Question

area of a triangle: law of sines!! help!! some tutors couldn’t do it so they referred me to other people!!

Answer 1

We will have that the area of the triangle is the following:

*First: We determine the missing value for the sides and angle:

Angle Y:

$Y=180-67-50\Rightarrow Y=63$

Side y:

$\frac{y}{\sin(63)}=\frac{44}{\sin(50)}\Rightarrow y=\frac{44\sin(63)}{\sin(50)}$$\Rightarrow y=51.17756211\ldots$

Side x:

$\frac{x}{\sin(67)}=\frac{44}{\sin(50)}\Rightarrow x=\frac{44\sin(67)}{\sin(50)}$$\Rightarrow x=52.8718848\ldots$

*Second: We determine the area of the triangle as follows [Heron's formula]:

$s=\frac{44+\frac{44\sin(63)}{\sin(50)}+\frac{44\sin(67)}{\sin(50)}}{2}\Rightarrow s\approx74.02472346$

Then:

$A=\sqrt[]{(74.02472346)(74.02472346-44)(74.02472346-\frac{44\sin(63)}{\sin(50)})(74.02472346-\frac{44\sin (67)}{\sin (50)})}$$\Rightarrow A\approx1874.8$

So, the area of the triangle is approximately 1874.8 mm^2.