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 $\angle W$

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

Step2: Use the Law of Sines to find side $x$

By the Law of Sines, $\frac{v}{\sin V}=\frac{x}{\sin X}$. Substituting the given values: $\frac{7.5}{\sin33^{\circ}}=\frac{x}{\sin99^{\circ}}$. Then $x = \frac{7.5\times\sin99^{\circ}}{\sin33^{\circ}}$.
We know that $\sin33^{\circ}\approx0.5446$ and $\sin99^{\circ}\approx0.9877$. So, $x=\frac{7.5\times0.9877}{0.5446}\approx13.6$.

Step3: Calculate the area of the triangle

The area formula for a triangle is $A=\frac{1}{2}vx\sin W$. Substitute $v = 7.5$, $x\approx13.6$, and $\sin W=\sin48^{\circ}\approx0.7431$.
$A=\frac{1}{2}\times7.5\times13.6\times0.7431$.
$A = 3.75\times13.6\times0.7431$.
$A=37.5486\approx37.5$ square - inches.

Answer:

$37.5$ square inches