Question

solve the triangle shown to the right.
a ≈ □°, b ≈ □°, and c ≈ □°
(round to the nearest tenth as needed.)

Explanation:

Step1: Identify the triangle type

This is an isosceles triangle with \( AB = AC = 11 \), \( BC = 13 \). So, \( \angle A=\angle B \)? Wait, no, wait: \( AC = 11 \), \( AB = 11 \), \( BC = 13 \). So sides: \( AC = 11 \), \( AB = 11 \), \( BC = 13 \). So angles opposite: \( \angle B \) opposite \( AC \), \( \angle A \) opposite \( BC \)? Wait, no, let's label correctly. In triangle \( ABC \), \( A \) is connected to \( B \) (length 11), \( A \) to \( C \) (length 11), \( B \) to \( C \) (length 13). So sides: \( AB = 11 \), \( AC = 11 \), \( BC = 13 \). So angle at \( A \) is between \( AB \) and \( AC \), angle at \( B \) is between \( AB \) and \( BC \), angle at \( C \) is between \( AC \) and \( BC \). So using the Law of Cosines to find angle \( A \) first.

Law of Cosines: \( BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos A \)

So \( 13^2 = 11^2 + 11^2 - 2 \cdot 11 \cdot 11 \cdot \cos A \)

Step2: Calculate \( \cos A \)

\( 169 = 121 + 121 - 242 \cos A \)

\( 169 = 242 - 242 \cos A \)

Subtract 242 from both sides: \( 169 - 242 = -242 \cos A \)

\( -73 = -242 \cos A \)

Divide both sides by -242: \( \cos A = \frac{73}{242} \approx 0.30165 \)

Step3: Find \( \angle A \)

\( \angle A = \arccos(0.30165) \approx 72.4^\circ \)

Step4: Find \( \angle B \)

Since \( AB = AC \), triangle is isosceles with \( \angle B = \angle A \)? Wait, no, wait: \( AB = AC = 11 \), so the base is \( BC \), so the base angles are \( \angle B \) and \( \angle C \)? Wait, I made a mistake earlier. Let's re-label: In triangle \( ABC \), \( AB = 11 \), \( AC = 11 \), \( BC = 13 \). So the two equal sides are \( AB \) and \( AC \), so the vertex is \( A \), and the base is \( BC \). Therefore, the base angles are \( \angle B \) and \( \angle C \), so \( \angle B = \angle C \). Wait, that's my mistake. So angle at \( A \) is the vertex angle, angles at \( B \) and \( C \) are equal.

So let's correct: Law of Cosines for angle \( A \):

\( BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos A \)

\( 13^2 = 11^2 + 11^2 - 2 \cdot 11 \cdot 11 \cdot \cos A \)

\( 169 = 121 + 121 - 242 \cos A \)

\( 169 = 242 - 242 \cos A \)

\( 242 \cos A = 242 - 169 = 73 \)

\( \cos A = \frac{73}{242} \approx 0.30165 \)

\( \angle A = \arccos(0.30165) \approx 72.4^\circ \)

Then, since the sum of angles in a triangle is \( 180^\circ \), \( \angle B + \angle C = 180 - 72.4 = 107.6^\circ \)

And since \( \angle B = \angle C \) (because \( AB = AC \)), each is \( \frac{107.6}{2} = 53.8^\circ \)

Wait, let's verify with Law of Sines. Law of Sines: \( \frac{\sin A}{BC} = \frac{\sin B}{AC} \)

\( \frac{\sin 72.4^\circ}{13} = \frac{\sin B}{11} \)

\( \sin B = \frac{11 \cdot \sin 72.4^\circ}{13} \)

\( \sin 72.4^\circ \approx 0.9535 \)

\( \sin B = \frac{11 \cdot 0.9535}{13} \approx \frac{10.4885}{13} \approx 0.8068 \)

\( \angle B = \arcsin(0.8068) \approx 53.8^\circ \), which matches. So \( \angle A \approx 72.4^\circ \), \( \angle B \approx 53.8^\circ \), \( \angle C \approx 53.8^\circ \)

Wait, but wait, the problem says "the triangle shown to the right" with \( AC = 11 \), \( BC = 13 \), \( AB = 11 \). So \( AB = AC = 11 \), so \( \angle B = \angle C \), and \( \angle A \) is the vertex angle. So that's correct.

Answer:

\( A \approx 72.4^\circ \), \( B \approx 53.8^\circ \), \( C \approx 53.8^\circ \)