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    How would u solve this problem

    if sin x = 0.6 and x is acute find the value of cos x and tan x

    im so stuck...
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    Write 0.6 as a fraction. Draw a right-angled triangle, and call one of the angles x. Can you now give values to the lengths of two of the sides, and work out the length of the other side? This should give you all of the information you need to finish off.
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    Remember that sin(x) is the ratio of the opposite side to the hypotenuse, therefore any scalar values will cancel out and you can assume the length of the hypotenuse is 1 and the length of the opposite side is 0.6

    This works because;

    For a triangle with hypotenuse a and opposite side c
    sin(x) = a/c = 0.6/1 = 0.6

    So if you draw out a triangle with these side lengths you should be able to see where to go from there to calculate cos(x) and tan(x)
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    (Original post by agrajaag)
    Remember that sin(x) is the ratio of the opposite side to the hypotenuse, therefore any scalar values will cancel out and you can assume the length of the hypotenuse is 1 and the length of the opposite side is 0.6.
    All correct, but I would have thought it easier to understand (and work with) to notice that 0.6 = 3/5 and to make the opposite side have length 3 and the hypotenuse length 5. The adjacent side then comes out very nicely - it's a bit more fiddly using 0.6 and 1.
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    (Original post by Pangol)
    All correct, but I would have thought it easier to understand (and work with) to notice that 0.6 = 3/5 and to make the opposite side have length 3 and the hypotenuse length 5. The adjacent side then comes out very nicely - it's a bit more fiddly using 0.6 and 1.
    Yeah I think you're right That's definitely simpler. Though I think it is easier to understand it as ratios as opposed to fixed values
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    (Original post by marzipan100)
    How would u solve this problem

    if sin x = 0.6 and x is acute find the value of cos x and tan x

    im so stuck...
    It states the angle is acute, which implies the angle lies in the first quadrant.

    You should also know that all trig functions are positive in the first quadrant.

    It's also given that  \sin(x) = 0.6 = \frac{6}{10}=\frac{3}{5}

    You can then form a right triangle and use the trig ratios
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    sohcahtoaaaaaaaa
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    Thank you so much x
 
 
 
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