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# Find the missing side lengths. Leave your answers as radicals in simplest form.

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tan 60=12sqrt(3)/b & sin60=12sqrt(3)/a

Guest May 10, 2017
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$$\tan60º=\frac{12\sqrt3}{b} \qquad|\tan 60º=\sqrt3 \\~\\ \sqrt3=\frac{12\sqrt3}{b} \\~\\ b\sqrt3=12\sqrt3\\~\\ b=12$$

....and....

$$\sin60º=\frac{12\sqrt3}{a} \qquad|\sin60º=\frac{\sqrt3}{2}\\~\\ \frac{\sqrt3}{2}=\frac{12\sqrt3}{a} \\~\\ a\sqrt3=(2)12\sqrt3 \\~\\ a=24$$

hectictar  May 10, 2017