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

In triangle ABC, side $AB = 7\text{ cm}$, side $AC = 10\text{ cm}$, and angle $BAC = 42^\circ$. Calculate the area of the triangle correct to 3 significant figures.

$\text{Area} = \frac{1}{2} \times 7 \times 10 \times \sin(42^\circ) \approx 23.419... \approx 23.4\text{ cm}^2$

Identify the two sides ($b=10, c=7$) and the included angle ($A=42^\circ$). Substitute these values into the formula $\frac{1}{2}bc \sin A$ and calculate.

A triangle has an area of $30\text{ cm}^2$. Two of its sides are $8\text{ cm}$ and $11\text{ cm}$. Find the size of the acute angle between these two sides.

$30 = \frac{1}{2} \times 8 \times 11 \times \sin \theta \Rightarrow 30 = 44 \sin \theta \Rightarrow \sin \theta = \frac{30}{44} \Rightarrow \theta = \sin^{-1}(0.6818) \approx 43.0^\circ$

Rearrange the area formula to solve for the unknown angle $\theta$. Divide the area by the product of $0.5$ and the two sides, then use the inverse sine ($\sin^{-1}$) function.

Calculate the area of an equilateral triangle with side lengths of $6\text{ cm}$.

$\text{Area} = \frac{1}{2} \times 6 \times 6 \times \sin(60^\circ) = 18 \times \frac{\sqrt{3}}{2} = 9\sqrt{3} \approx 15.6\text{ cm}^2$

In an equilateral triangle, all sides are equal ($6\text{ cm}$) and all internal angles are $60^\circ$. Using $a=6, b=6,$ and $C=60^\circ$ allows the use of the sine area formula.

$\text{Area} = \frac{1}{2} \times PQ \times QR \times \sin(Q)$ $\text{Area} = \frac{1}{2} \times 12 \times 15 \times \sin(58^\circ)$ $\text{Area} = 90 \times 0.8480...$ $\text{Area} \approx 76.32...$ $\text{Area} = 76.3 \text{ cm}^2$

Identify the two sides ($12$ and $15$) and the angle trapped between them ($58^\circ$). Substitute these into the formula $\frac{1}{2}ab \sin C$.

$25 = \frac{1}{2} \times 8 \times 9 \times \sin(Y)$ $25 = 36 \times \sin(Y)$ $\sin(Y) = \frac{25}{36}$ $Y = \arcsin\left(\frac{25}{36}\right) \approx 43.98^\circ$ $\text{Since angle is obtuse: } Y = 180^\circ - 43.98^\circ$ $Y \approx 136.02^\circ$ $Y = 136^\circ$

Rearrange the area formula to solve for $\sin Y$. After finding the inverse sine, subtract the result from $180^\circ$ because the question specifies the angle is obtuse.