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Math Help - i need trig help

  1. #1
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    Smile i need trig help

    Two sides of a right triangle ABC (C is the right angle) are given. Find the indicated trigonometric function of the given angle.

    5) Find cosA when a=7 an b=2

    6) Find tanA when a=7 and b=6

    thank you
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  2. #2
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    5)

    c = sqrt(53)

    CosA = 2/sqrt(53)


    6)

    tanA = 7/6
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  3. #3
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    Quote Originally Posted by tangerine_tomato View Post
    5) Find cosA when a=7 an b=2

    6) Find tanA when a=7 and b=6
    Hello again, tangerine. We usually label the sides of a right triangle with the lowercase letters corresponding to their opposite angles (as in my drawing).

    So,

    5. We know a and b, but we'll need to use the Pythagorean theorem to get the hypotenuse:

    \cos A = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac bc = \frac2{\sqrt{7^2 + 2^2}} = \frac2{\sqrt{53}} = \frac{2\sqrt{53}}{53}

    6. \tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac ab = \frac76
    Attached Thumbnails Attached Thumbnails i need trig help-right_triangle.png  
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  4. #4
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    Quote Originally Posted by Reckoner View Post
    Hello again, tangerine. We usually label the sides of a right triangle with the lowercase letters corresponding to their opposite angles (as in my drawing).

    So,

    5. We know a and b, but we'll need to use the Pythagorean theorem to get the hypotenuse:

    \cos A = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac bc = \frac2{\sqrt{7^2 + 2^2}} = \frac2{\sqrt{53}} = \frac{2\sqrt{53}}{53}

    6. \tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac ab = \frac76
    oooh o k i just i didnt understand the directions
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  5. #5
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    Quote Originally Posted by Reckoner View Post
    Hello again, tangerine. We usually label the sides of a right triangle with the lowercase letters corresponding to their opposite angles (as in my drawing).

    So,

    5. We know a and b, but we'll need to use the Pythagorean theorem to get the hypotenuse:

    \cos A = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac bc = \frac2{\sqrt{7^2 + 2^2}} = \frac2{\sqrt{53}} = \frac{2\sqrt{53}}{53}

    6. \tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac ab = \frac76
    how did you draw the triangle

    how did you know where the b and the a went
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  6. #6
    MHF Contributor Reckoner's Avatar
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    Quote Originally Posted by tangerine_tomato View Post
    how did you draw the triangle

    how did you know where the b and the a went
    Really, I didn't know for sure since the problem didn't specify. But as I said before, that is the usual convention: side a goes opposite angle A, side b opposite angle B, etc., and the hypotenuse is usually c.
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  7. #7
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    The right angle is always opposite the longest side (hypoteneuse).

    So in this case, the hypoteneuse would be c, right angle would be C.

    The other two angles A and B are oppsite sides a and b respectively. The positioning of these two sides is irrelevant because you're basically just flipping or turning the same triangle.
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