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February 15th, 2017, 04:14 AM  #1 
Newbie Joined: Feb 2017 From: America Posts: 2 Thanks: 0  Finding a side of a triangle
Please view the image provided for the question. Also please attempt to explain the working out of the question, thanks.

February 15th, 2017, 04:51 AM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,640 Thanks: 1319  Let the total side opposite the 70 degree angle be $y$ ... $y= \sqrt{15^2+30^22(15)(30)\cos(70)}$ So, you've got $y$ Let the top angle be $A$ ... $\dfrac{\sin{A}}{30} = \dfrac{\sin(70)}{y} \implies A = \arcsin\left[\dfrac{30\sin(70)}{y}\right]$ Top smaller triangle has angles $A$, $40$, and $(140A)$ $\dfrac{x}{\sin{A}}=\dfrac{15}{\sin(140A)}$ You should be able to get $x$ from here ... 

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