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    angle B is acute and tanB = 4/3.

    Find sinB
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    Use SOH CAH TOA

    If tan B is 4/3, that means that the side of the triangle opposite to the angle is of relative length 4 and the side of the triangle adjacent to the angle is of relative length 3.

    This means that by Pythagoras' the hypotenuse is root of 4^2 + 3^2, which is 5.

    Hence, sin B is opposite over hypotenuse, which is 4/5.
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    umm you can just use a 3,4,5 triangle...
    so if you draw that...
    tanB = 4/3 (opp/adj)
    cosB = 3/5 (adj/hyp)
    sinB = 4/5 (opp/hyp)
    so just inverse sin(4/5) and should give a number?

    (someone kick me if im doing it completely wrong :p: )
    • Thread Starter
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    omg... ignore me.
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    cant we just kick you instead?
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    ok then, let's try this...

    angle B is acute and tanB = 3/4

    Find sin(B/2)
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    (Original post by chewwy)
    angle B is acute and tanB = 4/3.

    Find sinB

    same route as the others, expressed slightly differently:
    tanb = 4/3
    sinb/cosb=4/3
    3sinb=4cosb
    9sin²b=16cos²b
    9sin²b=16(1-sin²b)
    9sin²b=16-16sin²b
    25sin²b=16
    sin²b=16/25
    sinb=4/5

    Aitch
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    (Original post by Aitch)
    same route as the others, expressed slightly differently:
    tanb = 4/3
    sinb/cosb=4/3
    3sinb=4cosb
    9sin²b=16cos²b
    9sin²b=16(1-sin²b)
    9sin²b=16-16sin²b
    25sin²b=16
    sin²b=16/25
    sinb=4/5

    Aitch
    There are two flaws in this argument that cancel each other out (who said that two wrongs make a right?). I've emboldened them.

    First one: Squaring will give extra solutions. You should check all solutions in the original equation.

    Second one: Rooting will give positive and negative values (i.e. ±4/5)
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    (Original post by chewwy)
    ok then, let's try this...

    angle B is acute and tanB = 3/4

    Find sin(B/2)
    ^^^
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    cos(A+B)=cosAcosB-sinAsinB

    Let A and B both=B/2, therefore cosB=cos^2(B/2)-sin^2(B/2).

    Everybody's favourite identity says cos^2(B/2)=1-sin^2(B/2).

    This means cosB=1-2sin^2(B/2). From the above posts, we can see that cosB is 3/5.

    So 3/5=1-2sin^2(B/2), sin^2(B/2)=1/5, sin(B/2)=+ or - 1/root5.

    Since it's acute, choose the + value.
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    (Original post by chewwy)
    ok then, let's try this...

    angle B is acute and tanB = 3/4

    Find sin(B/2)
    sin²(B/2) = ½(1 - cosB)
    sin²(B/2) = ½(1 - 3/5)
    sin²(B/2) = ½(2/5)
    sin²(B/2) = 1/5
    sin(B/2) = 1/√5
    ===========
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    Aaaarg! just beat
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    (Original post by JohnSPals)
    There are two flaws in this argument that cancel each other out (who said that two wrongs make a right?). I've emboldened them.

    First one: Squaring will give extra solutions. You should check all solutions in the original equation.

    Second one: Rooting will give positive and negative values (i.e. ±4/5)
    B is acute?

    Aitch
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    (Original post by Fermat)
    sin²(B/2) = ½(1 - cosB)
    sin²(B/2) = ½(1 - 3/5)
    sin²(B/2) = ½(2/5)
    sin²(B/2) = 1/5
    sin(B/2) = 1/√5
    ===========
    thankyou. i don't mean to be picky, but neither of you noticed that while tanB was 4/3 in my first question, it was 3/4 in the second, hence cosB was 4/5, and the answer is sqrt10/10
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    (Original post by chewwy)
    thankyou. i don't mean to be picky, but neither of you noticed that while tanB was 4/3 in my first question, it was 3/4 in the second, hence cosB was 4/5, and the answer is sqrt10/10
    good stuff, but a bit tricky at this time of night after 8 glasses of claret...

    Aitch
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    (Original post by chewwy)
    thankyou. i don't mean to be picky, but neither of you noticed that while tanB was 4/3 in my first question, it was 3/4 in the second, hence cosB was 4/5, and the answer is sqrt10/10
    We were just making sure you were checking the answers and not copying them down straight off TSR into your homework...

    I should be a teacher
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    (Original post by Aitch)
    B is acute?

    Aitch
    Ahh I missed that .
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    (Original post by sarforaz)
    cant we just kick you instead?
    nuff of the high school banter...
    Do you guys want a ****ing blue ticket? ^o)
    (visesh)
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    (Original post by Camford)
    nuff of the high school banter...
    Do you guys want a ****ing blue ticket? ^o)
    (visesh)
    ah man.
 
 
 
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