1. **Problem Statement:** Complete the missing angles and sides of the oblique triangles given partial data. 2. **Key Formula:** The sum of angles in any triangle is always $$180^\circ$$, so $$\angle X + \angle Y + \angle Z = 180^\circ$$. 3. **Law of Sines:** For sides $$x, y, z$$ opposite angles $$\angle X, \angle Y, \angle Z$$ respectively, $$\frac{x}{\sin \angle X} = \frac{y}{\sin \angle Y} = \frac{z}{\sin \angle Z}$$. 4. **Step-by-step solutions:** **Triangle 1:** Given $$\angle X = 140^\circ$$, $$y=15$$, $$z=20$$. - Find $$\angle Y + \angle Z = 180^\circ - 140^\circ = 40^\circ$$. - Without more info, assume $$\angle Y$$ and $$\angle Z$$ split 40°; but to find sides, use Law of Sines: $$\frac{y}{\sin \angle Y} = \frac{z}{\sin \angle Z}$$. - Let $$\angle Y = \theta$$, then $$\angle Z = 40^\circ - \theta$$. - So $$\frac{15}{\sin \theta} = \frac{20}{\sin(40^\circ - \theta)}$$. - Solve numerically for $$\theta$$: approximate $$\theta \approx 24.6^\circ$$, so $$\angle Z = 15.4^\circ$$. - Find side $$x$$ opposite $$140^\circ$$: $$x = \frac{y \sin 140^\circ}{\sin \theta} = \frac{15 \times \sin 140^\circ}{\sin 24.6^\circ} \approx 35.7$$. **Triangle 2:** Given $$\angle Y = 50^\circ$$, $$x=68$$, $$z=110$$. - Find $$\angle X + \angle Z = 130^\circ$$. - Use Law of Sines: $$\frac{68}{\sin \angle X} = \frac{110}{\sin \angle Z}$$. - Let $$\angle X = \alpha$$, then $$\angle Z = 130^\circ - \alpha$$. - So $$\frac{68}{\sin \alpha} = \frac{110}{\sin(130^\circ - \alpha)}$$. - Solve numerically for $$\alpha \approx 38.3^\circ$$, so $$\angle Z = 91.7^\circ$$. - Find side $$y$$ opposite $$50^\circ$$: $$y = \frac{68 \sin 50^\circ}{\sin 38.3^\circ} \approx 84.3$$. **Triangle 3:** Given $$\angle X = 50^\circ$$, $$\angle Z = 30^\circ$$, $$y=10$$. - Find $$\angle Y = 180^\circ - 50^\circ - 30^\circ = 100^\circ$$. - Use Law of Sines to find sides $$x$$ and $$z$$: $$\frac{y}{\sin \angle Y} = \frac{x}{\sin \angle X} = \frac{z}{\sin \angle Z}$$. - Calculate $$x = \frac{y \sin 50^\circ}{\sin 100^\circ} = \frac{10 \times 0.7660}{0.9848} \approx 7.78$$. - Calculate $$z = \frac{y \sin 30^\circ}{\sin 100^\circ} = \frac{10 \times 0.5}{0.9848} \approx 5.08$$. **Triangle 4:** Given $$\angle X = 45.5^\circ$$, $$\angle Y = 81.3^\circ$$, $$z=15$$. - Find $$\angle Z = 180^\circ - 45.5^\circ - 81.3^\circ = 53.2^\circ$$. - Use Law of Sines to find $$x$$ and $$y$$: $$x = \frac{z \sin 45.5^\circ}{\sin 53.2^\circ} = \frac{15 \times 0.7133}{0.7997} \approx 13.37$$. $$y = \frac{z \sin 81.3^\circ}{\sin 53.2^\circ} = \frac{15 \times 0.9890}{0.7997} \approx 18.53$$. **Triangle 5:** Given $$\angle Y = 20^\circ$$, $$\angle Z = 130^\circ$$, $$y=21$$. - Find $$\angle X = 180^\circ - 20^\circ - 130^\circ = 30^\circ$$. - Use Law of Sines to find $$x$$ and $$z$$: $$x = \frac{y \sin 30^\circ}{\sin 20^\circ} = \frac{21 \times 0.5}{0.3420} \approx 30.7$$. $$z = \frac{y \sin 130^\circ}{\sin 20^\circ} = \frac{21 \times 0.7660}{0.3420} \approx 47.1$$. **Final Table:** | # | $\angle X$ | $\angle Y$ | $\angle Z$ | $x$ | $y$ | $z$ | |---|--------------|--------------|--------------|-------|-------|-------| | 1 | 140° | 24.6° | 15.4° | 35.7 | 15 | 20 | | 2 | 38.3° | 50° | 91.7° | 68 | 84.3 | 110 | | 3 | 50° | 100° | 30° | 7.78 | 10 | 5.08 | | 4 | 45.5° | 81.3° | 53.2° | 13.37 | 18.53 | 15 | | 5 | 30° | 20° | 130° | 30.7 | 21 | 47.1 |