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Math Help - Triangle Angle Problem

  1. #1
    Newbie Tom G's Avatar
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    Triangle Angle Problem

    Hi everyone, I think that I have got the first part (a) of this question correct but would appreciate someone checking and pointing me in the right direction as I have a feeling that I am missing something as the question seems too simple to be true.

    In triangle ABC, angle A=90 and sec B=2

    a/ Find cos B - I have got:

    sec B=2
    1/cos B=2
    1=2cos B
    cosB=1/2

    b/ Find angles B and C

    I've got:

    cosB=1/2
    B=60,300

    can't use B=300 as angles in a triangle = 180 so B=60

    180-A-B=C so 180-90-60=C=30
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Tom G View Post
    Hi everyone, I think that I have got the first part (a) of this question correct but would appreciate someone checking and pointing me in the right direction as I have a feeling that I am missing something as the question seems too simple to be true.

    In triangle ABC, angle A=90 and sec B=2

    a/ Find cos B - I have got:

    sec B=2
    1/cos B=2
    1=2cos B
    cosB=1/2

    b/ Find angles B and C

    I've got:

    cosB=1/2
    B=60,300

    can't use B=300 as angles in a triangle = 180 so B=60

    180-A-B=C so 180-90-60=C=30
    Too simple, perhaps, but also correct.

    -Dan
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  3. #3
    Bar0n janvdl's Avatar
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    Quote Originally Posted by Tom G View Post
    Hi everyone, I think that I have got the first part (a) of this question correct but would appreciate someone checking and pointing me in the right direction as I have a feeling that I am missing something as the question seems too simple to be true.

    In triangle ABC, angle A=90 and sec B=2

    a/ Find cos B - I have got:

    sec B=2
    1/cos B=2
    1=2cos B
    cosB=1/2

    b/ Find angles B and C

    I've got:

    cosB=1/2
    B=60,300

    can't use B=300 as angles in a triangle = 180 so B=60

    180-A-B=C so 180-90-60=C=30
    Yes, seems fine
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  4. #4
    Newbie Tom G's Avatar
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    I've just looked at the next section of the question and am stuck (I have a feeling its really simple but my brain just doesn't seem to be working today).

    c/ Find tan B

    I've got tan B = Sin B / Cos B

    I've already worked out cos B, but could someone tell me how I find Sin B? Thanks
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  5. #5
    Bar0n janvdl's Avatar
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    Quote Originally Posted by Tom G View Post
    I've just looked at the next section of the question and am stuck (I have a feeling its really simple but my brain just doesn't seem to be working today).

    c/ Find tan B

    I've got tan B = Sin B / Cos B

    I've already worked out cos B, but could someone tell me how I find Sin B? Thanks
    You've just worked B out to be 30 degrees didn't you?

    Now use your special triangles to aid you.

    AND Sin B = cos (90 - B)
    Last edited by janvdl; September 22nd 2007 at 08:34 AM. Reason: My terrible English grammar :-D
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by janvdl View Post
    You've just worked B out to be 30 degrees didn't you?

    Now use your special triangles to aid you.

    AND Sin B = cos (90 - B)
    Actually, the more direct one to use would be either
    cos(B) = \pm \sqrt{1 - sin^2(B)}<-- Choose the "+" since cosine is positive for an acute angle.

    or
    cos(B) = sin(90 - B)

    -Dan
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  7. #7
    Newbie Tom G's Avatar
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    Hi, I'm still struggling with this one.

    I have now got;

    sin B = +Squareroot(1-cos^2 B)

    but am now confused as I know cos B x cos B = cos^2 B ... so does this mean that cos 60 x cos 60 = 1/2 x 1/2 = 1/4

    Then 1-1/4 = 3/4 so sin B = \pm \sqrt{3/4} = 0.866025........ and so B=59.99999 ( I already have worked out that B=60)

    I really have no clue what is going on and am going around in circles - would really appreciate someone showing me how its done - Thanks.
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  8. #8
    Bar0n janvdl's Avatar
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    Quote Originally Posted by Tom G View Post
    Hi, I'm still struggling with this one.

    I have now got;

    sin B = +Squareroot(1-cos^2 B)

    but am now confused as I know cos B x cos B = cos^2 B ... so does this mean that cos 60 x cos 60 = 1/2 x 1/2 = 1/4

    Then 1-1/4 = 3/4 so sin B = \pm \sqrt{3/4} = 0.866025........ and so B=59.99999 ( I already have worked out that B=60)

    I really have no clue what is going on and am going around in circles - would really appreciate someone showing me how its done - Thanks.
    Sin B = 0.866025 (which is actually \frac{ \sqrt{3} }{2} )

    And Cos B = \frac{1}{2}

    And Tan B = \frac{Sin B}{Cos B}

    Take it from here.
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