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, $\angle Z=180^{\circ}-\angle X - \angle Y=180^{\circ}-8^{\circ}-147^{\circ}=25^{\circ}$.

Step2: Use the sine - law to find side x

By the sine - law, $\frac{x}{\sin X}=\frac{y}{\sin Y}$. Substituting the given values, $\frac{x}{\sin8^{\circ}}=\frac{55}{\sin147^{\circ}}$. Then $x = \frac{55\sin8^{\circ}}{\sin147^{\circ}}$.
We know that $\sin8^{\circ}\approx0.1392$ and $\sin147^{\circ}\approx0.5446$. So, $x=\frac{55\times0.1392}{0.5446}=\frac{7.656}{0.5446}\approx14.1$.

Step3: Calculate the area of the triangle

The area of a triangle is given by $A=\frac{1}{2}xy\sin Z$. Substitute $x\approx14.1$, $y = 55$, and $\sin Z=\sin25^{\circ}\approx0.4226$.
$A=\frac{1}{2}\times14.1\times55\times0.4226=\frac{1}{2}\times14.1\times23.243 = 14.1\times11.6215\approx163.9$.

Answer:

$163.9$ square inches