We are given two problems involving triangles. Problem 1: In triangle $ABC$, we are given angle $C = 96.2^{\circ}$, side $b = 11.2$ cm, and side $c = 39.4$ cm. We need to solve the triangle completely. Problem 2: In triangle $XYZ$, we are given angle $Y = 29.8^{\circ}$, angle $Z = 51.4^{\circ}$, and side $x = 19.6$ cm. We need to solve the triangle completely.

1. Problem Description

We are given two problems involving triangles.

Problem 1: In triangle $ABC$, we are given angle $C = 96.2^{\circ}$, side $b = 11.2$ cm, and side $c = 39.4$ cm. We need to solve the triangle completely.

Problem 2: In triangle $XYZ$, we are given angle $Y = 29.8^{\circ}$, angle $Z = 51.4^{\circ}$, and side $x = 19.6$ cm. We need to solve the triangle completely.

2. Solution Steps

Problem 1:

We use the Law of Sines to find angle $B$.

$\frac{b}{\sin B} = \frac{c}{\sin C}$

$\sin B = \frac{b \sin C}{c}$

$\sin B = \frac{11.2 \sin(96.2^{\circ})}{39.4}$

$\sin B = \frac{11.2 \times 0.9941}{39.4}$

$\sin B = \frac{11.13392}{39.4} \approx 0.2825868$

$B = \arcsin(0.2825868) \approx 16.39^{\circ}$

Now, we find angle $A$ using the fact that the sum of the angles in a triangle is $180^{\circ}$.

$A = 180^{\circ} - B - C$

$A = 180^{\circ} - 16.39^{\circ} - 96.2^{\circ}$

$A = 67.41^{\circ}$

Now, we use the Law of Sines again to find side $a$.

$\frac{a}{\sin A} = \frac{c}{\sin C}$

$a = \frac{c \sin A}{\sin C}$

$a = \frac{39.4 \sin(67.41^{\circ})}{\sin(96.2^{\circ})}$

$a = \frac{39.4 \times 0.9234}{0.9941}$

$a = \frac{36.378}{0.9941} \approx 36.6$ cm

Problem 2:

First, we find angle $X$ using the fact that the sum of the angles in a triangle is $180^{\circ}$.

$X = 180^{\circ} - Y - Z$

$X = 180^{\circ} - 29.8^{\circ} - 51.4^{\circ}$

$X = 98.8^{\circ}$

Now, we use the Law of Sines to find side $y$.

$\frac{x}{\sin X} = \frac{y}{\sin Y}$

$y = \frac{x \sin Y}{\sin X}$

$y = \frac{19.6 \sin(29.8^{\circ})}{\sin(98.8^{\circ})}$

$y = \frac{19.6 \times 0.4972}{0.9882}$

$y = \frac{9.74512}{0.9882} \approx 9.86$ cm

Now, we use the Law of Sines to find side $z$.

$\frac{x}{\sin X} = \frac{z}{\sin Z}$

$z = \frac{x \sin Z}{\sin X}$

$z = \frac{19.6 \sin(51.4^{\circ})}{\sin(98.8^{\circ})}$

$z = \frac{19.6 \times 0.7821}{0.9882}$

$z = \frac{15.32916}{0.9882} \approx 15.51$ cm

3. Final Answer

Problem 1:

$A = 67.41^{\circ}$

$B = 16.39^{\circ}$

$a = 36.6$ cm

Problem 2:

$X = 98.8^{\circ}$

$y = 9.86$ cm

$z = 15.51$ cm