Problems from Law of sines and cosines

Given a triangle with $$2$$ sides of $$5$$ and $$6$$ centimeters that form an angle of $$60$$ degrees, find the third side.

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Development:

Let's draw the triangle:

imagen

In this case, we cannot use the theorem of the sine, since we do not know any angle opposite to the known sides. We can, however, apply the theorem of the cosine to find the unkown side of the triangle. So, from the theorem of the cosine, we have:

$$$a^2=b^2+c^2-2\cdot b \cdot c \cdot \cos(\alpha)=6^2+5^2-2\cdot6\cdot5\cdot \cos(60)=$$$ $$$36+25-60\cdot\dfrac{1}{2}=36+25-30=31$$$ $$$a=\sqrt{31}$$$

Solution:

$$a=\sqrt{31}$$ cm

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