Question

in $delta ijk$, $j = 8.2$ inches, $angle j = 49^{circ}$ and $angle k = 57^{circ}$. find the area of $delta ijk$, to the nearest 10th of an square inch.

Explanation:

Step1: Find angle $\angle I$

The sum of angles in a triangle is $180^{\circ}$. So $\angle I=180^{\circ}-\angle J - \angle K=180^{\circ}-49^{\circ}-57^{\circ}=74^{\circ}$.

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

By the Law of Sines $\frac{j}{\sin J}=\frac{k}{\sin K}$. So $k = \frac{j\sin K}{\sin J}=\frac{8.2\times\sin57^{\circ}}{\sin49^{\circ}}$. Since $\sin57^{\circ}\approx0.8387$ and $\sin49^{\circ}\approx0.7547$, then $k=\frac{8.2\times0.8387}{0.7547}\approx9.18$.

Step3: Calculate the area of the triangle

The area of a triangle $A=\frac{1}{2}jk\sin I$. Substitute $j = 8.2$, $k\approx9.18$ and $\sin I=\sin74^{\circ}\approx0.9613$. Then $A=\frac{1}{2}\times8.2\times9.18\times0.9613\approx36.5$.

Answer:

$36.5$ square - inches