Question

question 2 of 10
the law of cosines is ( a^{2}+b^{2}-2abcos c = c^{2}). find the value of ( 2abcos c).

a. -3.5
b. 7
c. -7
d. 3.5

Explanation:

Step1: Identify values of a, b, c

From the triangle, \(a = 2\), \(b = 2\), \(c = 1\).

Step2: Rearrange the law of cosines

Given \(a^{2}+b^{2}-2ab\cos C = c^{2}\), we can rearrange it to solve for \(2ab\cos C\):
\(2ab\cos C=a^{2}+b^{2}-c^{2}\)

Step3: Substitute the values

Substitute \(a = 2\), \(b = 2\), \(c = 1\) into the formula:
\(2\times2\times2\cos C=2^{2}+2^{2}-1^{2}\)
Wait, no, directly use \(2ab\cos C=a^{2}+b^{2}-c^{2}\):
\(a^{2}=2^{2} = 4\), \(b^{2}=2^{2}=4\), \(c^{2}=1^{2}=1\)
So \(2ab\cos C=4 + 4-1=7\)? Wait, no, wait. Wait, the rearrangement: from \(a^{2}+b^{2}-2ab\cos C=c^{2}\), we can move \(c^{2}\) to left and \(-2ab\cos C\) to right: \(a^{2}+b^{2}-c^{2}=2ab\cos C\)? Wait, no, let's do it again.

Original equation: \(a^{2}+b^{2}-2ab\cos C=c^{2}\)

Subtract \(c^{2}\) from both sides: \(a^{2}+b^{2}-c^{2}-2ab\cos C = 0\)

Add \(2ab\cos C\) to both sides: \(a^{2}+b^{2}-c^{2}=2ab\cos C\)

Wait, but let's check the values. \(a = 2\), \(b = 2\), \(c = 1\). So \(a^{2}+b^{2}-c^{2}=4 + 4-1=7\). So \(2ab\cos C=7\)? But wait, that seems conflicting with the options? Wait no, wait the triangle: the two sides are 2 and 2, base is 1. Wait, maybe I mixed up the labels. Wait, in the law of cosines, \(c\) is the side opposite angle \(C\). So in the triangle, angle \(C\) is between sides \(a\) and \(b\) (lengths 2 and 2), and side \(c\) is the base (length 1). So yes, \(a = 2\), \(b = 2\), \(c = 1\). Then \(a^{2}+b^{2}-2ab\cos C=c^{2}\)
So \(4 + 4-2\times2\times2\cos C=1\)
\(8-8\cos C=1\)
\(-8\cos C=1 - 8=-7\)
\(8\cos C=7\)
\(2ab\cos C=2\times2\times2\cos C=8\cos C=7\)? Wait, no, \(2ab\) is \(2\times2\times2\)? No, \(a = 2\), \(b = 2\), so \(2ab=2\times2\times2\)? No, \(a\) and \(b\) are the sides adjacent to angle \(C\), so \(a = 2\), \(b = 2\), so \(2ab=2\times2\times2\)? No, \(2ab\) is \(2\times a\times b=2\times2\times2 = 8\)? Wait, no, \(a = 2\), \(b = 2\), so \(2ab=2\times2\times2=8\)? Wait, no, \(a\) and \(b\) are two sides, so \(a = 2\), \(b = 2\), so \(2ab=2\times2\times2=8\)? Wait, no, \(2\times a\times b=2\times2\times2 = 8\). Then from the equation \(a^{2}+b^{2}-2ab\cos C=c^{2}\), so \(4 + 4-8\cos C=1\)
\(8-8\cos C=1\)
\(-8\cos C=-7\)
\(8\cos C=7\)
So \(2ab\cos C=8\cos C=7\). So the value is 7, which is option B.

Answer:

B. 7