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In a right triangle sinA= 3/5 and C=17. Find a and b.
Be sure to draw a right triangle and letter it properly.

how can i solve the math in this problem. i know i would have to use the Pythagoream Theorem but i don't know how to set the problem up first.

  • trigonometry. -

    SinA= b/C=3/4

    you are given C, so b= 3/4 * 17

    then a can be found from a^2+b^2=C^2

  • trigonometry. -

    your question makes no sense
    If you have a right-angled triangle where
    sinA = 3/5, then the opposite side to vertex A is 3 and the hypotenuse is 5,
    by Pythagoras it is easy to show that the third side is 4
    (you should have recognized the 3-4-5 right-angled triangle)

    if sinA = 3/5, then angle A is appro 37.9°

    what does C=17 mean?

  • trigonometry. -

    sinA= 3/5 and C=17

    Finding a and b

    would this work as an answer ..?

    sinA = opp/hyp = a/c = 3/5 = 3*3.4/5*3.4 = 10.2/17

    a = 10.2

    Applying Pythagoream Theorem a^2 + b^2 = c^2

  • trigonometry. -

    b= sqrt 17^2 - 10.2^2 = 13.6


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