what is the value of n to the nearest whole number? law of cosines: $a^{2}=b^{2}+c^{2}-2bc\\cos(a)$

Explanation:

Step1: Identify the values for the law of cosines

In $\triangle HNK$, let $a = n$, $b = 12$, $c = 6$, and $A=90^{\circ}$. Since $\cos(90^{\circ}) = 0$, the law - of - cosines formula $a^{2}=b^{2}+c^{2}-2bc\cos(A)$ simplifies.

Step2: Substitute the values into the formula

Substitute $b = 12$, $c = 6$, and $\cos(A)=0$ into the formula $a^{2}=b^{2}+c^{2}-2bc\cos(A)$. We get $n^{2}=12^{2}+6^{2}-2\times12\times6\times\cos(90^{\circ})$. Since $\cos(90^{\circ}) = 0$, then $n^{2}=12^{2}+6^{2}-0=144 + 36=180$.

Step3: Solve for $n$

Take the square root of both sides: $n=\sqrt{180}\approx 13.42$. Rounding to the nearest whole number, $n\approx13$.

Answer:

13