# 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.

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Hint: Heron's Formula. You can also use the Sine Formula if you want sines and cosines. Regards – Amzoti Jan 21 '13 at 2:55

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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