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AQA C4 -pythagoras obtuse angle question

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    This is from the Jan 12 C4 paper:

    "Angle a is acute and cos a = 3/5. Angle b is obtuse and sin b = 1/2.
    Find the value of tan b in surd form"

    I would use pythagoras to solve this if the angle was acute and therefore had a right angle, but seeing as it is obtuse, I'm not really sure what to do.
    I just don't understand where to go with this question, can anyone explain this to me?

    Thank you very much!
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    (Original post by musicmedic)
    This is from the Jan 12 C4 paper:

    "Angle a is acute and cos a = 3/5. Angle b is obtuse and sin b = 1/2.
    Find the value of tan b in surd form"

    I would use pythagoras to solve this if the angle was acute and therefore had a right angle, but seeing as it is obtuse, I'm not really sure what to do.
    I just don't understand where to go with this question, can anyone explain this to me?

    Thank you very much!
    cant you just type in tan(sin inverse(1/2) in your caculator which would give you the answer. Thats what i did in january
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    (Original post by musicmedic)
    This is from the Jan 12 C4 paper:

    "Angle a is acute and cos a = 3/5. Angle b is obtuse and sin b = 1/2.
    Find the value of tan b in surd form"

    I would use pythagoras to solve this if the angle was acute and therefore had a right angle, but seeing as it is obtuse, I'm not really sure what to do.
    I just don't understand where to go with this question, can anyone explain this to me?

    Thank you very much!
    Obtuse indicates that the angle is in the 2nd quadrant which means tan will be negative.

    Draw a triangle by using  \sin b = \dfrac12

    Find the missing side.

    Find tan, it will have a negative in front as tan is negative in the 2nd quadrant.
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    (Original post by musicmedic)
    This is from the Jan 12 C4 paper:

    "Angle a is acute and cos a = 3/5. Angle b is obtuse and sin b = 1/2.
    Find the value of tan b in surd form"

    I would use pythagoras to solve this if the angle was acute and therefore had a right angle, but seeing as it is obtuse, I'm not really sure what to do.
    I just don't understand where to go with this question, can anyone explain this to me?

    Thank you very much!
    Inverse sin 1/2, you'll get 30 degrees or pi/6 radians. Use this value along with whatever method you please (cast/general solution/graph/plain knowledge) to find another value of x - one which is obtuse. Tan this angle on your calculator.

    (Original post by darkshadow1111)
    cant you just type in tan(sin inverse(1/2) in your caculator which would give you the answer. Thats what i did in january
    You would have got the answer wrong, then - as your calculator would've given inverse sin (1/2) as pi/6 (or 30 degrees) which is acute.
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    (Original post by darkshadow1111)
    cant you just type in tan(sin inverse(1/2) in your caculator which would give you the answer. Thats what i did in january
    It won't give the right answer.

    Why use a calc for such a question?
    This question can be done without a calc.

    Doing what you did gives the wrong answer, it gives the acute angle we need the obtuse.
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    (Original post by hassi94)
    Inverse sin 1/2, you'll get 30 degrees or pi/6 radians. Use this value along with whatever method you please (cast/general solution/graph/plain knowledge) to find another value of x - one which is obtuse. Tan this angle on your calculator.
    Why use a calc for such a question? :confused:

    Why do we need to use cast/general solutions???

    We need to draw a triangle, then find the missing side using Pythagoras theorem. Then find tan, and put a negative in front of it.
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    (Original post by raheem94)
    Why use a calc for such a question? :confused:

    Why do we need to use cast/general solutions???

    We need to draw a triangle, then find the missing side using Pythagoras theorem. Then find tan, and put a negative in front of it.
    yeah that's fair - there's many ways of doing it. I did it my way in the exam in January (except I didn't use my calculator because I knew the value anyway) - took 5 or 6 seconds.
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    (Original post by raheem94)
    Why use a calc for such a question? :confused:

    Why do we need to use cast/general solutions???

    We need to draw a triangle, then find the missing side using Pythagoras theorem. Then find tan, and put a negative in front of it.
    if you put what i said in the calulator you would get 1/root3. then put a negative in front of it gets you the right answer.
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    (Original post by hassi94)
    yeah that's fair - there's many ways of doing it. I did it my way in the exam in January (except I didn't use my calculator because I knew the value anyway) - took 5 or 6 seconds.
    My method also hardly takes some time.

    The third side is  \sqrt3
    By taking tan we get  tan \alpha  = \dfrac1{\sqrt3}

    We know tan is negative in the 2nd quadrant, so the answer is  tan b = - \dfrac1{\sqrt3}

    Though any method is fine if you know what you are doing.
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    (Original post by raheem94)
    My method also hardly takes some time.

    The third side is  \sqrt3
    By taking tan we get  tan \alpha  = \dfrac1{\sqrt3}

    We know tan is negative in the 2nd quadrant, so the answer is  tan b = - \dfrac1{\sqrt3}

    Though any method is fine if you know what you are doing.
    As I said, that's fair and I have no problem with your method. But as I said it's not like my method took any time at all either
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    (Original post by darkshadow1111)
    if you put what i said in the calulator you would get 1/root3. then put a negative in front of it gets you the right answer.
    You didn't said to put a negative in front of the answer in your previous post.

    It looks that the OP knows how to do the questions for acute angles but not for obtuse. Not explaining the other part doesn't helps the OP in any way.
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    Thank you! I had the answer  \dfrac1{\sqrt3} written down but it was the negative which I was unsure about.

    Would I be correct in saying that for any obtuse angle it would be negative then as it is in the second quadrant? Thanks
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    (Original post by musicmedic)
    Thank you! I had the answer  \dfrac1{\sqrt3} written down but it was the negative which I was unsure about.

    Would I be correct in saying that for any obtuse angle it would be negative then as it is in the second quadrant? Thanks
    It would be negative for all except sin because sin is positive in the 2nd quadrant.
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    (Original post by raheem94)
    It would be negative for all except sin because sin is positive in the 2nd quadrant.
    I'm not entirely sure I understnad the quadrant thing then Would the angle I'm calculating always be in the first quadrant if it was acute and in the second if it were obtuse? Also, do I need to worry about the 3rd/4th quadrants?

    Sorry for bugging you with all these questions!
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    (Original post by musicmedic)
    I'm not entirely sure I understnad the quadrant thing then Would the angle I'm calculating always be in the first quadrant if it was acute and in the second if it were obtuse? Also, do I need to worry about the 3rd/4th quadrants?

    Sorry for bugging you with all these questions!
    If the angle is acute then all(sin, cos, tan) will be positive.

    If it is in the second quadrant, then only sin will be positive. Hence we wrote sin as 0.5 while tan had a negative sign in front of it.

    When something is in the 2nd quadrant then the angle is between 90 and 180deg, hence the angle is obtuse.

    If the angle is reflex(greater than 180deg) then it will be in the 3rd or 4th quadrant. To decide it is in which quadrant we will have to look at the sign.

    Example:
    If a question says that  \cos \theta = \dfrac12 and  \theta is reflex.

    So due to reflex we know that it should be in the 3rd or 4th quadrant.

    We know that only tan is positive in the 3rd quadrant while only cos is positive in the 4th quadrant.

    In the question cos is positive, hence we can deduce that  \theta is in the 4th quadrant.

    Does it makes sense?
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    If you ever get stuck just use your sin and cos rules.

    \sin(\pi/2 + \theta) = sin(\pi/2 -\theta)

    \cos(\pi/2 + \theta) = -cos(\pi/2 - \theta)

    (I got those by visualising the graphs)

    so

    \tan(\pi/2 + \theta) = -tan(\pi/2 - \theta)

    in your case, \pi/2 - \theta = \pi/6, so you know the answer is -\frac{0.5}{\sqrt{3}/2}=-1/\sqrt{3}
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    (Original post by raheem94)
    If the angle is acute then all(sin, cos, tan) will be positive.

    If it is in the second quadrant, then only sin will be positive. Hence we wrote sin as 0.5 while tan had a negative sign in front of it.

    When something is in the 2nd quadrant then the angle is between 90 and 180deg, hence the angle is obtuse.

    If the angle is reflex(greater than 180deg) then it will be in the 3rd or 4th quadrant. To decide it is in which quadrant we will have to look at the sign.

    Example:
    If a question says that  \cos \theta = \dfrac12 and  \theta is reflex.

    So due to reflex we know that it should be in the 3rd or 4th quadrant.

    We know that only tan is positive in the 3rd quadrant while only cos is positive in the 4th quadrant.

    In the question cos is positive, hence we can deduce that  \theta is in the 4th quadrant.

    Does it makes sense?
    Yes, that makes perfect sense! Thank you very much I will rep you at some point (but apparently I need to rate some other members before I can rate you again! )

    Seriously though, thanks for the help!
 
 
 
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