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
Morning kiendeperuse — let's get this done. Let H be the harbour. Ship X is 16 km from H on a bearing of 090^. So, HX = 16 km. Ship Y is 14 km from H on a bearing of 130^. So, HY = 14 km. Ship Z is 26.31 km to the west of Y and on a bearing of 240^ from H. So, HZ = 26.31 km. (Note: The problem states "Z is 26.31 Km to the west of Y and on a bearing of 240 from the harbour". This is contradictory. If Z is 26.31 km to the west of Y, its distance from H would depend on Y's position. However, the phrasing "and on a bearing of 240 from the harbour" implies HZ = 26.31 km. Given the context of bearing problems, we will assume HZ = 26.31 km and ignore "26.31 Km to the west of Y" as a potential typo or misstatement, as it makes the problem unsolvable with the given information. We will use the bearing from the harbour for Z.) a) The distance between X and Y Step 1: Find the angle XHY. The bearing of X from H is 090^. The bearing of Y from H is 130^. The angle between HX and HY is the difference in their bearings. XHY = 130^ - 090^ = 40^ Step 2: Use the Cosine Rule to find the distance XY. In XHY, we have HX = 16 km, HY = 14 km, and XHY = 40^. The Cosine Rule states c^2 = a^2 + b^2 - 2ab C. XY^2 = HX^2 + HY^2 - 2(HX)(HY) ( XHY) XY^2 = 16^2 + 14^2 - 2(16)(14) (40^) XY^2 = 256 + 196 - 448 (40^) XY^2 = 452 - 448(0.7660) XY^2 = 452 - 343.168 XY^2 = 108.832 XY = sqrt(108.832) XY ≈ 10.432 km The distance between X and Y is approximately 10.43 km. b) The distance from harbour This part of the question is incomplete. It asks "The distance from harbour:" without specifying which ship. Based on the previous parts, it's likely asking for the distance of Z from the harbour, which was given in the problem statement. The distance of Z from the harbour H is given as HZ = 26.31 km. The distance from the harbour to Z is 26.31 km. c) The distance between X and Z Step 1: Find the angle XHZ. The bearing of X from H is 090^. The bearing of Z from H is 240^. The angle between HX and HZ is the difference in their bearings. XHZ = 240^ - 090^ = 150^ Step 2: Use the Cosine Rule to find the distance XZ. In XHZ, we have HX = 16 km, HZ = 26.31 km, and XHZ = 150^. XZ^2 = HX^2 + HZ^2 - 2(HX)(HZ) ( XHZ) XZ^2 = 16^2 + (26.31)^2 - 2(16)(26.31) (150^) XZ^2 = 256 + 692.2261 - 841.92 (150^) XZ^2 = 948.2261 - 841.92(-0.8660) XZ^2 = 948.2261 + 729.00992 XZ^2 = 1677.23602 XZ = sqrt(1677.23602) XZ ≈ 40.954 km The distance between X and Z is approximately 40.95 km. Final Answers: a) The distance between X and Y is 10.43 km b) The distance from the harbour to Z is 26.31 km c) The distance between X and Z is 40.95 km Drop the next question.