior angle ABC. The bearing of B from A is 053^. This means the angle from the North line at A to AB is 53^. The bearing of C from B is 161^. This means the angle from the North line at B to BC is 161^. Since the North lines at A and B are parallel, the alternate interior angle to 53^ (angle NAB) is the angle between the line BA and the North line at B, which is 53^. Therefore, ABC = 161^ - 53^ = 108^. Step 2: Apply the Cosine Rule to triangle ABC. Given AB = 15\, m, BC = 18\, m, and ABC = 108^. AC^2 = AB^2 + BC^2 - 2(AB)(BC)( ABC) AC^2 = 15^2 + 18^2 - 2(15)(18)(108^) AC^2 = 225 + 324 - 540(-0.309017) AC^2 = 549 + 166.869 AC^2 = 715.869 AC = sqrt(715.869) AC = 26.7555\, m Correct to three significant figures, the length of AC is 26.8\, m. b) Calculate the bearing of C from A: Step 1: Use the Sine Rule to find BAC. (BC)/(( BAC)) = (AC)/(( ABC)) (18)/(( BAC)) = (26.7555)/((108^)) ( BAC) = (18 × (108^))/(26.7555) ( BAC) = (18 × 0.951056)/(26.7555) ( BAC) = 0.63908 BAC = (0.63908) BAC = 39.739^ Step 2: Calculate the bearing of C from A. The bearing of B from A is 053^. The bearing of C from A is the bearing of B from A plus BAC. Bearing of C from A = 53^ + 39.739^ = 92.739^ Correct to three significant figures, the bearing of C from A is 092.7^. c) Calculate the height of the pole BT: Step 1: Identify the right-angled triangle ABT. The pole BT is vertical, so ABT = 90^. Given AB = 15\, m and ATB = 58^. Step 2: Use the tangent trigonometric ratio. ( ATB) = OppositeAdjacent = (AB)/(BT) (58^) = (15)/(BT) BT = (15)/((58^)) BT = (15)/(1.60033) BT = 9.3730\, m Correct to three significant figures, the height of the pole BT is 9.37\, m. That's 2 down. 3 left today — send the next one.