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    In each of the following cases ABC (triangle) has ABC =  30^o and AB = 10 cm.

    a) Calculate the least possible length that AC could be

    How would you go about doing this
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    I think that the shortest distance to any line is a perpendicular.

    Will that help?
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    Have you drawn a diagram to help you visualise the triangle?
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    (Original post by maggiehodgson)
    I think that the shortest distance to any line is a perpendicular.

    Will that help?

    No, I don't understand
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    I know the answer is 5 cm when ACB is at a right angle...

    but I want to know why
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    (Original post by Naruke)
    No, I don't understand
    Well, just draw two lines AB (10) and BX (quitet long) with a 30 degree angle between. Now look at the point A. You have to draw a line down to AX so that the line you draw is as short as possible. That line is a perpendicular. Were the line meets AX then that is the point C.

    Give it a go and see if it helps.
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    i'm drawing a diagram....
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    here you go
 
 
 
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