Question

in $delta vwx, v = 7.5$ inches, $angle x = 99^{circ}$ and $angle v = 33^{circ}$. find the area of $delta vwx$, to the nearest 10th of an square inch.

Explanation:

Step1: Find angle $\angle W$

Since the sum of angles in a triangle is $180^{\circ}$, $\angle W=180^{\circ}-\angle X - \angle V$. So, $\angle W = 180^{\circ}-99^{\circ}-33^{\circ}=48^{\circ}$.

Step2: Use the sine - law to find side $w$

By the sine - law $\frac{v}{\sin V}=\frac{w}{\sin W}$. Substituting the given values: $\frac{7.5}{\sin33^{\circ}}=\frac{w}{\sin48^{\circ}}$. Then $w=\frac{7.5\times\sin48^{\circ}}{\sin33^{\circ}}$. We know that $\sin33^{\circ}\approx0.5446$ and $\sin48^{\circ}\approx0.7431$. So $w=\frac{7.5\times0.7431}{0.5446}\approx10.27$.

Step3: Calculate the area of the triangle

The area of a triangle is $A = \frac{1}{2}vw\sin X$. Substituting $v = 7.5$, $w\approx10.27$, and $\sin X=\sin99^{\circ}\approx0.9877$. Then $A=\frac{1}{2}\times7.5\times10.27\times0.9877\approx38.8$.

Answer:

$38.8$ square inches