We are given three triangle problems. 1. In triangle $ABC$, $B = 135.5^\circ$, $c = 8$ cm, and $a = 5$ cm. We need to find the length of side $b$.

1. Problem Description

We are given three triangle problems.

1. In triangle $ABC$, $B = 135.5^\circ$, $c = 8$ cm, and $a = 5$ cm. We need to find the length of side $b$.

2. In triangle $ABC$, $A = 125.4^\circ$, $b = 29.4$ cm, and $c = 5$ cm. We need to find the length of side $a$.

3. In triangle $ABC$, $C = 47.8^\circ$, $a = 13.1$ m, and $b = 24.2$ m. We need to find the length of side $c$.

2. Solution Steps

Problem 1:

We will use the Law of Cosines to find side $b$. The Law of Cosines states:

$b^2 = a^2 + c^2 - 2ac\cos(B)$

Plugging in the given values, we have:

$b^2 = 5^2 + 8^2 - 2(5)(8)\cos(135.5^\circ)$

$b^2 = 25 + 64 - 80\cos(135.5^\circ)$

$b^2 = 89 - 80(-0.71325)$

$b^2 = 89 + 57.06$

$b^2 = 146.06$

$b = \sqrt{146.06}$

$b \approx 12.0855$ cm

Problem 2:

We will 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, we have:

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

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

$a^2 = 889.36 - 294(-0.57877)$

$a^2 = 889.36 + 169.95$

$a^2 = 1059.31$

$a = \sqrt{1059.31}$

$a \approx 32.547$ cm

Problem 3:

We will 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, we have:

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

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

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

$c^2 = 757.25 - 425.11$

$c^2 = 332.14$

$c = \sqrt{332.14}$

$c \approx 18.2247$ m

3. Final Answer

1. $b \approx 12.09$ cm

2. $a \approx 32.55$ cm

3. $c \approx 18.22$ m