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Math Help - Addition Formula Question

  1. #1
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    Addition Formula Question

    Hi there,
    I have a problem involving a right angled triangle ABC. It is right angled at C.
    The hypoteneuse has a value of 13 units.
    A line stretches across the triangle from B to side AC at D. Side AC is divided into AD and DC. DC has the value 4 units.
    Angle ABD equals x degrees and angle DBC equals 30 degrees.
    I have to find the exact value of sin x.

    I have worked out that side BC has a value of 6.928 units. Hypoteneuse BD has a value of 7.9 units and side AD equals 7 units.
    Can anyone give me help with this problem please?

    Thank you,

    Cromlix
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  2. #2
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    Quote Originally Posted by cromlix View Post
    Hi there,
    I have a problem involving a right angled triangle ABC. It is right angled at C.
    The hypoteneuse has a value of 13 units.
    A line stretches across the triangle from B to side AC at D. Side AC is divided into AD and DC. DC has the value 4 units.
    Angle ABD equals x degrees and angle DBC equals 30 degrees.
    I have to find the exact value of sin x.

    I have worked out that side BC has a value of 6.928 units. Hypoteneuse BD has a value of 7.9 units and side AD equals 7 units.
    Can anyone give me help with this problem please?

    Thank you,

    Cromlix

    In triangle BCD

    tan30^o=\frac{4}{|BC|}\Rightarrow\ |BC|=4\sqrt{3}


    By Pythagoras' Theorem on triangle ABC

    \left[4\sqrt{3}\right]^2+|AC|^2=13^2=169\Rightarrow\ |AC|=11\Rightarrow\ |AD|=7


    Now you can use the Sine Law on triangle ABD

    \frac{Sin\;x}{7}=\frac{Sin120^o}{13}

    Angle ADB is 120^o because angle DCB is 90^o-30^o
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  3. #3
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    Too late...

    I think the easiest thing is to use the law of sines for the triangle ABD. The angle ADB is 90+30=120^\circ. Therefore, AD/(\sin x)=AB/(\sin 120).

    A longer way is to say \cos(30+x)=BC/AB and rewrite \cos(30+x) as \cos30\cos x-\sin 30\sin x. Since x<90, we have \cos x>0 and so \cos x=\sqrt{1-\sin^x}. This way we get a quadratic equation on \sin x.
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  4. #4
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    Quote Originally Posted by cromlix View Post
    I have a problem involving a right angled triangle ABC. It is right angled at C.
    The hypotenuse has a value of 13 units.
    A line stretches across the triangle from B to side AC at D. Side AC is divided into AD and DC. DC has the value 4 units.
    Angle ABD equals x degrees and angle DBC equals 30 degrees.
    I have to find the exact value of sin x.

    I have worked out that side BC has a value of 6.928 units. Hypotenuse BD has a value of 7.9 units and side AD equals 7 units.
    The values you have calculated for BC and BD are approximately correct. But the question is asking for exact values. So instead of using a calculator, you should have used the fact that \tan 30^\circ = 1/\sqrt3 to get the exact value BC = 4\sqrt3, as in Archie Meade's comment. Also, BD is exactly 8 (not 7.9). But you are correct to say that AD = 7, and this is exact (because AC = 11 by Pythagoras, and when you subtract 4 for DC you are left with exactly 7 for AD). Now you can complete the question as in Archie's comment.
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