The problem consists of two parts. (2) In triangle ABC, we are given angle $A = 125.4^{\circ}$, side $b = 29.4$ cm, and side $c = 5$ cm. We need to find the length of side $a$. (3) In triangle ABC, we are given angle $C = 47.8^{\circ}$, side $a = 13.1$ m, and side $b = 24.2$ m. We need to find the length of side $c$.

1. Problem Description

The problem consists of two parts.

(2) In triangle ABC, we are given angle $A = 125.4^{\circ}$, side $b = 29.4$ cm, and side $c = 5$ cm. We need to find the length of side $a$.

(3) In triangle ABC, we are given angle $C = 47.8^{\circ}$, side $a = 13.1$ m, and side $b = 24.2$ m. We need to find the length of side $c$.

2. Solution Steps

(2) We can use the Law of Cosines to find side $a$. The Law of Cosines states:

$a^2 = b^2 + c^2 - 2bc \cos{A}$

Plugging in the given values:

$a^2 = (29.4)^2 + (5)^2 - 2(29.4)(5) \cos{(125.4^{\circ})}$

$a^2 = 864.36 + 25 - 294 \cos{(125.4^{\circ})}$

$a^2 = 889.36 - 294(-0.5787)$

$a^2 = 889.36 + 170.13$

$a^2 = 1059.49$

$a = \sqrt{1059.49}$

$a \approx 32.55$ cm

(3) We can use the Law of Cosines to find side $c$. The Law of Cosines states:

$c^2 = a^2 + b^2 - 2ab \cos{C}$

Plugging in the given values:

$c^2 = (13.1)^2 + (24.2)^2 - 2(13.1)(24.2) \cos{(47.8^{\circ})}$

$c^2 = 171.61 + 585.64 - 634.84 \cos{(47.8^{\circ})}$

$c^2 = 757.25 - 634.84(0.6713)$

$c^2 = 757.25 - 425.15$

$c^2 = 332.10$

$c = \sqrt{332.10}$

$c \approx 18.22$ m

3. Final Answer

(2) $a \approx 32.55$ cm

(3) $c \approx 18.22$ m