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hello,
i am having a problem with law of cosines. i am using the C^2 = A^2-2ABcosc+B^2
what do i do if i dont have a value for A?
i have tried to rearrange to find A but with no luck.
any help would be great.
i get to here usually,
2ABcosc = A^2+B^2-C^2,
i then work out 2Bcosc, as i would have those values but no A
sometimes i think square root both sides. this would remove the exponents on the right and square everything on the left but that does not work.
i am using a book called "trigonometry: essentials practice workbook with answers" by chris mcmullen. it uses only certain values that mean you do not require a calculator. the values you use are 0, 30, 45 and 90 degrees. sine, cosine and tangent for these values and only those values.
any help would be great thanks
simon
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hi ninjaman
There are two ways I can think of.
(1) As you know angle C, and sides c and b, you could use the sine rule instead to get angle B. Once you know C and B you can work out A and hence use the sine rule again to get side a.
(2) Write the cosine equation as a quadratic in side a, and use the quadratic formula to solve for a.
note: (1) When you so asin() to get angle B, be prepared for two answers, one acute and one obtuse; hence two answers for a.
(2) If you choose the quadratic method you'll get two answers for your quadratic.
If you try a ruler and protractor construction, you'll see why this happens.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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thanks bob, unfortunately i dont understand the ruler/protracto method. the book i am using doesnt go into that sort of thing. thanks for your help.
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In method (1) described by bob bundy sometimes you may get sinB>1. Nothing to bother: the triangleisimpossible. If sinB<1 , two values of B and 2 values of a . Both triangles are possible. Only if sinB=1 B=90 degrees and only one triangle is possible.
Same thing with method(2) if both values of a are positive there are two possible triangles whereas if one value is negative and one positive positive one is the valid one.If both are imaginary there is no solution.
{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
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hi ninjaman
i dont understand the ruler/protractor method.
I just meant try to draw it. You could start with any line length b. Make an angle of C at one end. Label that end with letter C and the other with letter A. At this stage you won't know how long to make CB so just make it a long line and don't write on B yet. Set a compass to radius c and draw an arc from A to cut CB. Three things could happen: (1) c isn't long enough to reach CB. That means the triangle is impossible. (2) c just touches CB once. In that case the triangle is right angled at B. (3) Most likely case; the protractor cuts CB in two places. Either place makes a possible triangle so you'll have two answers, B' and B''.
Construction is not an acceptable way to get the answer if the question was to test your ability with trig. but it helps to at least imagine it as it gives you the 'tip off' that two answers are possible. Both calculation methods will yield two answers, so there's a sensible match between construction and calculation. (Just as well, or we'd have to reject the calculation method as an inadequate way to get an answer. )
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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If you have C as obtuse angle there will be unique solution or no solution at all if c is too small.
{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
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