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Question: In triangle ABC, angle A = 60°, AC = 21, BC = 19. The circle inscribed in triangle ABC touches the sides AB, BC, and CA at points P, Q, R respectively. Find all possible lengths of all AB, BQ and the radius of the inscribed circle.

Using the sine law, I have found that angle $B = 73.17355^\circ$ and angle $C = 46.82645^\circ$. I also found that the missing side length $C$ is $16$.

However, I don't understand how I can somehow use this knowledge to find the radius of the circle inscribed within the triangle. In addition, how can there be more than one length for $AB$ and $BQ$?

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From the Cosine Law,

$$AB^2+AC^2-BC^2=2\cdot AB\cdot AC\cdot \cos A \implies AB = 5 \text{ or } 16$$ The area of $\triangle ABC$ can be written as the following two ways: $$S_{ABC}=\frac{1}{2}\cdot AB\cdot AC\cdot \sin A = \frac{1}{2}\cdot(AB+AC+BC)\cdot r$$ where $r$ is the radius of the incircle of $\triangle ABC$. This will give us $$r=\frac{AB\cdot AC\cdot \sin60^\circ}{AB+AC+BC}$$ By plugging in the values of $AB, AC, BC$ and $\sin A$, we can get $r$. Further, because $BP=BQ,\ CQ=CR,\ AR=AP$, and $$BP+AP=AB\\BQ+CQ=BC\\AR+RC=AC$$ we know that $$BQ = \frac{BA+BC-AC}{2}$$Therefore, the following two cases are:

Case 1: $AB=5$, then $r=\frac{7\sqrt{3}}{6}$ and $BQ=\frac{3}{2}$.

Case 2: $AB = 16$, then $r = 3\sqrt{3}$ and $BQ = 7$.

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  • $\begingroup$ Thank you so much for the solution. However, I am confused about why these two expressions are equal: SABC=1/2⋅AB⋅AC⋅sinA=1/2⋅(AB+AC+BC)⋅r and why BQ = (BA+BC-AC/2) $\endgroup$ – jenboo12138 Apr 10 '17 at 3:12

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