Question

in $delta xyz$, $y = 55$ inches, $angle x = 8^{circ}$ and $angle y=147^{circ}$. find the area of $delta xyz$, to the nearest 10th of an square inch.

Explanation:

Step1: Find angle Z

The sum of angles in a triangle is 180°. So, $\angle Z=180^{\circ}-\angle X - \angle Y$.
$\angle Z = 180^{\circ}-8^{\circ}-147^{\circ}=25^{\circ}$

Step2: Use the sine - law to find side x

By the sine - law, $\frac{x}{\sin X}=\frac{y}{\sin Y}$.
$x=\frac{y\sin X}{\sin Y}$.
Substituting $y = 55$, $\angle X = 8^{\circ}$, and $\angle Y=147^{\circ}$, we have $x=\frac{55\sin8^{\circ}}{\sin147^{\circ}}$.
$\sin8^{\circ}\approx0.1392$, $\sin147^{\circ}\approx0.5446$.
$x=\frac{55\times0.1392}{0.5446}=\frac{7.656}{0.5446}\approx14.06$.

Step3: Calculate the area of the triangle

The area of a triangle is given by $A=\frac{1}{2}xy\sin Z$.
Substitute $x\approx14.06$, $y = 55$, and $\angle Z = 25^{\circ}$ (and $\sin25^{\circ}\approx0.4226$).
$A=\frac{1}{2}\times14.06\times55\times0.4226$.
$A=\frac{1}{2}\times14.06\times23.243$.
$A=\frac{326.79658}{2}\approx163.4$.

Answer:

$163.4$ square inches