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July 21st, 2015, 04:30 AM  #1 
Newbie Joined: Mar 2015 From: South Africa Posts: 6 Thanks: 0  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 
July 21st, 2015, 05:07 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra 
I feel for you. Try reading your post as if you were sitting 5,000 miles from you school as I am. 
July 21st, 2015, 05:19 AM  #3 
Newbie Joined: Mar 2015 From: South Africa Posts: 6 Thanks: 0 
The exercise problem is on the file that I attached to the thread.. Can you see the attachment?

July 21st, 2015, 05:23 AM  #4 
Newbie Joined: Mar 2015 From: South Africa Posts: 6 Thanks: 0 
Here is the file again..

July 21st, 2015, 09:38 PM  #5 
Math Team Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 
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 $PQ = PR$. 
July 21st, 2015, 11:07 PM  #6 
Global Moderator Joined: Dec 2006 Posts: 21,020 Thanks: 2256 
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, $\displaystyle \frac{\small\text{PR}}{\sin(y)} = \frac{15\text{m}}{\sin(z)}$. 
July 21st, 2015, 11:29 PM  #7 
Newbie Joined: Mar 2015 From: South Africa Posts: 6 Thanks: 0 
Thanks guys!!! I got it right 

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2 or 3, dimensional, problem, trig 
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