Triangle trig

[peep1963]

All right I undestand conceptual word problems but I don't get this.
It's geometry AHHHHHhhhhhhhhhhhhh!!!!!!!!!!!!!

Okay here's the problem.
(Sorry typing verbatum, don't know how to give or draw a visual.)

In the diagram at the right (there is none), line segment AC is tangent to a semicircle with radius OA=12 cm. If the measure of angle AOB is 28 degrees, find to the nearest square centimeter the area of the region that is inside triangle AOC and outside the circle.

I'll try to explain the visual.

There's a semicircle. O is at the bottom in the middle. A is at one edge of the semicircle. B is the edge of the 28 degree angle. C aligns directly above the A but is outside of the smicircle. Sorry that's the best I can do. If someome knows how I can draw a visual of this, please let me know.

 

[pka]

Go to the top of this webpage.
Click on the tab ‘Forum Help’.
Then pull down ‘Inserting Images’.
I think this one needs a diagram.

 

[galactus]

I took a long shot here. Is this it?.

You need the area of that portion of the triangle which lies outside the

circle?.

If so, find the area of the triangle by using \(\displaystyle 12tan({\theta})\)

to get the length of segment AC.

Find the area of the sector of the circle by using \(\displaystyle \frac{1}{2}r^{2}{\theta}\)

Subtract them.

Remember to first change the 28 degrees to radians.

 

[soroban]

I saw a different diagram.

      C
      *
      ::\
      ::::\ D
      ::::::* * *
      :::*    \    *
      :*        \    *
      :           \
      *          28°\ *
      *-------*-------*
      A   12  O   12  B

Draw radius OD.
Note that: \(\displaystyle \,\angle DOA\,=\,56^o,\;\angle DOB\,=\,124^o\)


\(\displaystyle \text{Shaded region }=\\Delta CAB)\,-\,(\text{sector }DOA)\,-\,(\Delta DOB)\)

\(\displaystyle \;\;\Delta CAB\:=\:\frac{1}{2}\cdot24(24\cdot\tan28^o) \:=\:288\cdot\tan28^o\)

\(\displaystyle \text{sector }DOA\:=\:12^2\pi\cdot\frac{56}{360}\;=\;\frac{112\pi}{5}\)

\(\displaystyle \;\;\Delta DOB\:=\:\frac{1}{2}\cdot12^2\cdot\sin124^o\;=\;72\cdot\sin124^o\)


My answer: \(\displaystyle \,23.069935165\;\approx\;23\) cm².