ABC ত্রিভুজের ক্ষেত্রে cosA+cosC= sinB হলে, C কোণের মান কত?

Explanation:

Another Explanation (5): ```html ABC ত্রিভুজে, \(A + B + C = \pi\) ∵ ত্রিভুজের তিন কোণের সমষ্টি \(180^\circ\) বা \(\pi\)। দেওয়া আছে, \(cosA + cosC = sinB\) \(cosA + cosC = sin(\pi - (A+C))\) [∵ \(B = \pi - (A+C)\)] \(cosA + cosC = sin(A+C)\) [∵ \(sin(\pi - \theta) = sin\theta\)] \(2cos(\frac{A+C}{2})cos(\frac{A-C}{2}) = 2sin(\frac{A+C}{2})cos(\frac{A+C}{2})\) [∵ \(cosX + cosY = 2cos(\frac{X+Y}{2})cos(\frac{X-Y}{2})\) এবং \(sin2X = 2sinXcosX\)] \(2cos(\frac{A+C}{2})[cos(\frac{A-C}{2}) - sin(\frac{A+C}{2})] = 0\) হয়, \(cos(\frac{A+C}{2}) = 0\) অথবা, \(cos(\frac{A-C}{2}) = sin(\frac{A+C}{2})\) যদি, \(cos(\frac{A+C}{2}) = 0\) হয়, \(\frac{A+C}{2} = \frac{\pi}{2}\) \(A+C = \pi\) \(B = \pi - (A+C) = \pi - \pi = 0\), যা সম্ভব নয়। ❌ সুতরাং, \(cos(\frac{A-C}{2}) = sin(\frac{A+C}{2})\) \(cos(\frac{A-C}{2}) = cos(\frac{\pi}{2} - \frac{A+C}{2})\) [∵ \(sin\theta = cos(\frac{\pi}{2} - \theta)\)] \(\frac{A-C}{2} = \frac{\pi}{2} - \frac{A+C}{2}\) \(A-C = \pi - A - C\) \(2A = \pi\) \(A = \frac{\pi}{2}\) এখন, \(cosA + cosC = sinB\) সমীকরণে \(A = \frac{\pi}{2}\) বসিয়ে পাই, \(cos(\frac{\pi}{2}) + cosC = sinB\) \(0 + cosC = sin(\pi - (A+C))\) \(cosC = sin(A+C)\) \(cosC = sin(\frac{\pi}{2} + C)\) \(cosC = cosC\), যা \(C\) এর যেকোনো মানের জন্য সত্য। 🤔 অন্যভাবে, \(cos(\frac{A-C}{2}) = sin(\frac{A+C}{2})\) \(cos(\frac{A-C}{2}) = cos(\frac{\pi}{2} - \frac{A+C}{2})\) \(\frac{A-C}{2} = \pm (\frac{\pi}{2} - \frac{A+C}{2})\) (+) চিহ্নের জন্য, \(\frac{A-C}{2} = \frac{\pi}{2} - \frac{A+C}{2}\) \(A - C = \pi - A - C\) \(2A = \pi\) \(A = \frac{\pi}{2}\) (-) চিহ্নের জন্য, \(\frac{A-C}{2} = -(\frac{\pi}{2} - \frac{A+C}{2})\) \(\frac{A-C}{2} = -\frac{\pi}{2} + \frac{A+C}{2}\) \(A - C = -\pi + A + C\) \(-2C = -\pi\) \(C = \frac{\pi}{2}\) 🎉 ```