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:

`(y)/(\sin(63))=(44)/(\sin(50))\Rightarrow y=(44\sin(63))/(\sin(50))`
`\Rightarrow y=51.17756211\ldots`

Side x:

`(x)/(\sin(67))=(44)/(\sin(50))\Rightarrow x=(44\sin(67))/(\sin(50))`
`\Rightarrow x=52.8718848\ldots`

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

`s=(44+(44\sin(63))/(\sin(50))+(44\sin(67))/(\sin(50)))/(2)\Rightarrow s\approx74.02472346`

Then:

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

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