Law of Cosines (SAS) Quesc: $$ 0 / 1 \quad $$ Penaltr: 0.5 off In $$ \Delta V W X, v=42 $$ inches, $$ w=76 $$ inches and $$ \angle \mathrm{X}=126^{\circ} $$. Find the length of $$ x $$, to the nearest inch.

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We use the Law of Cosines for a triangle with sides $$ v $$, $$ w $$, and $$ x $$ opposite the corresponding vertices. Since $$ \angle X $$ is opposite side $$ x $$, the formula is

$$
x^2 = v^2 + w^2 - 2vw\cos(\angle X)
$$

Substitute the given values $$ v = 42 $$, $$ w = 76 $$, and $$ \angle X = 126^\circ $$:

$$
x^2 = 42^2 + 76^2 - 2(42)(76)\cos(126^\circ)
$$

Calculate each term:

1. $$ 42^2 = 1764 $$
2. $$ 76^2 = 5776 $$
3. Sum: $$ 1764 + 5776 = 7540 $$
4. $$ 2(42)(76) = 6384 $$

Thus, the expression becomes

$$
x^2 = 7540 - 6384\cos(126^\circ)
$$

Since $$ \cos(126^\circ) $$ is negative (because $$126^\circ$$ is in the second quadrant), we note that

$$
\cos(126^\circ) \approx -0.5878
$$

Substitute this value:

$$
x^2 = 7540 - 6384(-0.5878) = 7540 + 6384(0.5878)
$$

Now, compute $$ 6384 \times 0.5878 $$:

$$
6384 \times 0.5878 \approx 3752.5
$$

Thus,

$$
x^2 \approx 7540 + 3752.5 = 11292.5
$$

Taking the square root of both sides gives

$$
x \approx \sqrt{11292.5} \approx 106.3 \text{ inches}
$$

Rounding to the nearest inch, we have

$$
x \approx 106 \text{ inches}
$$
The length of $$ x $$ is approximately 106 inches.

Law of Cosines (SAS)
Quesc: Penaltr: 0.5 off
In inches, inches and . Find the length of , to the nearest inch.

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Law of Cosines (SAS) Quesc: Penaltr: 0.5 off In inches, inches and . Find the length of , to the nearest inch.