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    Any mathematicians here?

    Need help working out the following question please.

    See post below.
    Rep given

    18 Comments

    Original Poster

    http://www.freeimagehosting.net/uploads/b453019c80.jpg

    The answer is 0

    Original Poster

    shirishtiwari01;8806824

    The answer is 0



    That's short for Opposite side.
    Need an answer in Metres

    I got 34.7m. Not 100% sure though.

    I get 34m but hey Im only a rocket scientist! :whistling:

    i believe You need to work out the middle line length using cosine rule on bottom triangle then use the sine rule to calculate the lenght you want.
    Could be easier way but not sure.

    all i get is a headache looking at it, so my answer is painkillers lol

    Original Poster

    Spammed;8806915

    I got 34.7m. Not 100% sure though.


    scottydont;8806927

    I get 34m but hey Im only a rocket scientist! :whistling:


    Did you guys use cosine and sine rule?

    Use cosine rule for lower triangle:

    QS= root (27 squared +50 squared - (2*27*50*cos82)

    This gives QS= 53.4m

    Now use sine rule for upper triangle:

    PS= (53.4*sin38)/sin118

    This gives PS=37.2m


    Hope this helps.

    Original Poster

    ErrorOperator;8806937

    i believe You need to work out the middle line length using cosine rule … i believe You need to work out the middle line length using cosine rule on bottom triangle then use the sine rule to calculate the lenght you want.Could be easier way but not sure.



    I was trying to use sohcahtoa :thinking: ..will try again

    Is that gcse?

    SQ = sqrt(SR^2 + QR^2 - 2* SR * QR * cos(82)) = sqrt ( 50^2 + 27^2 -2 *50*27*cos(82)) ~= 53.416m using Cosine rule

    Then use the sine rule
    Sin(38)/PS = Sin(118)/53.416
    => PS = 53.416 * Sin(38) / Sin(118) = 37.246m

    Edit: Doh, too slow

    Yes sorry It's 37. I accidentally used sin 110 instead on 118. I don't think you can use soh cah toa for this as there's no right angle in either triangle.

    I was trying to use sohcahtoa ..will try again


    thats for right angle triangles

    if you to do the question yourself, the equation they are using is :
    b^2=a^2+c^2-2acCosB

    you can count me out ;-) :whistling:

    Original Poster

    Met-Cast;8807025

    Is that gcse?



    Yes. How did you guess?

    Original Poster

    Very well explained guys repped you! And thanks
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