The problem consists of two parts. Part 1: In triangle $ABC$, given $A = 125.4^\circ$, $b = 29.4$ cm, and $c = 5$ cm, find $a$. Part 2: In triangle $ABC$, given $C = 47.8^\circ$, $a = 13.1$ m, and $b = 24.2$ m, find $c$.

1. Problem Description

The problem consists of two parts.

Part 1: In triangle $ABC$, given $A = 125.4^\circ$, $b = 29.4$ cm, and $c = 5$ cm, find $a$.

Part 2: In triangle $ABC$, given $C = 47.8^\circ$, $a = 13.1$ m, and $b = 24.2$ m, find $c$.

2. Solution Steps

Part 1:

We can use the Law of Cosines to find $a$:

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

Plugging in the values, we get:

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

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

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

$a^2 = 889.36 + 170.14$

$a^2 = 1059.5$

$a = \sqrt{1059.5}$

$a \approx 32.55$ cm

Part 2:

We can use the Law of Cosines to find $c$:

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

Plugging in the values, we get:

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

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

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

$c^2 = 757.25 - 425.14$

$c^2 = 332.11$

$c = \sqrt{332.11}$

$c \approx 18.22$ m

3. Final Answer

Part 1: $a \approx 32.55$ cm

Part 2: $c \approx 18.22$ m