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Math Help - Using Tangent: Word Problem

  1. #1
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    Post Using Tangent: Word Problem

    Hello everyone, today I am faced with a confusing trigonometry world problem. I need help in building a visual image of the world problem and it is been very hard for me to do that. I know the answer is 21.3m I need to know how to get there. All right here its:

    From the top of a cliff 32m high it is noted that the angles of depression of two boats lying in the line due east of the cliff are 21degrees and 17degrees. How far are the boats apart?

    Thank You For the Help!
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Attached is the required image:
    Attached Thumbnails Attached Thumbnails Using Tangent: Word Problem-trig.bmp  
    Last edited by alexmahone; June 16th 2009 at 06:36 AM.
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  3. #3
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    Draw a diagram with 2 triangles .

    tan21 =32/x

    tan 17=32/y

    y-x = ans
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  4. #4
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by mathaddict View Post
    Draw a diagram with 2 triangles .

    tan21 =32/x

    tan 17=32/y

    y-x = ans
    Alternative method:

    tan 69^o=\frac{OB_1}{32}
    tan 73^o=\frac{OB_2}{32}

    Distance= OB_2-OB_1
    Last edited by alexmahone; June 17th 2009 at 02:07 AM.
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  5. #5
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    Quote Originally Posted by alexmahone View Post
    Incorrect.

    tan 69^o=\frac{OB_1}{32}
    tan 73^o=\frac{OB_2}{32}

    Distance= OB_2-OB_1
    can you please elaborate more?!
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  6. #6
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by flutterby View Post
    can you please elaborate more?!
    Refer attachment in the second post.

    Note that the complements of the angles 21degrees and 17degrees are 69degrees and 73degrees respectively.
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  7. #7
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    so is it like you switch to the angle of elevation and then solve from there?
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  8. #8
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by flutterby View Post
    so is it like you switch to the angle of elevation and then solve from there?
    Not exactly.

    Refer this link: Angles of Elevation and Depression
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  9. #9
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    Quote Originally Posted by alexmahone View Post
    Not exactly.

    Refer this link: Angles of Elevation and Depression

    mmm can you please explain how it is more like?
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  10. #10
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    Quote Originally Posted by alexmahone View Post
    Incorrect.

    tan 69^o=\frac{OB_1}{32}
    tan 73^o=\frac{OB_2}{32}

    Distance= OB_2-OB_1
    Hi Alex .

    Angle PCB2=angle CB2O ( from the diagram you posted )

    so tan 21 = 32/x and same goes to the other one .

    Look at example 21 in the link , i did what exactly it says .
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  11. #11
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by mathaddict View Post
    Hi Alex .

    Angle PCB2=angle CB2O ( from the diagram you posted )

    so tan 21 = 32/x and same goes to the other one .

    Look at example 21 in the link , i did what exactly it says .
    Your method works as well. Fair enough. Sorry.
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  12. #12
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    Quote Originally Posted by mathaddict View Post
    Hi Alex .

    Angle PCB2=angle CB2O ( from the diagram you posted )
    Do you mean \angle OCB_2 = \angle CB_2O? If so, this is incorrect. \angle OCB_2 is not the angle that is labeled as 17. However, \angle CB_2O = 17^{\circ} is correct.

    Anyway, it looks like both you and alexmahone are correct in your working out this problem. It also looks like the y that mathaddict assigns is the same as the side OB_2.

    According to alexmahone:
    \tan 73^{\circ} = \frac{OB_2}{32}.

    According to mathaddict:
    \tan 17^{\circ} = \frac{32}{OB_2}.

    But,
    \tan (90^{\circ} - \theta) = \cot \theta

    So,
    \begin{aligned}<br />
\tan 73^{\circ} &= \tan (90^{\circ} - 17^{\circ}) \\<br />
\frac{OB_2}{32} &= \cot 17^{\circ} \\<br />
\frac{32}{OB_2} &= \tan 17^{\circ}<br />
\end{aligned}
    ... meaning that both of you are saying the same thing.


    01


    EDIT: Beaten to it
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