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 \( AB = AC = 14 \), so \( \angle B=\angle A \)? Wait, no, \( AB = 14 \), \( AC = 14 \), \( BC = 17 \). So sides: \( AB = c = 14 \), \( AC = b = 14 \), \( BC = a = 17 \). Wait, standard notation: in triangle \( ABC \), side \( a \) is opposite \( \angle A \), side \( b \) opposite \( \angle B \), side \( c \) opposite \( \angle C \). Wait, maybe better to use the Law of Cosines. Let's label: \( AB = 14 \), \( AC = 14 \), \( BC = 17 \). So \( \angle A \) is between \( AB \) and \( AC \), so sides \( AB = 14 \), \( AC = 14 \), \( BC = 17 \). So to find \( \angle A \), use Law of Cosines: \( BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos A \)
So \( 17^2 = 14^2 + 14^2 - 2 \cdot 14 \cdot 14 \cdot \cos A \)
Calculate: \( 289 = 196 + 196 - 392 \cos A \)
\( 289 = 392 - 392 \cos A \)
Subtract 392: \( 289 - 392 = -392 \cos A \)
\( -103 = -392 \cos A \)
So \( \cos A = \frac{103}{392} \approx 0.262755 \)
Then \( \angle A \approx \arccos(0.262755) \approx 74.7^\circ \)

Step2: Find \( \angle B \) and \( \angle C \)

Since \( AB = AC \), triangle is isosceles with \( \angle B = \angle C \). The sum of angles in a triangle is \( 180^\circ \), so \( \angle B + \angle C + \angle A = 180^\circ \). Since \( \angle B = \angle C \), \( 2\angle B = 180^\circ - \angle A \)
\( 2\angle B = 180 - 74.7 = 105.3 \)
\( \angle B = \angle C = \frac{105.3}{2} \approx 52.65^\circ \approx 52.7^\circ \) Wait, wait, maybe I mixed up the labels. Wait, no: if \( AB = AC = 14 \), then the base is \( BC = 17 \), so the base angles are \( \angle B \) and \( \angle C \)? Wait, no: in triangle \( ABC \), if \( AB = AC \), then \( \angle B = \angle C \), and \( \angle A \) is the vertex angle. Wait, maybe I labeled the angles wrong. Let's re-express:

Let me define: \( A \) is the vertex, \( B \) and \( C \) are the base vertices. So \( AB = AC = 14 \), so \( \angle B = \angle C \), and \( BC = 17 \) is the base. Then Law of Cosines on \( \angle A \):

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

\( 17^2 = 14^2 + 14^2 - 2 \cdot 14 \cdot 14 \cdot \cos A \)

\( 289 = 196 + 196 - 392 \cos A \)

\( 289 = 392 - 392 \cos A \)

\( 392 \cos A = 392 - 289 = 103 \)

\( \cos A = 103 / 392 ≈ 0.262755 \)

\( \angle A ≈ \arccos(0.262755) ≈ 74.7^\circ \)

Then \( \angle B = \angle C = (180 - 74.7)/2 = 105.3/2 ≈ 52.65^\circ ≈ 52.7^\circ \). Wait, but maybe I made a mistake in the angle labels. Let's check again.

Wait, maybe the sides are: \( AB = 14 \), \( BC = 17 \), \( AC = 14 \). So vertices: \( A \), \( B \), \( C \), with \( AB = 14 \), \( AC = 14 \), \( BC = 17 \). So angle at \( A \) is between \( AB \) and \( AC \), so sides \( AB \) and \( AC \) are equal, so angle at \( A \) is the vertex angle, and angles at \( B \) and \( C \) are equal. So that's correct.

Alternatively, maybe the triangle is labeled with \( A \) at the top, \( B \) and \( C \) at the base. So \( AB = 14 \), \( AC = 14 \), \( BC = 17 \). So angles: \( \angle A \) is the top angle, \( \angle B \) and \( \angle C \) are the base angles.

So recalculating \( \angle A \):

\( \cos A = (14^2 + 14^2 - 17^2)/(2 \cdot 14 \cdot 14) = (196 + 196 - 289)/392 = (392 - 289)/392 = 103/392 ≈ 0.262755 \)

\( \angle A ≈ \arccos(0.262755) ≈ 74.7^\circ \)

Then \( \angle B = \angle C = (180 - 74.7)/2 ≈ 52.65^\circ ≈ 52.7^\circ \). Wait, but maybe the problem has \( AB = 14 \), \( BC = 17 \), \( AC = 14 \), so angle at \( B \) and angle at \( C \) are equal? Wait, no, if \( AB = AC \), then angles at \( B \) and \( C \) are equal. Wait, maybe I had the labels wrong. Let's check the diagram: the triangle has \( A \) at the top, \( B \) and \( C \) at the bottom, with \( AB = 14 \), \( AC = 14 \), \( BC = 17 \). So yes, \( \angle B = \angle C \), \( \angle A \) is the vertex angle.

So final angles: \( \angle A \approx 74.7^\circ \), \( \angle B \approx 52.7^\circ \), \( \angle C \approx 52.7^\circ \)

Wait, but let's verify with Law of Sines. \( \sin B / AC = \sin A / BC \)

\( \sin B = (AC \cdot \sin A) / BC = (14 \cdot \sin 74.7^\circ) / 17 \)

\( \sin 74.7^\circ \approx 0.964 \)

\( \sin B = (14 \cdot 0.964)/17 ≈ 13.496 / 17 ≈ 0.794 \)

\( \angle B \approx \arcsin(0.794) ≈ 52.6^\circ \), which matches. So that's correct.

Answer:

\( A \approx 74.7^\circ \), \( B \approx 52.7^\circ \), \( C \approx 52.7^\circ \)