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Math Help - Trigonometry

  1. #1
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    Trigonometry

    Can anyone help me out with this question? It's really the third part that I'm having trouble with, but I'll type out the whole thing here. It asks;

    In the figure, ABCD is a rhombus in which AB = 60cm and angle A = 60 degrees. P is a point on AD and AP = AQ = 2cm. Angle PQC = x degrees. P is joined to C.



    a) Write down the length of PQ, giving reasons.

    b) Show that PC = QC

    c) Yse the cosine rule for the triangle QBC and, using part b), show that x^o = \frac{\sqrt{76}}{76}.
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  2. #2
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    Quote Originally Posted by Flay View Post
    Can anyone help me out with this question? It's really the third part that I'm having trouble with, but I'll type out the whole thing here. It asks;

    In the figure, ABCD is a rhombus in which AB = 60cm and angle A = 60 degrees. P is a point on AD and AP = AQ = 2cm. Angle PQC = x degrees. P is joined to C.



    a) Write down the length of PQ, giving reasons.

    b) Show that PC = QC

    c) Yse the cosine rule for the triangle QBC and, using part b), show that x^o = \frac{\sqrt{76}}{76}.
    so
    we know that triangle APQ is an equilateral triangle because it has a 60 degrees angle.
    a)which means that PQ=AP=AQ=2cm.
    b)you can use congreunce of triangles PDC and CBQ. where you have 2 equal lines DP and QB
    BC=DC
    and 2 equal opposite angles in the rhombus.
    then you got b)PC=QC
    c) looking at triangle QBC:
    QB=60-2=58
    BC=60
    angle B=120
    then you get QC with the cosine rule.
    then you use the cosine rule in triangle PQC
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  3. #3
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    Thanks, but my problem is that whenever I try and figure out the length of QC, I get \sqrt{10444} instead of \sqrt{76}. Can someone please give me a worked solution to this?
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  4. #4
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    Hello, Flay!

    I'm getting the same answer . . .

    And are you sure of the exact wording of the problem?

    That last statement doesn't make sense: . x^o \:=\:\frac{\sqrt{76}}{76}

    First of all, \sqrt{76} can be simplified: . 2\sqrt{19}

    Second, that can't be the answer: . x \:=\:0.1147^o ?
    . . We can see that x is nearly a right angle.

    They may have meant: . \cos x \:=\:\frac{\sqrt{76}}{76}
    . . But that is not the correct answer either.


    . . . . . ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    As Tom pointed out, APQ is an equilateral triangle: . PQ = 2

    We find that: . \angle D = 120^o,\;PD = 58,\;DC = 60


    Using the Law of Cosines on \Delta PDC, we have:

    . . PC^2 \;=\;58^2 + 60^2 - 2(58)(60)\cos120^o \;=\;3364 + 3600 - 6960\left(\text{-}\frac{1}{2}\right) \;=\;10,444

    Hence: . PC \;=\;\sqrt{10,444} \;=\;2\sqrt{2611}


    Use the Law of Cosines on \Delta PQC

    . . \cos x \;=\;\frac{PQ^2 + QC^2 - PC^2}{2(PQ)(QC)} \;=\;\frac{2^2 + (2\sqrt{2611})^2 - (2\sqrt{2611})^2}{2(2)(2\sqrt{2611})}

    . . \cos x \;=\;\frac{1}{2\sqrt{2611}} \;=\;0.009785129


    Therefore: . x \;=\;89.4393444 \;\approx\;89.44^o

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  5. #5
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    Yeah, that's what I was getting. I think the question paper must be wrong, because I certainly haven't written down the problem incorrectly.
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