geometry

What is the radius of the circle inscribed in triangle ABC if AB = 12, AC=14, BC=18? Express your answer in simplest radical form.

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The law of cosines:

\(18^2=14^2+12^2-2\cdot 14\cdot 12\cdot cos\ \alpha\\ cos\ \alpha = \frac{14^2+12^2-18^2}{2\cdot 14\cdot 12}=0.\overline{047619}\)

\(\alpha=87.271°\\ \)

\(cos\ \beta =\frac{18^2+12^2-14^2}{2\cdot 12\cdot 18}=0.\overline{629}\)

\(\beta=50.977°\)

The functions of the bisector in A and B:

\(f_A(x)=tan\ (\frac{\alpha}{2})\cdot x=0.95346\ x\)

\(f_B(x)=-tan\ (\frac{\beta}{2})\cdot x+12\ tan\ (\frac{\beta}{2}) \\ f_B(x)=-tan\ (25.4886°)\cdot x+12\ tan\ (25.4886°) \\ \color{blue}f_B(x)=-0.47673\ x+5.72078\)

Equate the functions

\(0.95346\ x=-0.47673\ x+5.72078\\ 1.43019\ x=5.72078\\ x=4\\ \color{blue}y=3.81385\)

The radius of the circle inscribed in triangle ABC is 3.81385.

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