Question

law of cosines
diagram: triangle hnk with ( hn = 20 ), ( nk = 9 ), ( angle n = 90^circ ), ( 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?
options: 22, 41, 18, partially visible

Explanation:

Step1: Identify values for Law of Cosines

In triangle \( HNK \), we have \( \angle N = 90^\circ \), \( HN = 20 \), \( NK = 9 \), and we need to find \( HK = n \). Using the Law of Cosines formula \( a^2 = b^2 + c^2 - 2bc\cos(A) \), here \( A = 90^\circ \), \( b = 20 \), \( c = 9 \), and \( a = n \). Since \( \cos(90^\circ)=0 \), the formula simplifies to \( n^2 = 20^2 + 9^2 - 2\times20\times9\times\cos(90^\circ) \).

Step2: Substitute and calculate

Substitute \( \cos(90^\circ) = 0 \) into the formula: \( n^2 = 20^2 + 9^2 - 2\times20\times9\times0 \). Calculate \( 20^2 = 400 \), \( 9^2 = 81 \), and the last term is 0. So \( n^2 = 400 + 81 = 481 \). Then take the square root: \( n = \sqrt{481} \approx 21.93 \), which rounds to 22.

Answer:

22