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 \( AB = AC = 11 \), so \( \angle B=\angle A \)? Wait, no, \( AB = 11 \), \( AC = 11 \), \( BC = 14 \). So triangle \( ABC \) has \( AB = AC = 11 \), \( BC = 14 \). So we can use the Law of Cosines to find angle \( A \) first.

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

Substitute \( AB = 11 \), \( AC = 11 \), \( BC = 14 \):

\( 14^{2}=11^{2}+11^{2}-2\cdot11\cdot11\cdot\cos A \)

\( 196 = 121 + 121 - 242\cos A \)

\( 196 = 242 - 242\cos A \)

\( -46 = -242\cos A \)

\( \cos A=\frac{46}{242}=\frac{23}{121}\approx0.19008 \)

\( A=\arccos(0.19008)\approx79.0^{\circ} \) (rounded to nearest tenth)

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

Since \( AB = AC \), triangle is isosceles, so \( \angle B=\angle C \). The sum of angles in a triangle is \( 180^{\circ} \), so \( \angle B=\angle C=\frac{180 - A}{2} \)

Substitute \( A\approx79.0^{\circ} \):

\( \angle B=\angle C=\frac{180 - 79.0}{2}=\frac{101}{2}=50.5^{\circ} \)

Wait, wait, let's recalculate the Law of Cosines step. Wait, \( AB = 11 \), \( AC = 11 \), \( BC = 14 \). So sides: \( AB = c = 11 \), \( AC = b = 11 \), \( BC = a = 14 \). Wait, maybe I mixed up the labels. Let's correct the labels: Let’s denote \( A \) as the vertex between \( AB \) and \( AC \), so \( AB = c = 11 \), \( AC = b = 11 \), \( BC = a = 14 \). Then Law of Cosines for angle \( A \): \( a^{2}=b^{2}+c^{2}-2bc\cos A \)

So \( 14^{2}=11^{2}+11^{2}-2\times11\times11\times\cos A \)

\( 196 = 121 + 121 - 242\cos A \)

\( 196 = 242 - 242\cos A \)

\( 242\cos A = 242 - 196 = 46 \)

\( \cos A=\frac{46}{242}\approx0.19008 \)

\( A=\arccos(0.19008)\approx79.0^{\circ} \) (using calculator: \( \arccos(0.19008)\approx79.0 \))

Then angles \( B \) and \( C \): since \( AB = AC \), triangle is isosceles with \( AB = AC \), so base is \( BC \), so base angles \( B \) and \( C \) are equal. So \( B = C=\frac{180 - A}{2}=\frac{180 - 79.0}{2}=50.5^{\circ} \)

Wait, but let's check with Law of Sines to verify. Law of Sines: \( \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} \)

\( \frac{14}{\sin 79.0^{\circ}}=\frac{11}{\sin B} \)

\( \sin B=\frac{11\sin 79.0^{\circ}}{14} \)

\( \sin 79.0^{\circ}\approx0.9816 \)

\( \sin B=\frac{11\times0.9816}{14}=\frac{10.7976}{14}\approx0.77126 \)

\( B=\arcsin(0.77126)\approx50.5^{\circ} \), which matches. So that's correct.

Answer:

\( A\approx79.0^{\circ} \), \( B\approx50.5^{\circ} \), \( C\approx50.5^{\circ} \)