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AS C2 Bearings
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danbomb123
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http://www.ocr.org.uk/Images/78547-q...hematics-2.pdf
I'm stuck on q4iii, I have no idea on finding the shortest path. I have tried splitting the triangle at point C so there is a right angle triangle and working it out from there. I have looked at the mark scheme and I have no idea how they got d=2*sin40
I'm stuck on q4iii, I have no idea on finding the shortest path. I have tried splitting the triangle at point C so there is a right angle triangle and working it out from there. I have looked at the mark scheme and I have no idea how they got d=2*sin40
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midgemeister7
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#2
(Original post by danbomb123)
http://www.ocr.org.uk/Images/78547-q...hematics-2.pdf
I'm stuck on q4iii, I have no idea on finding the shortest path. I have tried splitting the triangle at point C so there is a right angle triangle and working it out from there. I have looked at the mark scheme and I have no idea how they got d=2*sin40
http://www.ocr.org.uk/Images/78547-q...hematics-2.pdf
I'm stuck on q4iii, I have no idea on finding the shortest path. I have tried splitting the triangle at point C so there is a right angle triangle and working it out from there. I have looked at the mark scheme and I have no idea how they got d=2*sin40
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danbomb123
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(Original post by midgemeister7)
Draw a straight line from C to the line AB. Now you have a new right angled triangle, with longest side 2km. Sin = opp/hyp so the length is 2 (hyp) multiplied by sin40
Draw a straight line from C to the line AB. Now you have a new right angled triangle, with longest side 2km. Sin = opp/hyp so the length is 2 (hyp) multiplied by sin40
therefore 2*sin40=opp?
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midgemeister7
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