My Math Forum Finding a side of a triangle

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February 15th, 2017, 04:14 AM   #1
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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.
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 February 15th, 2017, 04:51 AM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,751 Thanks: 1401 Let the total side opposite the 70 degree angle be $y$ ... $y= \sqrt{15^2+30^2-2(15)(30)\cos(70)}$ 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 $(140-A)$ $\dfrac{x}{\sin{A}}=\dfrac{15}{\sin(140-A)}$ You should be able to get $x$ from here ...

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