Question

in triangle abc, side ab is 10 in., side ac is 7 in., angle c is 112°, and angle at a is x. (the image shows triangle abc with labels: a at the top, b at the bottom left, c at the bottom right, ab = 10 in., ac = 7 in., angle c = 112°, angle a = x.)

Explanation:

Step1: Identify the Law to Use

We have a triangle with two sides and the included angle? Wait, no, here we have two sides (10 in, 7 in) and the angle opposite? Wait, no, the angle at C is 112°, and we need to find side x? Wait, no, wait the labels: side AB is 10 in, side AC is 7 in, angle at C is 112°? Wait, no, maybe it's the Law of Cosines. Wait, let's check: in triangle ABC, we have sides AB = 10, AC = 7, angle at C is 112°? Wait, no, maybe angle at A is x, and we need to find side BC? Wait, no, the diagram: vertex A, with sides AB = 10 in, AC = 7 in, angle at A is x? Wait, no, the angle at C is 112°, side BC is x? Wait, maybe I misread. Wait, the triangle has vertices A, B, C. Side AB is 10 in, side AC is 7 in, angle at C is 112°, and we need to find side BC (x)? Wait, no, maybe it's the Law of Cosines. Wait, Law of Cosines: \( c^2 = a^2 + b^2 - 2ab \cos(C) \). Wait, if we have sides a and b, and angle C between them, then c is the side opposite. Wait, in this case, if angle at C is 112°, and sides AC = 7, BC = x, AB = 10? Wait, no, AB is 10, AC is 7, angle at A is x? Wait, maybe the angle at C is 112°, sides AC = 7, BC = x, AB = 10. Then using Law of Cosines: \( AB^2 = AC^2 + BC^2 - 2 \cdot AC \cdot BC \cdot \cos(C) \). Wait, no, \( AB^2 = AC^2 + BC^2 - 2 \cdot AC \cdot BC \cdot \cos(C) \) would be if angle C is between AC and BC. So AC = 7, BC = x, angle C = 112°, AB = 10. Then:

\( 10^2 = 7^2 + x^2 - 2 \cdot 7 \cdot x \cdot \cos(112°) \)

Wait, but that would be a quadratic. Alternatively, maybe it's the Law of Sines. Wait, Law of Sines: \( \frac{AB}{\sin(C)} = \frac{AC}{\sin(B)} = \frac{BC}{\sin(A)} \). Wait, AB = 10, angle C = 112°, AC = 7. So \( \frac{10}{\sin(112°)} = \frac{7}{\sin(B)} \), but maybe we need to find angle x (angle at A). Wait, maybe I messed up the labels. Let's re-express:

Wait, the triangle: vertex A, with sides AB = 10, AC = 7, angle at A is x, and angle at C is 112°. Wait, no, the angle at C is 112°, so angle at C is 112°, sides AC = 7, BC = x, AB = 10. Then using Law of Cosines:

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

So \( 10^2 = 7^2 + x^2 - 2 \cdot 7 \cdot x \cdot \cos(112°) \)

\( 100 = 49 + x^2 - 14x \cos(112°) \)

\( x^2 - 14x \cos(112°) + 49 - 100 = 0 \)

\( x^2 - 14x \cos(112°) - 51 = 0 \)

But maybe it's the Law of Cosines with angle at A. Wait, no, maybe the angle at A is x, and we have sides AB = 10, AC = 7, and side BC is... Wait, maybe I made a mistake. Wait, the problem is to find x, which is a side? Wait, the diagram shows angle x at A, sides AB = 10, AC = 7, angle at C = 112°. Wait, no, perhaps the correct approach is Law of Cosines. Wait, let's confirm:

In triangle ABC, if we have sides:

- AB = 10 in

- AC = 7 in

- Angle at C = 112°

We need to find side BC (x).

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

Wait, no, Law of Cosines is \( c^2 = a^2 + b^2 - 2ab \cos(C) \), where C is the angle between sides a and b. So if angle C is between AC and BC, then sides AC = b = 7, BC = a = x, angle C = 112°, then AB = c = 10. So:

\( 10^2 = 7^2 + x^2 - 2 \cdot 7 \cdot x \cdot \cos(112°) \)

\( 100 = 49 + x^2 - 14x \cos(112°) \)

\( x^2 - 14x \cos(112°) - 51 = 0 \)

But maybe it's the Law of Sines. Wait, Law of Sines: \( \frac{AB}{\sin(C)} = \frac{AC}{\sin(B)} = \frac{BC}{\sin(A)} \)

So \( \frac{10}{\sin(112°)} = \frac{7}{\sin(B)} \)

But we need to find x, which is BC. Wait, maybe x is angle A? Wait, the label x is at angle A. Oh! Wait, I misread. The angle at A is x, side AB is 10 in, side AC is 7 in, angle at C is 112°. So we have triangle ABC with:

- Side AB = 10 in (opposite angle C)

- Side AC = 7 in (opposite angle B)

- Angle at C = 112°

- Angle at A = x (opposite side BC)

So using Law of Sines: \( \frac{AB}{\sin(C)} = \frac{AC}{\sin(B)} = \frac{BC}{\sin(A)} \)

First, find angle B:

\( \frac{AB}{\sin(C)} = \frac{AC}{\sin(B)} \)

\( \frac{10}{\sin(112°)} = \frac{7}{\sin(B)} \)

\( \sin(B) = \frac{7 \sin(112°)}{10} \)

Calculate \( \sin(112°) = \sin(180° - 68°) = \sin(68°) ≈ 0.9272 \)

So \( \sin(B) = \frac{7 \times 0.9272}{10} ≈ \frac{6.4904}{10} ≈ 0.64904 \)

So angle B ≈ arcsin(0.64904) ≈ 40.4°

Then angle A (x) = 180° - angle B - angle C = 180° - 40.4° - 112° = 27.6°

Wait, but let's check with Law of Cosines for angle A.

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

But we don't know BC. Wait, but using Law of Sines, we found angle B, then angle A.

Wait, let's recalculate:

\( \sin(112°) = \sin(180° - 68°) = \sin(68°) ≈ 0.9271838546 \)

\( \sin(B) = (7 / 10) \times 0.9271838546 ≈ 0.6490286982 \)

Angle B ≈ arcsin(0.6490286982) ≈ 40.4 degrees (since sin(40°) ≈ 0.6428, sin(41°) ≈ 0.6561, so 40.4°)

Then angle A = 180 - 112 - 40.4 = 27.6°

So x ≈ 27.6°

Wait, but let's confirm with Law of Cosines for angle A.

First, find BC using Law of Sines: \( BC = \frac{AB \cdot \sin(A)}{\sin(C)} \)

But we know angle A is 27.6°, so BC = (10 * sin(27.6°)) / sin(112°)

sin(27.6°) ≈ 0.4648, sin(112°) ≈ 0.9272

BC ≈ (10 * 0.4648) / 0.9272 ≈ 4.648 / 0.9272 ≈ 5.01 in

Then using Law of Cosines for angle A:

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

AB = 10, AC = 7, BC ≈ 5.01

\( \cos(A) = (100 + 49 - 25.1001) / (2 * 10 * 7) = (123.8999) / 140 ≈ 0.884999 \)

arccos(0.885) ≈ 27.8°, which is close to 27.6°, considering rounding errors. So angle A (x) is approximately 27.6 degrees, or about 28 degrees.

Wait, but maybe the problem is to find side x, but the label x is at angle A. So the correct approach is Law of Sines.

Step1: Apply Law of Sines

In triangle \( ABC \), \( \frac{AB}{\sin(C)} = \frac{AC}{\sin(B)} \)

Given \( AB = 10 \) in, \( AC = 7 \) in, \( \angle C = 112^\circ \)

\( \frac{10}{\sin(112^\circ)} = \frac{7}{\sin(B)} \)

Solve for \( \sin(B) \):

\( \sin(B) = \frac{7 \cdot \sin(112^\circ)}{10} \)

Step2: Calculate \( \sin(112^\circ) \)

\( \sin(112^\circ) = \sin(180^\circ - 68^\circ) = \sin(68^\circ) \approx 0.9272 \)

Step3: Find \( \sin(B) \)

\( \sin(B) = \frac{7 \cdot 0.9272}{10} \approx 0.6490 \)

Step4: Find \( \angle B \)

\( \angle B \approx \arcsin(0.6490) \approx 40.4^\circ \)

Step5: Find \( \angle A (x) \)

Sum of angles in a triangle: \( \angle A + \angle B + \angle C = 180^\circ \)

\( x = 180^\circ - 40.4^\circ - 112^\circ = 27.6^\circ \) (approx)

Answer:

\( x \approx 27.6^\circ \) (or approximately 28°)