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secondary 3 | E Maths
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Hi, pls help on this E math question, thanks.
Now, mark a third point which is nowhere along the line (even if it is not extended). Call this point C.
Now, draw lines AC and BC to complete a triangle. Are you able to tell me the shortest path from A to B? It must be the direct path AB and not AC followed by CB.
This is because the shortest distance between any two points is a straight line. If you place a string following the path A —> C —> B and mark the start and end points, you will realise that when you straighten the string, the length would be more than that of AB.
Howsoever the angle at A is, the length AB must be shorter than the sum of the other lengths. This is what we call the triangle inequality.
Sadly, the shortest possible length has to be more than the difference between the two other lengths. Using the same principle, you cannot have the third length x being 4 cm, because 8 cm would then be the longest length and no 3 cm and 4 cm lines can be connected to form an 8 cm line.
Candice, if the angle between the lengths 3 cm and 8 cm is A, then we can use the cosine rule to say that
x = sqrt (3^2 + 8^2 - 2(3)(8) cos A)
= sqrt (73 - 48 cos A)
We can play around with the expression 73 - 48 cos A. You have also learnt that the maximum and minimum values of cos A are 1 and -1 (when A = 0 and A = 180 respectively) in A Maths. But here the function contains the negative of cos A, so the roles are reversed. The maximum happens when A = 180 while the minimum happens when A = 0.
Putting this extreme values,
max x^2 = 73 - 48 (-1) = 121 when A = 180 so max x = 11
min x^2 = 73 - 48 (1) = 25 when A = 0 so min x = 5
Let me know if you need more explanation.
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