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Math Help - Please Check My Answer.

  1. #1
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    Please Check My Answer.

    I need to calculate the unknown quantity x and express answer in degrees, minutes & seconds. Pls advise if my calculations are correct and if not, could please assist me - thanks again.

    tan^-1 (20/18)
    x = 48.0128

    x = 48^o, 0' & 46"
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    Quote Originally Posted by Risrocks View Post
    I need to calculate the unknown quantity x and express answer in degrees, minutes & seconds. Pls advise if my calculations are correct and if not, could please assist me - thanks again.

    tan^-1 (20/18)
    x = 48.0128

    x = 48^o, 0' & 46"
    No. Be very careful here. You can only use the trigonometric ratios when dealing with right triangles, and this is not a right triangle.

    Here, we use the law of cosines:

    c^2 = a^2 + b^2 - 2ab\cos C (where the angle C is opposite to side c)

    So, we have,

    18^2 = 20^2 + 33^2 - 2(20)(33)\cos X

    \Rightarrow324 = 1489 - 1320\cos X

    \Rightarrow\cos X = \frac{233}{264}

    \Rightarrow X = \arccos\frac{233}{264} (we don't have to worry about other solutions, since 0 < X < \pi)

    X\approx0.4895\text{ rad}\approx28.045346^\circ

    Now see if you can convert to deg-min-sec.
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  3. #3
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    Thanks - I should have dealt with it as an obtuse triangle.

    Converted = 28^o 2' 43"

    Thanks for taking the time to assist me.
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    Quote Originally Posted by Risrocks View Post
    Thanks - I should have dealt with it as an obtuse triangle.

    Converted = 28^o 2' 43"

    Thanks for taking the time to assist me.
    Correct!

    28.045346^\circ = 28^\circ + 0.045346^\circ\left(\frac{60'}{1^\circ}\right) (60' per 1)

    = 28^\circ + \left(0.045346\cdot60\right)' = 28^\circ + 2.72076' (canceling units)

    = 28^\circ2' + 0.72076'\left(\frac{60''}{1'}\right) (again, multiplying by conversion factor)

    = 28^\circ2' + (0.72076\cdot60)'' (canceling units)

    \approx28^\circ2'43.25'' or about \boxed{28^\circ2'43''}.
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