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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

1 Answer 1

up vote 2 down vote accepted

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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