trig

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Ido not understand why the rt triangel was in ratio of 1:√3:2 in Sue's question
where in triangle abc c isthert angle and sin a √3/2 - find csc b-please explain where the 1 comes from

  • trig -

    (We know that in rt triangle
    Sin=P/H)

    given sinA=_/3/2

    SO P=_/3 , H=2

    ACB is rt triangle
    by pythagoras theorem

    H^2=P^2+B^2

    B^2=H^2-P^2

    B^2=(2)^2-(_/3)^2

    B^2=4-3

    B^2=1

    B=_/1

    B=1

    (where P=perpendicular,
    B=base,H=hypotenuse)

  • trig -

    thank you.

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