The image presents two geometry problems related to triangles. Problem 1: In triangle $ABC$, angle $B = 135.5^\circ$, side $c = 8$ cm, and side $a = 5$ cm. Find the length of side $b$. Problem 2: In triangle $ABC$, angle $A = 125.4^\circ$, side $b = 29.4$ cm, and side $c = 5$ cm. Find the length of side $a$.

1. Problem Description

The image presents two geometry problems related to triangles.

Problem 1: In triangle $ABC$, angle $B = 135.5^\circ$, side $c = 8$ cm, and side $a = 5$ cm. Find the length of side $b$.

Problem 2: In triangle $ABC$, angle $A = 125.4^\circ$, side $b = 29.4$ cm, and side $c = 5$ cm. Find the length of side $a$.

2. Solution Steps

Problem 1:

We can use the Law of Cosines to find the length of side $b$. The Law of Cosines states that:

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

Plugging in the given values:

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

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

$b^2 = 89 - 80(-0.7133)$

$b^2 = 89 + 57.064$

$b^2 = 146.064$

$b = \sqrt{146.064}$

$b \approx 12.0857$

Problem 2:

We can use the Law of Cosines to find the length of side $a$. The Law of Cosines states that:

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

Plugging in the given values:

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

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

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

$a^2 = 889.36 + 170.1378$

$a^2 = 1059.4978$

$a = \sqrt{1059.4978}$

$a \approx 32.5499$

3. Final Answer

Problem 1: $b \approx 12.09$ cm

Problem 2: $a \approx 32.55$ cm