Questions: The angle bisectors of triangle ABC are AV, BV, and CV. They meet at a single point V. (In other words, V is the incenter of triangle ABC) Suppose SV=8,8V=21, m angle S8T=32 degrees, and m angle UCV=26 degrees, Find the following measures.

Solution Steps

Since $CV$ is the angle bisector of $\angle ACB$, we have $m\angle UCV = m\angle TCV$. We are given that $m\angle UCV = 26^\circ$. Therefore, $m\angle TCV = 26^\circ$. Since $m\angle UCT = m\angle UCV + m\angle TCV$, we have $m\angle UCT = 26^\circ + 26^\circ = 52^\circ$.

Since $BV$ is the angle bisector of $\angle ABC$, we have $m\angle SBV = m\angle TBV$. We are given that $m\angle SBT = 32^\circ$, so $m\angle SBV = m\angle TBV = \frac{1}{2} m\angle SBT = \frac{1}{2}(32^\circ) = 16^\circ$.

We are also given that $SV = 8$ and $BV = 21$. Since $AV$, $BV$, and $CV$ are angle bisectors, $V$ is the incenter of $\triangle ABC$. The incenter is equidistant from all sides of the triangle. Let $r$ be the inradius of $\triangle ABC$. Thus, $SU = TU = UV = r$. We are given $UV = 8$, so $r=8$. Thus $SV = TU = UV = 8$.

Since $AV$ is the angle bisector of $\angle BAC$, we have $m\angle SAV = m\angle UAV$. In $\triangle SAV$, we have $\angle ASV = 90^\circ$ since $S$ is the foot of the perpendicular from the incenter $V$ to $AB$. Similarly, $\angle ATB = 90^\circ$.

In right $\triangle SBV$, $\tan(\angle SBV) = \frac{SV}{BS} = \frac{8}{BS}$. We have $\angle SBV = 16^\circ$, so $\tan(16^\circ) = \frac{8}{BS}$, which gives $BS = \frac{8}{\tan(16^\circ)} \approx \frac{8}{0.2867} \approx 27.9$.

We are given $SV=8$. Since the segments from incenter to sides are perpendicular, $\angle ASV = 90^\circ$, so $\triangle SAV$ is a right triangle.

Final Answer