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Dodgy trig question 
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#1
Jan806, 10:21 AM

P: n/a

in the diagram (attached), ABCD is a quadrilateral where AD parallel to BC. it is given that AB=9, BC=6, CA=5 and CD=15.
show that cos BCA=1/3 and hence find the value of sin BCA i can do the 'show that' bit but i don't know how to find sinBCA. any help?? also when evaluating log15 + log20  log12 (all to the base 5) do you do the addition to multiplication conversion first or the subtraction to division first? 


#2
Jan806, 10:35 AM

P: 400

i cant see the triangle yet... beu about the logs... what does it matter what order you do it?



#3
Jan806, 12:00 PM

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re: logs
HINT 1: Multiplication is commutative HINT 2: Division is the mulitiplicative inverse. 


#4
Jan806, 01:17 PM

HW Helper
P: 876

Dodgy trig question
For the first part.
You know that [tex]\theta =\cos^{1}(1/3)[/tex] where [tex]\theta = B\hat CA[/tex] [tex]\mbox{Let } \phi = \pi /2  \theta[/tex] then, [tex]\cos\phi = \cos\theta = (1/3) = 1/3[/tex] [tex]\sin\phi = \sin\theta[/tex] Draw a small rightangled triangle, with one of the angles as [tex]\phi[/tex]. Mark the adjacent side as 1, and mark the hypotenuse as [tex]3[/tex]. So, what is the length of the opposite side? What is the value of [tex]\sin\phi[/tex] ? What is the value of [tex]\sin\theta[/tex] ? 


#5
Jan806, 05:13 PM

P: n/a

i didn't really follow that. i don't get this bit: [tex]\cos\phi = \cos\theta = (1/3) = 1/3[/tex] cos i don't think we've done stuff like that, at least we haven't been taught it, i can't find it on the syllabus and the exam is in 1 week. is there another way?? also, that post about the logs with all the long words in it, i didn't understand that either. i'm only 17. in english, possibly?? 


#6
Jan806, 05:34 PM

P: 400

hehe, comutative means you can rearange the operators in what order youd like and youd get the same answer...
so if multiplication is comutative a*b*c=c*a*b and if division by b is just multiplication by 1/b then division is also comutative... so theres no point in asking what to do first: add the logs or substruct them... 


#7
Jan806, 06:03 PM

Math
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Thanks
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#8
Jan806, 06:19 PM

P: 400

hmmm... i wonder where i went wrong here:
I. [tex]CB*BAcos(CBA)+BA*ACcos(BAC)=9=45cos(CBA)+54cos(BAC)[/tex] II. [tex]BC*CAcos(BCA)+BA*ACcos(BAC)=6=30cos(BCA)+54cos(BAC)[/tex] III. [tex]BC*CAcos(BCA)+CB*BAcos(CBA)=5=30cos(BCA)+45cos(CBA)[/tex] from IIIIII i get [tex]965=60cos(BCA)[/tex]... which means [tex]cos(BCA)=\frac{1}{30}[/tex]... hmm... you said you proved its [tex]\frac{1}{3}[/tex]? 


#9
Jan806, 06:22 PM

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#10
Jan806, 06:24 PM

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#11
Jan906, 01:44 AM

P: 400

anyway, it should give the right result... but i got [tex]cos(BCA)=\frac{1}{30}[/tex] while in the question they mentioned [tex]cos(BCA)=\frac{1}{3}[/tex] 


#12
Jan906, 02:08 AM

P: 400

heh, oops
to find the projectio of BC for example on CA you only need BCcos(BCA)... thats what ive done wrong, i multiplied the two vectors instead of just projecting them... what a silly mistake, guess thats because it was really late at night when i did it... and the floor was in a slope too 


#13
Jan1006, 12:48 PM

P: n/a

all that stuff with the lines in it looks really confusing...i just used the cosine rule...
anyway thanks for all the help i got it in the end 


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