To find the angle measure at vertex C in triangle ABC, we can use the cosine rule or simply use trigonometry because we know the lengths of all three sides.
Using the cosine rule:
\[ \cos(C) = \frac{AB^2 + AC^2 - BC^2}{2 \cdot AB \cdot AC} \]
Substituting the lengths of the sides:
\[ AB = 9, \quad AC = 40, \quad BC = 41 \]
We can plug these values into the formula:
\[ \cos(C) = \frac{9^2 + 40^2 - 41^2}{2 \cdot 9 \cdot 40} \]
Calculating \(9^2\), \(40^2\), and \(41^2\):
\[ 9^2 = 81, \quad 40^2 = 1600, \quad 41^2 = 1681 \]
Now substituting these values into the cosine formula:
\[ \cos(C) = \frac{81 + 1600 - 1681}{2 \cdot 9 \cdot 40} \] \[ \cos(C) = \frac{81 + 1600 - 1681}{720} \] \[ \cos(C) = \frac{0}{720} = 0 \]
Since \(\cos(C) = 0\), it implies that angle C is \(90°\). However, this cannot be correct since \(C\) must be an angle in a right triangle and cannot exceed \(90°\). Let's double-check using the sine formula instead, due to an apparent earlier mistake.
Using the sine approach instead:
\[ \sin(C) = \frac{opposite}{hypotenuse} = \frac{AB}{BC} = \frac{9}{41} \]
Calculating the angle C:
\[ C = \arcsin\left(\frac{9}{41}\right) \]
Calculating this using a scientific calculator, we find \(C\):
\[ C \approx \arcsin(0.2195) \approx 12.6° \]
Let's check other options. Continuing the calculation reinforces:
\[ \sin^{-1}\left(\frac{9}{41}\right) \approx 13° \]
Thus, the angle at vertex C is approximately:
13°