Question

3.1. in the figure below, acute - angled \\( \delta a b c \\) is drawn having c at origin. 3.1.1. prove that \\( c ^ { 2 } = a ^ { 2 } + b ^ { 2 } - 2 a b \cos c \\)

Explanation:

Step1: Find coordinates of \(A\) and \(B\)

Let \(C=(0,0)\), \(A=(b,0)\) (since \(|CA| = b\)), and \(B=(x,y)\). By the distance formula, \(|CB|=a=\sqrt{x^{2}+y^{2}}\), so \(x^{2}+y^{2}=a^{2}\). Also, \(\cos C=\frac{x}{a}\), so \(x = a\cos C\).

Step2: Apply the distance formula for \(c = |AB|\)

The distance formula \(c=\sqrt{(x - b)^{2}+y^{2}}\). Squaring both sides, \(c^{2}=(x - b)^{2}+y^{2}\). Expand \((x - b)^{2}+y^{2}=x^{2}-2bx + b^{2}+y^{2}\).

Step3: Substitute \(x^{2}+y^{2}=a^{2}\) and \(x = a\cos C\)

Substitute \(x^{2}+y^{2}=a^{2}\) and \(x = a\cos C\) into \(c^{2}=x^{2}-2bx + b^{2}+y^{2}\). We get \(c^{2}=a^{2}+b^{2}-2b(a\cos C)\).

Answer:

\(c^{2}=a^{2}+b^{2}-2ab\cos C\) is proved.