Question

in $\triangle def, d = 2.7$ cm, $angle e = 6^{circ}$ and $angle f = 67^{circ}$. find the area of $\triangle def$, to the nearest 10th of a square centimeter.

Explanation:

Step1: Find angle D

The sum of angles in a triangle is 180°. So, $\angle D=180^{\circ}-\angle E - \angle F=180^{\circ}-6^{\circ}-67^{\circ}=107^{\circ}$.

Step2: Use the sine - law to find side e

By the sine - law, $\frac{d}{\sin D}=\frac{e}{\sin E}$. Substituting the known values: $e = \frac{d\sin E}{\sin D}=\frac{2.7\times\sin6^{\circ}}{\sin107^{\circ}}$.
We know that $\sin6^{\circ}\approx0.1045$ and $\sin107^{\circ}\approx0.9563$. Then $e=\frac{2.7\times0.1045}{0.9563}=\frac{0.28215}{0.9563}\approx0.295$.

Step3: Calculate the area of the triangle

The area of a triangle is given by $A=\frac{1}{2}de\sin F$.
Substitute $d = 2.7$, $e\approx0.295$, and $\sin F=\sin67^{\circ}\approx0.9205$.
$A=\frac{1}{2}\times2.7\times0.295\times0.9205=\frac{1}{2}\times2.7\times0.2715475\approx0.367$.

Answer:

$0.4$ cm²