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July 21st, 2015, 04:30 AM   #1
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2/3 Dimensional Trig problem

Hi,

I have tried, but failed to solve question 4.2.2 and 4.2.3 of the exercise problem.

Any help please!!

I tried to use the sine rule and cosine rule but can't get side PR.

Thanks,
Alexeia
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July 21st, 2015, 05:07 AM   #2
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I feel for you.

Try reading your post as if you were sitting 5,000 miles from you school as I am.
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July 21st, 2015, 05:19 AM   #3
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The exercise problem is on the file that I attached to the thread.. Can you see the attachment?
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July 21st, 2015, 05:23 AM   #4
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Here is the file again..
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File Type: jpg Trig2dProb.jpg (102.9 KB, 13 views)
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July 21st, 2015, 09:38 PM   #5
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For 4.2.2, note that $\angle QPR = 180 - (x + y)$. Now use the sine rule on $\triangle PQR$, focusing on $\angle QPR$ and $y$.

For 4.2.3, note that when $x = y$, then $PR = \dfrac{15\sin y}{\sin 2y}.$ Simplify that. Also, note that if $x = y$, $\triangle PQR$ is isosceles, so $PQ = PR$.
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July 21st, 2015, 11:07 PM   #6
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4.2.1 $\ {\small\text{PW} = \text{PQ}} \tan(\alpha)$

4.2.2 $\ $If ${\small\text{Q}\hat{\text{P}}\text{R}} = z,\ \sin(z) = \sin(x + y)$.

$\ \ \ \ \ \ \ \ \,\,$ By the sine rule, $\displaystyle \frac{\small\text{PR}}{\sin(y)} = \frac{15\text{m}}{\sin(z)}$.
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July 21st, 2015, 11:29 PM   #7
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Thanks guys!!! I got it right
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