Weekly Problem 33

6263 /www/nrich/html/content/id/6263/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">In a village the pub is $300 \; \text{m}$ due north of the school, and the church is $300 \; \text{m}$ from the pub at a bearing of $060$. What is the bearing of the school from the church? If you liked this problem, <a href="http://nrich.maths.org/1860">here is an NRICH task</a> which challenges you to use similar mathematical ideas. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> We are given: <mdo:image height="298" width="250" alt="Diagram of information given in question" src="solution1.png"></mdo:image> We want to find the bearing of the School from the Church, which is the green angle marked on the diagram. We can use the fact that the angles in a triangle sum to $180^\circ$ to mark in some other angles <mdo:image height="298" width="250" src="solution2.png" alt="More angles marked in"></mdo:image> We know the triangle linkingthe three buildings is isosceles, so the two missing angles are equal, so are both $30^\circ$.So: <mdo:image height="298" width="250" src="solution.png" alt="solution"></mdo:image> So the green angle is $180^\circ+30^\circ = 210^\circ$, so the bearing of the School from the Church is $210$. </mdoxml> 2 3 0 0 1 0 0 Weekly Problem 33 - 2009 In a village the pub is 300m due north of the school, and the church is 300m from the pub at a bearing of 060. What is the bearing of the school from the church? 2D Geometry, Shape and Space Compass directions and bearings Secondary Mapping Document Angles and polygons