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April 17th, 2013, 03:58 AM   #1
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Cosine and Sine laws

Some help here please. Don't get these four questions.
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 April 17th, 2013, 04:54 AM #2 Member   Joined: Jan 2012 Posts: 80 Thanks: 0 Re: Cosine and Sine laws For the first one you need to use the cosine law. There is a simple explanation here http://www.mathsisfun.com/algebra/trig-cosine-law.html So we can write: $d^{2}= 50^{2} + 52.4^{2} - 2\times50\times52.4\times\cos(25)$ $d^{2} \approx 464.7$ $d \approx \sqrt{464.7} \approx 22.3km$
 April 17th, 2013, 05:04 AM #3 Member   Joined: Jan 2012 Posts: 80 Thanks: 0 Re: Cosine and Sine laws For the second one you need to rearrange the cosine law. $\cos(\theta)= \frac{a^2 + b^2 - c^2}{2ab}$ Plugging in the values: $\cos(\theta)= \frac{4.7^2 + 3.9^2 - 2.9^2}{2\times4.7\times3.9}$ $\cos(\theta) \approx 0.788$ $\theta \approx \cos^{-1}(0.78" /> $\theta \approx 38.0^{\circ}$
 April 17th, 2013, 05:14 AM #4 Member   Joined: Jan 2012 Posts: 80 Thanks: 0 Re: Cosine and Sine laws The third one you need to use the sine law for: $\frac{\sin(A)}{a}= \frac{\sin(B)}{b} = \frac{\sin(C)}{c}$ We know that all the internal angles of a triangle add up to 180 degrees so: $A\hat{D}C= 180 - 43 -83 = 54^{\circ}$ We know the length AC and the angle opposite it and we know the angle opposite the side we wish to calculate. $\frac{\sin(43)}{CD}= \frac{\sin(54)}{31.2}$ $CD= \frac{31.2\sin(43)}{sin(54)} \approx 26.3km$
 April 17th, 2013, 05:15 AM #5 Member   Joined: Jan 2012 Posts: 80 Thanks: 0 Re: Cosine and Sine laws The fourth one is very similar to the third one.
 April 17th, 2013, 06:25 AM #6 Member   Joined: Apr 2013 Posts: 30 Thanks: 0 Re: Cosine and Sine laws Thank you SO MUCH !!! thank you, again ! Question resolved. ~

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