Question

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 values for law of cosines

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

Step2: Substitute values into the formula

Substitute $b = 12$ and $c = 6$ into $a^{2}=b^{2}+c^{2}$. We get $n^{2}=12^{2}+6^{2}=144 + 36=180$.

Step3: Solve for $n$

$n=\sqrt{180}\approx 13.42$. Rounding to the nearest whole number, $n\approx13$.

Answer:

13