Question

consider a triangle abc, with positive angle measures m∠a, m∠b, and m∠c and side lengths of a, b, and c with the relative positions as shown in the diagram below. select all the equalities that are true. (in other words, select all of the following equations that represent either form of the law of cosines.)
options:
□ ( c^2 = a^2 + b^2 - 2ab cdot cos(mangle c) )
□ ( a^2 = b^2 + c^2 - 2bc cdot cos(mangle a) )
□ ( cos(mangle c) = \frac{c^2 - a^2 - b^2}{-2ab} )
□ ( cos(mangle a) = \frac{cos(mangle c)}{c} cdot a ) (note: original ocr might have formatting, corrected to ( cos(mangle a) = \frac{c cdot cos(mangle c)}{a} )? wait, original option: ( cos(mangle a) over a = cos(mangle c) over c ) → ( \frac{cos(mangle a)}{a} = \frac{cos(mangle c)}{c} ))
□ ( \frac{cos(mangle a)}{a} = \frac{cos(mangle b)}{b} )
□ ( cos(mangle a) = \frac{a^2 - b^2 - c^2}{-2bc} )
□ ( b^2 = a^2 + c^2 - 2ac cdot cos(mangle b) )
□ ( cos(mangle b) = \frac{b^2 - a^2 - c^2}{-2ac} )
□ ( \frac{cos(mangle b)}{b} = \frac{cos(mangle c)}{c} )
(diagram of triangle abc with vertices a, b, c; side opposite a is a, opposite b is b, opposite c is c? wait, diagram: a at top, b at right, c at bottom; side ab is c, ac is b, bc is a.)

Explanation:

Step1: Recall Law of Cosines

The Law of Cosines states that for a triangle with sides \(a\), \(b\), \(c\) and opposite angles \(A\), \(B\), \(C\) respectively, the formulas are:
\(c^{2}=a^{2}+b^{2}-2ab\cos(C)\)
\(a^{2}=b^{2}+c^{2}-2bc\cos(A)\)
\(b^{2}=a^{2}+c^{2}-2ac\cos(B)\)
And the Law of Sines (not Cosines) is \(\frac{\sin(A)}{a}=\frac{\sin(B)}{b}=\frac{\sin(C)}{c}\), so the ones with cosine ratios are not Law of Cosines.

Step2: Check each equation

• \(c^{2}=a^{2}+b^{2}-2ab\cos(m\angle C)\): Matches Law of Cosines (side \(c\), angle \(C\)).
• \(a^{2}=b^{2}+c^{2}-2bc\cos(m\angle A)\): Matches Law of Cosines (side \(a\), angle \(A\)).
• \(\cos(m\angle C)=\frac{c^{2}-a^{2}-b^{2}}{-2ab}\): Rearranged from \(c^{2}=a^{2}+b^{2}-2ab\cos(C)\) (divide by \(-2ab\) and rearrange terms).
• \(\cos(m\angle A)=\frac{a^{2}-b^{2}-c^{2}}{-2bc}\): Rearranged from \(a^{2}=b^{2}+c^{2}-2bc\cos(A)\) (divide by \(-2bc\) and rearrange terms).
• \(b^{2}=a^{2}+c^{2}-2ac\cos(m\angle B)\): Matches Law of Cosines (side \(b\), angle \(B\)).
• \(\cos(m\angle B)=\frac{b^{2}-a^{2}-c^{2}}{-2ac}\): Rearranged from \(b^{2}=a^{2}+c^{2}-2ac\cos(B)\) (divide by \(-2ac\) and rearrange terms).

The equations with \(\frac{\cos(m\angle A)}{a}=\frac{\cos(m\angle C)}{c}\) or \(\frac{\cos(m\angle A)}{a}=\frac{\cos(m\angle B)}{b}\) or \(\frac{\cos(m\angle B)}{b}=\frac{\cos(m\angle C)}{c}\) are not Law of Cosines (Law of Cosines has no such cosine - side ratios; that would be a misapplication, likely confusing with Law of Sines which uses sine).

Answer:

The true equalities (Law of Cosines or rearranged) are:
\(c^{2}=a^{2}+b^{2}-2ab\cdot\cos(m\angle C)\)
\(a^{2}=b^{2}+c^{2}-2bc\cdot\cos(m\angle A)\)
\(\cos(m\angle C)=\frac{c^{2}-a^{2}-b^{2}}{-2ab}\)
\(\cos(m\angle A)=\frac{a^{2}-b^{2}-c^{2}}{-2bc}\)
\(b^{2}=a^{2}+c^{2}-2ac\cdot\cos(m\angle B)\)
\(\cos(m\angle B)=\frac{b^{2}-a^{2}-c^{2}}{-2ac}\)

(To mark the checkboxes: The first, second, third, fourth, sixth, seventh options (assuming order: 1. \(c^{2}=...\), 2. \(a^{2}=...\), 3. \(\cos(C)=...\), 4. \(\cos(A)=...\), 5. \(\cos(A)/a=...\) (false), 6. \(b^{2}=...\), 7. \(\cos(B)=...\), 8. \(\cos(B)/b=...\) (false), 9. \(\cos(A)/a=...\) (false), 10. \(\cos(B)/b=...\) (false))