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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: squance on January 18, 2009, 10:17:15 pm

Title: Sine and Cosine rules
Post by: squance on January 18, 2009, 10:17:15 pm: Hi! I'm squance's sister and i have question about the sine and cosine rule that i would like some help with.

Question:
The adjacent sides of a parallelogram are 9 cm and 11 cm. One of its angles is 67*. Find the length of the longer diagonal, correct to two decimal places.

i keep getting close to the answer but i can't actually get the answer.

Question 2
In triangle ABC, <ABC=35*, AB=10cm and BC=4.7. Find, correct to two decimal places:
<ACB.

this one, i'm not sure what i'm doing wrong because i keep getting answers that are not alike to the answers in the back of the book.

thanks :)
Title: Re: Sine and Cosine rules
Post by: squance on January 18, 2009, 11:02:45 pm: oops...um..i posted the wrong question! sorry..i could solve that problem but i can't solve this question:

in the triangle ABC, <ACB=34*, AC=8.5cm and AB=5.6cm. Find, correct to two decimal places:
BC in each case.

sorry about that...i guess i was a bit tired.

oh. and i still can't solve the first question :)

thanks for trying the other question..but umm..its not right..but thanks anyway ;D
Title: Re: Sine and Cosine rules
Post by: Flaming_Arrow on January 18, 2009, 11:11:45 pm: is BC = 10 cm?
Title: Re: Sine and Cosine rules
Post by: squance on January 18, 2009, 11:17:31 pm: yep thats one of the answers!! Thanks :) How did you get it??? oh and how would you go about finding the other answer??? :)
Title: Re: Sine and Cosine rules
Post by: Flaming_Arrow on January 18, 2009, 11:24:17 pm: first workout $\angle ABC$

$\frac {5.6}{\sin 34^o} = \frac{8.5}{\sin \theta}$

$\angle ABC = 58.08^o$

then workout $\angle BAC= 180^o-(58.08^o + 34^o) = 87.92^o$

$\frac {5.6}{\sin 34^o} = \frac{BC}{\sin 87.92^o}$

$AC = 10cm$
Title: Re: Sine and Cosine rules
Post by: squance on January 18, 2009, 11:29:07 pm: thanks
Title: Re: Sine and Cosine rules
Post by: Flaming_Arrow on January 18, 2009, 11:30:41 pm: $a^2 = 202 - 2 \cdot 9 \cdot 11 \cos 113^p$

$a = 16.71cm$
Title: Re: Sine and Cosine rules
Post by: squance on January 18, 2009, 11:36:17 pm: thanks..but where did you get the 113 degrees from???
Title: Re: Sine and Cosine rules
Post by: Flaming_Arrow on January 18, 2009, 11:38:52 pm: opposites angles of a parallelogram are equal, so $360 - 134 = \frac {226}{2} = 113^o$
Title: Re: Sine and Cosine rules
Post by: squance on January 18, 2009, 11:41:39 pm: oh! thanks!!! i get it now :)