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  1. #1
    Member classicstrings's Avatar
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    Compound Angle Forumla

    If A and B are angles between 0 and 90, and sinA = 3/5 and tanB = 7/24, find the value of

    sin(A-B).

    How do I tackle this question? I just need to know the pathway. Thanks.
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  2. #2
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    Hello, classicstrings!

    If A and B are angles between 0^i and 90^o,
    and \sin A = \frac{3}{5} and \tan B = \frac{7}{24},
    find the value of \sin(A-B)

    We have the identity: . \sin(A - B) \:=\:\sin(A)\cos(B) - \sin(B)\cos(A)

    We know that: . \sin(A) = \frac{3}{5}

    We need the values of: . \cos(A),\:\sin(B),\:\cos(B)


    From \sin(A) = \frac{3}{5} = \frac{opp}{hyp} and Pythagorus, we find that: . adj = 4
    . . Now you can find \cos(A).

    From \tan(B) = \frac{7}{24} = \frac{opp}{adj} and Pythagorus, we find that: . hyp = 25
    . . Now you can find \sin(B) and \cos(B).


    Got it?

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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by classicstrings View Post
    If A and B are angles between 0 and 90, and sinA = 3/5 and tanB = 7/24, find the value of

    sin(A-B).

    How do I tackle this question? I just need to know the pathway. Thanks.
    First you need to know that:

    sin(A-B)=sin(A)cos(B)-cos(A)sin(B).

    Then as A and B are both in the range 0 to 90 degrees,

    if sin(A)=3/5, then cos(A)=4/5 (there are a number of ways of getting this
    result one of which is recognising that A is an angle of a 3,4,5 right triangle
    which imediatly tells you that cos(A)=4/5).

    if tan(B)=7/24, then sin(B)=7/25, and cos(B)=24/25 (as 25 is the hypotenuse
    of a right triangle who's other two sides are 7 and 24).

    So:

    sin(A-B)=(3/5)(24/25)-(4/5)(7/25)=44/125

    RonL
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