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February 15th, 2017, 05:14 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, 05:51 AM   #2
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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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