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Irregular quadrilateral finding path/angle watch

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    Specimen paper 2015 question 6 is split into 3 parts and I'm stuck on the 3rd part.

    It starts with this quadrilateral, http://i.imgur.com/H8mnPyu.jpg

    (a) - to which I proved the diagonal's length is 135 by using the law of cosine
    120^{2} + 95^{2} - 2 (120)(95) cos 77
    \sqrt(ans) = 135.26cm

    (b) - Then it asks to calculate the obtuse angle of ABC, to which I used the sine rule.
    angle B = sin^{-1}(\frac{sin 26}{79} \cdot 135) = 48.5

    Here's where it gets tricky for me. An obtuse angle is 90<x<180, right? So obviously I had to draw to extend the line and then subtract 180 out of 48.5 to get 131.5. Which is right in the markscheme.

    (c) - Straight path is built from B to nearest point on AC. Calculate the length.

    I tried solving it by first finding angle A by simply adding angle B and C. Like so,
    http://i.imgur.com/rKUxmEl.jpg

    But it's wrong. In the markscheme it says angle A is 22.5 and it doesn't make sense to me AT ALL. Because 22.5 + 26 (given angle) + 48.5 (found using sine rule plus the fact it's right in (b) does not equal 180!

    I just don't get it. Help anyone?


    markscheme: http://puu.sh/l954n/652c230831.png
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    (Original post by throwaway800)
    Specimen paper 2015 question 6 is split into 3 parts and I'm stuck on the 3rd part.

    It starts with this quadrilateral, http://i.imgur.com/H8mnPyu.jpg

    (a) - to which I proved the diagonal's length is 135 by using the law of cosine
    120^{2} + 95^{2} - 2 (120)(95) cos 77
    \sqrt(ans) = 135.26cm

    (b) - Then it asks to calculate the obtuse angle of ABC, to which I used the sine rule.
    angle B = sin^{-1}(\frac{sin 26}{79} \cdot 135) = 48.5

    Here's where it gets tricky for me. An obtuse angle is 90<x<180, right? So obviously I had to draw to extend the line and then subtract 180 out of 48.5 to get 131.5. Which is right in the markscheme.

    (c) - Straight path is built from B to nearest point on AC. Calculate the length.

    I tried solving it by first finding angle A by simply adding angle B and C. Like so,
    http://i.imgur.com/rKUxmEl.jpg

    But it's wrong. In the markscheme it says angle A is 22.5 and it doesn't make sense to me AT ALL. Because 22.5 + 26 (given angle) + 48.5 (found using sine rule plus the fact it's right in (b) does not equal 180!

    I just don't get it. Help anyone?


    markscheme: http://puu.sh/l954n/652c230831.png
    You've worked out angle ABC (131.486...) and you've got angle ACB (26) so angle BAC is 180-131.486-26=22.514... Remember that angle ABC is 131.486... not 48.5. As you may or may not know, because of the shape of the sine graph, there are actually two possible solutions when you use the sine rule but the sine function on your calculator only gives you the acute answer. This is the reason why in part (b) you had to subtract your answer (48.5) from 180 to get the real answer (131.5).
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    Thank you. I tried to search up on more answers based on your 'sine graph... two possible solutions' etc and found there's an 'ambiguous case' when it comes to it. Sadly because I'm self-teaching, I think I totally missed that part.

    For anyone anyone who'd ever stumble in here, this video explained it perfectly clear https://youtu.be/URmhII_IyPE?list=PL...Djo7fb2a&t=509
 
 
 
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