<s>[INST] How would I find the area of a triangle given three sides and using either the sine/cosine laws? Triangle ABC has sides $8.5m$ (a), $7.1$ (b), and $9$ (c). I have been asked to find the area of the triangle using trigonometry. 
 [/INST] If you must use Trigonometry, we can use this formula:
$$ K = \frac{1}{2} ab \sin C $$
In order to find the $ \sin C $, we can use the law of cosines:
$$ c^2 = a^2 + b^2 - 2ab \cos C $$
$$ 9^2 = 8.5^2 + 7.1^2 - 2 \cdot 8.5 \cdot 7.1 \cos C $$
$$ -41.66 = -120.7 \cos C $$
$$ 0.34515327257 = \cos C $$
$$ C \approx 69.80^{\circ} $$
Now, we have:
$$ K = \frac{1}{2} \cdot 8.5 \cdot 7.1 \sin 69.8 $$
$$ K \approx 30.175 \sin 69.8 $$
$$ \color{green}{K \approx 28.32} $$
If you only have a scientific calculator, you can also avoid the inverse cosine by using:
$$ \sin x = \sqrt{1 - \cos^2 x}$$
So:
$$ K = \frac{1}{2} \cdot 8.5 \cdot 7.1 \sqrt{1 - 0.34515327257^2} $$
$$ K = \frac{1}{2} \cdot 8.5 \cdot 7.1 \cdot 0.9385463325985667 $$
$$ K = 28.3206355862 $$
Applying Heron's Formula verifies this result. We have $s = 12.3$.
$$ \sqrt{12.3(12.3 - 8.5)(12.3 - 7.1)(12.3 - 9)} $$
$$ \sqrt{802.058} \approx 28.3206 $$
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