We are given that $m\angle ACT = 15a - 8$ and $m\angle ACB = 74$. Also, S is the incenter of $\triangle ABC$. We need to find $a$ and $m\angle ACT$.

1. Problem Description

We are given that $m\angle ACT = 15a - 8$ and $m\angle ACB = 74$. Also, S is the incenter of $\triangle ABC$. We need to find $a$ and $m\angle ACT$.

2. Solution Steps

Since S is the incenter of $\triangle ABC$, CT is the angle bisector of $\angle ACB$.

Therefore, $m\angle ACT = \frac{1}{2} m\angle ACB$.

Substitute the given values:

$15a - 8 = \frac{1}{2} \times 74$

$15a - 8 = 37$

$15a = 37 + 8$

$15a = 45$

$a = \frac{45}{15}$

$a = 3$

Now, substitute $a = 3$ into the expression for $m\angle ACT$:

$m\angle ACT = 15a - 8$

$m\angle ACT = 15(3) - 8$

$m\angle ACT = 45 - 8$

$m\angle ACT = 37$

3. Final Answer

$a = 3$ and $m\angle ACT = 37$.