ABC ত্রিভুজে a⁴ + b⁴ + c⁴ = 2c²(a² + b²) হলে ∠C =? (In the triangle ABC, if a⁴ + b⁴ + c⁴ = 2c²(a² + b²) then ∠C =?)

Explanation:

Another Explanation (5): দেওয়া আছে, \( a^4 + b^4 + c^4 = 2c^2(a^2 + b^2) \) \( \implies a^4 + b^4 + c^4 = 2a^2c^2 + 2b^2c^2 \) \( \implies a^4 + b^4 + c^4 - 2a^2c^2 - 2b^2c^2 = 0 \) \( \implies a^4 + b^4 + 2a^2b^2 - 2a^2b^2 + c^4 - 2a^2c^2 - 2b^2c^2 = 0 \) \( \implies (a^2 + b^2)^2 + c^4 - 2c^2(a^2 + b^2) - 2a^2b^2 = 0 \) \( \implies (a^2 + b^2)^2 - 2c^2(a^2 + b^2) + c^4 = 2a^2b^2 \) \( \implies (a^2 + b^2 - c^2)^2 = 2a^2b^2 \) \( \implies a^2 + b^2 - c^2 = \pm \sqrt{2}ab \) আমরা জানি, \( \cos C = \frac{a^2 + b^2 - c^2}{2ab} \) সুতরাং, \( \cos C = \frac{\pm \sqrt{2}ab}{2ab} = \pm \frac{\sqrt{2}}{2} = \pm \frac{1}{\sqrt{2}} \) যদি \( \cos C = \frac{1}{\sqrt{2}} \) হয়, তবে \( C = 45^\circ \) 🤓 যদি \( \cos C = -\frac{1}{\sqrt{2}} \) হয়, তবে \( C = 135^\circ \)🥳 যেহেতু ত্রিভুজের কোণ \( 135^\circ \) হতে পারে, তাই \( \angle C = 135^\circ \) 👍