Question

law of cosines
diagram: triangle hnk with vertex n (90°), hn=12, nk=6, hk=n
law of cosines: ( a^2 = b^2 + c^2 - 2bc cos(a) )
what is the value of n to the nearest whole number?
18
13
10
21 (partially visible)

Explanation:

Step1: Identify values for Law of Cosines

In triangle \( HNK \), we have \( \angle N = 90^\circ \), \( HN = 12 \), \( NK = 6 \), and we need to find \( HK = n \). Using the Law of Cosines formula \( a^2 = b^2 + c^2 - 2bc\cos(A) \), here \( a = n \), \( b = 12 \), \( c = 6 \), and \( A = 90^\circ \). Since \( \cos(90^\circ) = 0 \), the formula simplifies to \( n^2 = 12^2 + 6^2 - 2 \times 12 \times 6 \times \cos(90^\circ) \).

Step2: Substitute and calculate

Substitute \( \cos(90^\circ) = 0 \) into the formula:
\( n^2 = 12^2 + 6^2 - 2 \times 12 \times 6 \times 0 \)
\( n^2 = 144 + 36 - 0 \)
\( n^2 = 180 \)
Then, take the square root of 180: \( n = \sqrt{180} \approx 13.416 \). Rounding to the nearest whole number, \( n \approx 13 \).

Answer:

13