Question

Kaylee draws a right triangle ABC on a paper. The right angle is located at vertex A with side length BC as the hypotenuse. The side length measurements are AB¯¯¯¯¯¯¯¯=9 units, BC¯¯¯¯¯¯¯¯=41 units, and CA¯¯¯¯¯¯¯¯=40 units. What is the approximate angle measure at vertex C ?

(1 point)
Responses

55°
55°

29°
29°

77°
77°

13°

Answer 1

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°**