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 \(a = n\), \(b = 20\), \(c = 9\), 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

Substitute \(b = 20\) and \(c = 9\) into the formula \(a^{2}=b^{2}+c^{2}\). We get \(n^{2}=20^{2}+9^{2}\).
\[n^{2}=400 + 81\]
\[n^{2}=481\]

Step3: Solve for \(n\)

Take the square - root of both sides: \(n=\sqrt{481}\approx21.93\).

Step4: Round to the nearest whole number

Rounding \(21.93\) to the nearest whole number gives \(n = 22\).

Answer:

22