Question

in $delta klm$, $m = 12$ cm, $angle k = 76^{circ}$ and $angle l = 20^{circ}$. find the area of $delta klm$, to the nearest square centimeter.

Explanation:

Step1: Find angle $\angle M$

The sum of angles in a triangle is $180^{\circ}$. So $\angle M=180^{\circ}-\angle K - \angle L=180^{\circ}-76^{\circ}-20^{\circ}=84^{\circ}$.

Step2: Use the sine - law to find side $k$

By the sine - law $\frac{k}{\sin K}=\frac{m}{\sin M}$. We know $m = 12$ cm, $\angle K = 76^{\circ}$, and $\angle M=84^{\circ}$. So $k=\frac{m\sin K}{\sin M}=\frac{12\times\sin76^{\circ}}{\sin84^{\circ}}$. Since $\sin76^{\circ}\approx0.9703$ and $\sin84^{\circ}\approx0.9945$, then $k=\frac{12\times0.9703}{0.9945}\approx11.7$ cm.

Step3: Calculate the area of the triangle

The area of a triangle is given by $A=\frac{1}{2}km\sin L$. Substitute $k\approx11.7$ cm, $m = 12$ cm, and $\angle L = 20^{\circ}$ (and $\sin20^{\circ}\approx0.3420$). Then $A=\frac{1}{2}\times11.7\times12\times0.3420$.
$A = 6\times11.7\times0.3420=6\times4.0014 = 24.0084\approx24$ $cm^{2}$.

Answer:

$24$