Inscribed circle: calculate area of triangle formed in botto

[Mitch251]

Given the picture above, I calculate the inscribed circle to be (rounding to the nearest hundred) 18.200. I so far have had it verified that this is correct. I am now being asked to calculate the area of the triangle formed in the bottom right corner. I have absolutely no idea how to go about this.

 

[Mitch251]

please note that all angles are unknown

 

[pka]

I do not know if this is the entire problem as it was given.
But you must have found the lengths of each tangent, say AN & AM for example.
I assume the triangle is something like ANM?
The segment NM is a cord of the circle. Find its length.
Then with the three sides use Heron’s Theorem.

 

[soroban]

Hello, Mitch!

I found that the radius of the inscribed circle is about $\displaystyle 76.064$


Using Heron's Formula: $\displaystyle \,A\:=\:\sqrt{s(s\,-\,a)(s\,-\,b)(s\,-\,c)}$

We have: $\displaystyle \,s\:=\:\frac{330\,+\,270\,+\,240}{2}\:=\:420$

Hence: $\displaystyle \,A\:=\:\sqrt{420(90)(150)(180)} \:=\:5400\sqrt{35}$


The radius $\displaystyle (r)$ of the inscribed circle is given by: .$\displaystyle A \:=\:s\cdot r$

Therefore: \(\displaystyle \L\,r\:=\:\frac{A}{s}\:=\:\frac{5400\sqrt{35}}{420}\:=\:\frac{90\sqrt{35}}{7} \:\approx\:76.064\)

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Let $\displaystyle \theta \,=\,\angle CAB$

From the Law of Cosines: $\displaystyle \,\cos\theta \:=\:\frac{240^2\,+\,270^2\,-\,330^2}{2(240)(270)} \:=\:\frac{1}{6}$

Hence: $\displaystyle \,\theta \:=\:\cos^{-1}\left(\frac{1}{6}\right) \:\approx\:80.406^o\;\;\Rightarrow\;\;\frac{1}{2}\theta\,=\,40.203^o$

In the lower-right, we have this right triangle:

      *
      |   * 
      |       *
    r |           *
      |               *
      |               ½θ  *
      *- - - - - - - - - - - *
                  x

We have: $\displaystyle \,\tan\left(\frac{1}{2}\theta\right) \:=\:\frac{r}{x}\;\;\Rightarrow\;\;x \:=\:\frac{r}{\tan\left(\frac{1}{2}\theta\right)} \:=\:\frac{76.064}{\tan40.203^o}\:\approx\:90$

The lower-right triangle is an isosceles triangle with equal sides $\displaystyle x$ and vertex angle $\displaystyle \theta$.

The area is: \(\displaystyle \L\,A\:=\:\frac{1}{2}x^2\sin\theta\:=\:\frac{1}{2}(90^2)\sin80.4065^o \:\approx\:3993.35\)