Trigonometry - Area of a Triangle (1/2 ab sin C)

In triangle ABC, side $AB = 8$ cm, side $BC = 11$ cm, and angle $ABC = 35^\circ$. Calculate the area of the triangle correct to 3 significant figures.

$\text{Area} = \frac{1}{2} \times 8 \times 11 \times \sin(35^\circ) = 44 \times 0.5735... \approx 25.2$ cm$^2$

Identify the two sides ($a=11, c=8$) and the included angle ($B=35^\circ$). Substitute these values into the formula $\text{Area} = \frac{1}{2}ac \sin B$.

The area of a triangle is $30$ cm$^2$. Two of its sides are $12$ cm and $10$ cm. Find the possible values of the included angle $\theta$.

$30 = \frac{1}{2} \times 12 \times 10 \times \sin \theta \Rightarrow 30 = 60 \sin \theta \Rightarrow \sin \theta = 0.5$. Therefore, $\theta = \sin^{-1}(0.5) = 30^\circ$ or $\theta = 180^\circ - 30^\circ = 150^\circ$.

Rearrange the formula to solve for $\sin \theta$. Remember that $\sin \theta$ is positive in both the first and second quadrants, so for a given area, the triangle could be acute ($30^\circ$) or obtuse ($150^\circ$).

An equilateral triangle has side lengths of $6$ cm. Find its area in surd form.

$\text{Area} = \frac{1}{2} \times 6 \times 6 \times \sin(60^\circ) = 18 \times \frac{\sqrt{3}}{2} = 9\sqrt{3}$ cm$^2$

In an equilateral triangle, all sides are equal ($6$ cm) and all angles are $60^\circ$. Using $a=6, b=6, C=60^\circ$ and the exact value $\sin 60^\circ = \frac{\sqrt{3}}{2}$ gives the result.