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    I'll attach an image of the question below:
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    Question 8, part a.

    https://imgur.com/a/n2DbL
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    (Original post by Illidan2)
    Question 8, part a.

    https://imgur.com/a/n2DbL
    Determine the length AD in terms of theta, first.
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    (Original post by RDKGames)
    Determine the length AD in terms of theta, first.
    How do I determine length in terms of theta? I understand the significance of theta as an angle measure but I don't know how to determine length in theta.
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    (Original post by Illidan2)
    How do I determine length in terms of theta? I understand the significance of theta as an angle measure but I don't know how to determine length in theta.
    There is a very obvious right-angled triangle in the diagram which allows you to relate AD to \theta
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    (Original post by RDKGames)
    There is a very obvious right-angled triangle in the diagram which allows you to relate AD to \theta
    Yes, ABD is a right-angled triangle. I just don't understand how to express length as theta.I can see that Angle BAC= theta, and that DAB= theta. Could I use the right-angled-triangle rule cos theta= 6/AD, therefore AD= 6/cos theta? Would this be along the right lines?

    Then, to find CD, i'd need to find AC and subtract from CD?
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    (Original post by Illidan2)
    Yes, ABD is a right-angled triangle. I just don't understand how to express length as theta.I can see that Angle BAC= theta, and that DAB= theta. Could I use the right-angled-triangle rule cos theta= 6/AD, therefore AD= 6/cos theta? Would this be along the right lines?
    you can use soh cah toh and then pythagoras to get AD in terms of theta.

    later on you can use right-angled-triangle rule to find AC in terms of theta.
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    (Original post by e^x)
    you can use soh cah toh and then pythagoras to get AD in terms of theta
    Yeah. That's what I just tried to do just then. Whether or not i've done it correct is another matter, but as cah means cos= adj/hyp, I can only assume i've done that part correctly.
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    (Original post by Illidan2)
    Yes, ABD is a right-angled triangle. I just don't understand how to express length as theta.I can see that Angle BAC= theta, and that DAB= theta. Could I use the right-angled-triangle rule cos theta= 6/AD, therefore AD= 6/cos theta? Would this be along the right lines?

    Then, to find CD, i'd need to find AC and subtract from CD?
    Exactly. Though last line is not quite correct, almost, it's CD = AD - AC

    For the length CD, consider drawing an extra line connecting C and B.
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    (Original post by RDKGames)
    Exactly. Though last line is not quite correct, almost, it's CD = AD - AC

    For the length CD, consider drawing an extra line connecting C and B.
    AD yeah. That's what I meant instead of CD. My bad! :P
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    (Original post by Illidan2)
    Yes, ABD is a right-angled triangle. I just don't understand how to express length as theta.I can see that Angle BAC= theta, and that DAB= theta. Could I use the right-angled-triangle rule cos theta= 6/AD, therefore AD= 6/cos theta?
    Yes. You might find it helpful to rewrite \dfrac{6}{\cos \theta} in terms of \sec \theta. (as it will then be more obvious how it relates to the given answer).
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    Awesome. So now I have AD= 6sec theta, and as I need to show that CD in the form mentioned in the question, AC must be 6 cos theta, in order to take out the factor of 6 as i'm supposed to show. I assume I can use right-angled trig to deduce that this is the case, too. Give me a moment... :P
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    Ignore previous post. I wasn't looking at angle ACB. I was looking at angle BAC. >.> Fail. Give me a sec again... :L
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    (Original post by Illidan2)
    I've managed to confuse myself, it seems. Because ACB=90 degrees, it is therefore the right angle, and therefore CB must be the hypotenuse, correct?
    No, AB is the hypotenuse.

    When was the last time you saw a right-angled triangle with the hypotenuse being adjacent to the right-angle? It is always opposite the right-angle.
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    Yeah, I know it's opposite the right angle, as I stated myself. However, for some reason, I thought angle BAC was angle ACB, and have been thinking this for quite a few minutes. My apologies. I'll do it again now :L
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    Yeah i've done it now, it makes sense now. I was just looking at the wrong angle and therefore basing my assumptions on the sides of the triangle in relation to that angle. But now that i've noticed, I finished it in about 30 seconds. Thanks very much for your help on this!
 
 
 
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