Question

use the diagram to derive the law of cosines. fill in the missing statements in the derivation.

1. ( a^{2}=h^{2}+x^{2} )
2. ( h^{2}= )
3. ( c^{2}=h^{2}+(b - x)^{2} )
4. ( c^{2}=h^{2}+ )
5. ( c^{2}=a^{2}-x^{2}+b^{2}-2bx + x^{2} )
6. ( c^{2}=a^{2}+b^{2} )
7. ( x = acos c )
8. ( c^{2}=a^{2}+b^{2}2abcos c )

Explanation:

Step1: Apply Pythagorean theorem to left - hand right - triangle

In the right - triangle with side \(a\) (hypotenuse), \(h\) (one leg) and \(x\) (the other leg), by the Pythagorean theorem \(a^{2}=h^{2}+x^{2}\), so \(h^{2}=a^{2}-x^{2}\).

Step2: Expand \((b - x)^{2}\)

Using the formula \((m - n)^{2}=m^{2}-2mn + n^{2}\), where \(m = b\) and \(n=x\), we have \((b - x)^{2}=b^{2}-2bx+x^{2}\).

Step3: Simplify the equation for \(c^{2}\)

Substitute \(h^{2}=a^{2}-x^{2}\) into \(c^{2}=h^{2}+b^{2}-2bx + x^{2}\). The \(x^{2}\) terms \(-x^{2}+x^{2}=0\), so \(c^{2}=a^{2}+b^{2}-2bx\).

Answer:

1. \(a^{2}-x^{2}\); 4) \(b^{2}-2bx + x^{2}\); 6) \(-2bx\)