(a + b + c) (b + c - a) = 3bc হলে, ∠A = কত?

দেওয়া আছে, \( (a + b + c) (b + c - a) = 3bc \)।

আমরা জানি, \( (x + y)(x - y) = x^2 - y^2 \)। ধরি, \( x = b + c \) এবং \( y = a \)।

তাহলে, \( (b + c + a) (b + c - a) = (b + c)^2 - a^2 \)।

সুতরাং, \( (b + c)^2 - a^2 = 3bc \)

\(\implies b^2 + 2bc + c^2 - a^2 = 3bc \)

\(\implies b^2 + c^2 - a^2 = bc \)

আমরা জানি, কোসাইন সূত্রানুসারে, \( a^2 = b^2 + c^2 - 2bc \cos A \)

\(\implies 2bc \cos A = b^2 + c^2 - a^2 \)

অতএব, \( 2bc \cos A = bc \)

\(\implies \cos A = \frac{bc}{2bc} = \frac{1}{2} \)

\(\implies A = \cos^{-1} \left(\frac{1}{2}\right) \)

\(\implies A = \frac{\pi}{3} \) 🥳🥳🥳

অতএব, \( \angle A = \frac{\pi}{3} \)।