Question

find ab
b
a
69°
12°
18 km
c
a) 2 km b) 9 km
c) 7 km d) 4 km

Explanation:

Step1: Find angle at C

In triangle \(ABC\), sum of angles is \(180^\circ\). So \(\angle C = 180^\circ - 69^\circ - 12^\circ = 99^\circ\)? Wait, no, wait. Wait, the angles at A and B? Wait, the diagram: angle at A is... Wait, no, the triangle has angle at B? Wait, the labels: B, A, C. So angle at A: wait, the angle between B, A, C? Wait, the given angles: \(69^\circ\) at A? Wait, no, the angle between B and A is \(69^\circ\), angle at C is \(12^\circ\)? Wait, no, the triangle: vertices B, A, C. So sides: AC is 18 km. Angles: at A? Wait, the angle between BA and CA is \(69^\circ\)? Wait, no, the angle inside the triangle: \(\angle BAC = 69^\circ\), \(\angle ACB = 12^\circ\), so \(\angle ABC = 180 - 69 - 12 = 99^\circ\)? Wait, no, maybe I misread. Wait, the problem is to find AB. Let's use the Law of Sines. Law of Sines: \(\frac{AB}{\sin C} = \frac{AC}{\sin B}\). Wait, let's identify the angles. Let's denote:

• \(AC = 18\) km (side opposite angle B)
• \(AB\) is the side we need, opposite angle C (which is \(12^\circ\))
• Angle at B: let's see, the angle at B: wait, the angle between BA and BC is... Wait, the given angle is \(69^\circ\) at A? Wait, maybe the angles are: \(\angle BAC = 69^\circ\), \(\angle ACB = 12^\circ\), so \(\angle ABC = 180 - 69 - 12 = 99^\circ\)? No, that can't be. Wait, maybe the angle at C is \(12^\circ\), angle at A is \(69^\circ\), so angle at B is \(180 - 69 - 12 = 99^\circ\). Then by Law of Sines: \(\frac{AB}{\sin \angle ACB} = \frac{AC}{\sin \angle ABC}\). So \(AB = \frac{AC \cdot \sin \angle ACB}{\sin \angle ABC}\). But \(\sin 99^\circ \approx \sin(90 + 9) = \cos 9 \approx 0.9877\), \(\sin 12^\circ \approx 0.2079\), \(AC = 18\). So \(AB = \frac{18 \cdot 0.2079}{0.9877} \approx \frac{3.7422}{0.9877} \approx 3.79\), which is approximately 4 km? Wait, but the options are 2,9,7,4. So D) 4 km? Wait, maybe I messed up the angles. Wait, maybe the angle at B is \(12^\circ\), angle at A is \(69^\circ\), so angle at C is \(180 - 69 - 12 = 99^\circ\). No, that still gives the same. Wait, maybe the angle at C is \(12^\circ\), angle at B is \(69^\circ\), so angle at A is \(180 - 69 - 12 = 99^\circ\). Then Law of Sines: \(\frac{AB}{\sin C} = \frac{AC}{\sin B}\). So \(AB = \frac{AC \cdot \sin C}{\sin B} = \frac{18 \cdot \sin 12^\circ}{\sin 69^\circ}\). \(\sin 12^\circ \approx 0.2079\), \(\sin 69^\circ \approx 0.9336\). So \(AB = \frac{18 \cdot 0.2079}{0.9336} \approx \frac{3.7422}{0.9336} \approx 4.01\), which is approximately 4 km. So the answer is D) 4 km.

Step2: Apply Law of Sines

Law of Sines: \(\frac{AB}{\sin \angle ACB} = \frac{AC}{\sin \angle ABC}\)

Given:

• \(AC = 18\) km
• \(\angle ACB = 12^\circ\)
• \(\angle ABC = 69^\circ\) (wait, maybe I had the angles reversed. Let's check the diagram again. The triangle: B, A, C. So side AC is 18 km. The angle at B is \(69^\circ\), angle at C is \(12^\circ\), so angle at A is \(180 - 69 - 12 = 99^\circ\). No, but if we take angle at B as \(69^\circ\), angle at C as \(12^\circ\), then side AC is opposite angle B, and side AB is opposite angle C. So Law of Sines: \(\frac{AB}{\sin C} = \frac{AC}{\sin B}\)

So \(AB = \frac{AC \cdot \sin C}{\sin B} = \frac{18 \cdot \sin 12^\circ}{\sin 69^\circ}\)

Calculate \(\sin 12^\circ \approx 0.2079\), \(\sin 69^\circ \approx 0.9336\)

\(AB \approx \frac{18 \times 0.2079}{0.9336} \approx \frac{3.7422}{0.9336} \approx 4\) km

Answer:

D) 4 km