Question

assume that q, r, and s are the angles of a triangle, with opposite sides q, r, and s respectively. select all of the following that represent the law of cosines. q² = r² + s² - 2rscos(q) q² = r² + s² - 2qscos(r) r² = q² + s² - 2qscos(q) r² = q² + s² - 2qscos(r) s² = q² + r² - 2qrcos(s) done

Explanation:

Step1: Recall the Law of Cosines formula

The Law of Cosines states that for a triangle with sides \(a,b,c\) and opposite angles \(A,B,C\) respectively, \(a^{2}=b^{2}+c^{2}-2bc\cos(A)\), \(b^{2}=a^{2}+c^{2}-2ac\cos(B)\), \(c^{2}=a^{2}+b^{2}-2ab\cos(C)\)

Step2: Analyze each option

• For \(q^{2}=r^{2}+s^{2}-2r s\cos(Q)\):

Here, side \(q\) is opposite angle \(Q\), and using the Law of Cosines formula \(a^{2}=b^{2}+c^{2}-2bc\cos(A)\) (where \(a = q\), \(b = r\), \(c = s\), \(A=Q\)), this formula is correct.

• For \(q^{2}=r^{2}+s^{2}-2q s\cos(R)\):

By the Law of Cosines, if we consider the formula for side \(q\), the coefficient should be \(2rs\) (not \(2qs\)) related to the adjacent sides of angle \(R\). So this formula is incorrect.

• For \(r^{2}=q^{2}+s^{2}-2q s\cos(Q)\):

Using the Law of Cosines formula \(b^{2}=a^{2}+c^{2}-2ac\cos(B)\) (where \(b = r\), \(a = q\), \(c = s\), \(B = R\) (not \(Q\))), this formula is incorrect.

• For \(r^{2}=q^{2}+s^{2}-2q s\cos(R)\):

Using the Law of Cosines formula \(b^{2}=a^{2}+c^{2}-2ac\cos(B)\) (where \(b = r\), \(a = q\), \(c = s\), \(B = R\)), this formula is correct.

• For \(s^{2}=q^{2}+r^{2}-2q r\cos(S)\):

Using the Law of Cosines formula \(c^{2}=a^{2}+b^{2}-2ab\cos(C)\) (where \(c = s\), \(a = q\), \(b = r\), \(C = S\)), this formula is correct.

Answer:

\(q^{2}=r^{2}+s^{2}-2r s\cos(Q)\), \(r^{2}=q^{2}+s^{2}-2q s\cos(R)\), \(s^{2}=q^{2}+r^{2}-2q r\cos(S)\)