In ΔABC, if bc2−a2+cc2−b2=0 then

Given bc2−a2+ac2−b2=0⇒bc2−b3+ac2−a3=0⇒c2(a+b)−b3+a3=0⇒c2(a+b)−(a+b)a2+b2−ab=0⇒c2(a+b)=(a+b)a2+b2−ab⇒(c2−a2+b2−ab)=0⇒c2=a2+b2−ab⇒a2+b2−c2=ab⇒a2+b2−c22ab=12⇒cos⁡C=12⇒C=600

In ΔABC, if bc2−a2+cc2−b2=0 then

Given bc2−a2+ac2−b2=0⇒bc2−b3+ac2−a3=0⇒c2(a+b)−b3+a3=0⇒c2(a+b)−(a+b)a2+b2−ab=0⇒c2(a+b)=(a+b)a2+b2−ab⇒(c2−a2+b2−ab)=0⇒c2=a2+b2−ab⇒a2+b2−c2=ab⇒a2+b2−c22ab=12⇒cos⁡C=12⇒C=600

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In $ΔABC, if \frac{b}{c^{2} - a^{2}} + \frac{c}{c^{2} - b^{2}} = 0$ then

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A = 60°

C = 90°

C = 60°

B = 60°

answer is C.

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Detailed Solution

$\begin{array}{l} G i v e n \frac{b}{c^{2} - a^{2}} + \frac{a}{c^{2} - b^{2}} = 0 \\ \Rightarrow {bc}^{2} - b^{3} + {ac}^{2} - a^{3} = 0 \\ \Rightarrow c^{2} (a + b) - (b^{3} + a^{3}) = 0 \\ \Rightarrow c^{2} (a + b) - (a + b) (a^{2} + b^{2} - ab) = 0 \\ \Rightarrow c^{2} (a + b) = (a + b) (a^{2} + b^{2} - ab) \\ \Rightarrow (c^{2} - (a^{2} + b^{2} - ab)) = 0 \\ \Rightarrow c^{2} = (a^{2} + b^{2} - ab) \\ \Rightarrow a^{2} + b^{2} - c^{2} = ab \\ \Rightarrow \frac{a^{2} + b^{2} - c^{2}}{2 ab} = \frac{1}{2} \\ \Rightarrow \cos C = \frac{1}{2} \\ \Rightarrow C = 60^{0} \end{array}$