Question

in $delta xyz$, $y = 3.6$ cm, $angle x = 158^{circ}$ and $angle y = 16^{circ}$. find the area of $delta xyz$, to the nearest 10th of a square centimeter.

Explanation:

Step1: Find angle Z

The sum of angles in a triangle is 180°. So, $\angle Z=180^{\circ}-\angle X - \angle Y=180^{\circ}-158^{\circ}-16^{\circ}=6^{\circ}$.

Step2: Use the sine - rule to find side z

By the sine - rule $\frac{y}{\sin Y}=\frac{z}{\sin Z}$. Substituting $y = 3.6$ cm, $\angle Y = 16^{\circ}$, and $\angle Z=6^{\circ}$, we get $z=\frac{y\sin Z}{\sin Y}=\frac{3.6\times\sin6^{\circ}}{\sin16^{\circ}}$.
Since $\sin6^{\circ}\approx0.1045$ and $\sin16^{\circ}\approx0.2756$, then $z=\frac{3.6\times0.1045}{0.2756}\approx1.37$ cm.

Step3: Calculate the area of the triangle

The area of a triangle is given by $A=\frac{1}{2}yz\sin X$.
Substitute $y = 3.6$ cm, $z\approx1.37$ cm, and $\angle X = 158^{\circ}$ (and $\sin158^{\circ}\approx0.3746$) into the formula.
$A=\frac{1}{2}\times3.6\times1.37\times0.3746$.
$A = 0.5\times3.6\times1.37\times0.3746=0.92$ square cm.

Answer:

$0.9$ square cm