SOLUTION: Hi, could you please help me with the question (10ii) in the following exam paper. This is the link, thank you. http://papers.xtremepapers.com/CIE/Cambridge%20International%20A

Algebra ->  Parallelograms -> SOLUTION: Hi, could you please help me with the question (10ii) in the following exam paper. This is the link, thank you. http://papers.xtremepapers.com/CIE/Cambridge%20International%20A      Log On


   

Question 865766: Hi, could you please help me with the question (10ii) in the following exam paper. This is the link, thank you.
http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_s10_qp_11.pdf
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
NOTE: Check my calculations, because it is way past my bedtime.

The diagonals are
%22=%22%22%2B%22%22=%22%28i%2B3j%2B3k%29%2B%283i-j%2Bk%29=4i%2B2j%2B4k
AND
%22=%22%22-%22%22=%22%283i-j%2Bk%29-%28i%2B3j%2B3k%29=2i-4j-2k%29
So OB=sqrt%284%5E2%2B2%5E2%2B4%5E2%29=sqrt%2816%2B4%2B16%29=sqrt%2836%29=6 and
AC=sqrt%282%5E2%2B4%5E2%2B2%5E2%29=sqrt%284%2B16%2B4%29=sqrt%2824%29=2sqrt%286%29
The diagonals bisect each other and split parallelogram OABC into 4 triangles.
Two sides of each triangle will be the halves of the two diagonals.
That is, two sides will have lengths of 6%2F2=3 and 2sqrt%286%29%2F2=sqrt%286%29 .
The remaining side of each triangle will be one side of the parallelogram, and not all sides of the parallelogram are the same length. (It is not a rhombus).
The parallelogram's side lengths are:
OA=sqrt%281%5E2%2B3%5E3%2B3%5E2%29=sqrt%281%2B9%2B9%29=sqrt%2819%29
OC=sqrt%283%5E3%2B1%5E2%2B1%5E2%29=sqrt%289%2B1%2B1%29=sqrt%2811%29
The angle between the sides measuring 3 and sqrt%286%29 will be smaller in the two triangles where it is opposite the shorter parallelogram side measuring sqrt%2811%29 .
We need the measure of that angle, that I will call theta .
The law of cosines, applied to that triangle says that

11=9%2B6-6sqrt%286%29cos%28theta%29
6sqrt%286%29cos%28theta%29=9%2B6-11
6sqrt%286%29cos%28theta%29=4
cos%28theta%29=4%2F6sqrt%286%29
cos%28theta%29=2%2F3sqrt%286%29
cos%28theta%29=2sqrt%286%29%2F%283%2A6%29
cos%28theta%29=sqrt%286%29%2F9 ---> theta=74.2%5Eo