We are given a triangle formed by a stove, a sink, and a refrigerator. The distance between the stove and the sink ($AB$) is 1.3 meters, the distance between the sink and the refrigerator ($BC$) is 2.2 meters, and the angle at the stove ($BAC$) is 31.9 degrees. We need to find the missing measurements (the distance between the stove and refrigerator $AC$, angle $ABC$, and angle $BCA$) of this triangle. Then, determine if a unique triangle is possible with this set of measurements.

1. Problem Description

We are given a triangle formed by a stove, a sink, and a refrigerator. The distance between the stove and the sink ($AB$) is 1.3 meters, the distance between the sink and the refrigerator ($BC$) is 2.2 meters, and the angle at the stove ($BAC$) is 31.9 degrees. We need to find the missing measurements (the distance between the stove and refrigerator $AC$, angle $ABC$, and angle $BCA$) of this triangle. Then, determine if a unique triangle is possible with this set of measurements.

2. Solution Steps

We have a triangle with two sides and one angle known. We can use the Law of Cosines to find the third side.

The Law of Cosines states that for any triangle:

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

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

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

In our case, let $a = BC = 2.2$, $b = AC$, and $c = AB = 1.3$. Let $A = BAC = 31.9^{\circ}$. We want to find $b$. Using the Law of Cosines:

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

$2.2^2 = b^2 + 1.3^2 - 2(b)(1.3) \cos(31.9^{\circ})$

$4.84 = b^2 + 1.69 - 2.6b \cos(31.9^{\circ})$

$4.84 = b^2 + 1.69 - 2.6b(0.8495)$

$4.84 = b^2 + 1.69 - 2.2087b$

$b^2 - 2.2087b - 3.15 = 0$

Now, we can solve for $b$ using the quadratic formula:

$b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$

$b = \frac{2.2087 \pm \sqrt{(-2.2087)^2 - 4(1)(-3.15)}}{2(1)}$

$b = \frac{2.2087 \pm \sqrt{4.8784 + 12.6}}{2}$

$b = \frac{2.2087 \pm \sqrt{17.4784}}{2}$

$b = \frac{2.2087 \pm 4.1807}{2}$

We have two possible solutions:

$b_1 = \frac{2.2087 + 4.1807}{2} = \frac{6.3894}{2} = 3.1947$

$b_2 = \frac{2.2087 - 4.1807}{2} = \frac{-1.972}{2} = -0.986$

Since the length of a side cannot be negative, we take the positive value:

$b = AC = 3.1947 \approx 3.19$ meters.

Now, we can use the Law of Sines to find the angle $ABC$.

$\frac{\sin(A)}{a} = \frac{\sin(B)}{b} = \frac{\sin(C)}{c}$

$\frac{\sin(31.9)}{2.2} = \frac{\sin(ABC)}{3.19}$

$\sin(ABC) = \frac{3.19 \sin(31.9)}{2.2} = \frac{3.19(0.5284)}{2.2} = \frac{1.6856}{2.2} = 0.7662$

$ABC = \arcsin(0.7662) = 49.99^{\circ} \approx 50.0^{\circ}$

Now, we can find the remaining angle $BCA$.

$BAC + ABC + BCA = 180^{\circ}$

$31.9^{\circ} + 50.0^{\circ} + BCA = 180^{\circ}$

$BCA = 180^{\circ} - 31.9^{\circ} - 50.0^{\circ} = 98.1^{\circ}$

Since we are given two sides and an included angle, there is only one possible triangle (Side-Angle-Side).

3. Final Answer

The missing measurements are:

$AC = 3.19$ meters

$ABC = 50.0^{\circ}$

$BCA = 98.1^{\circ}$

Only one triangle is possible because we were given Side-Angle-Side (SAS).