My Math Forum 2/3 Dimensional Trig problem

 Trigonometry Trigonometry Math Forum

 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: 7,635 Thanks: 2620 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

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Here is the file again..
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 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: 20,469 Thanks: 2038 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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