ক) AI-তৈরি নমুনা উত্তর
আমরা জানি,c 2 = a 2 + b 2 − 2 a b cos C c^2 = a^2 + b^2 - 2ab \cos C c 2 = a 2 + b 2 − 2 ab cos C a = 3 + 1 a = \sqrt{3} + 1 a = 3 + 1 b = 3 − 1 b = \sqrt{3} - 1 b = 3 − 1 C = 60 ∘ C = 60^\circ C = 6 0 ∘
মান বসিয়ে পাই,c 2 = ( 3 + 1 ) 2 + ( 3 − 1 ) 2 − 2 ( 3 + 1 ) ( 3 − 1 ) cos 60 ∘ c^2 = (\sqrt{3} + 1)^2 + (\sqrt{3} - 1)^2 - 2(\sqrt{3} + 1)(\sqrt{3} - 1) \cos 60^\circ c 2 = ( 3 + 1 ) 2 + ( 3 − 1 ) 2 − 2 ( 3 + 1 ) ( 3 − 1 ) cos 6 0 ∘ = 2 [ ( 3 ) 2 + 1 2 ] − 2 [ ( 3 ) 2 − 1 2 ] ⋅ 1 2 = 2[(\sqrt{3})^2 + 1^2] - 2[(\sqrt{3})^2 - 1^2] \cdot \frac{1}{2} = 2 [( 3 ) 2 + 1 2 ] − 2 [( 3 ) 2 − 1 2 ] ⋅ 2 1 = 2 ( 3 + 1 ) − ( 3 − 1 ) = 2(3 + 1) - (3 - 1) = 2 ( 3 + 1 ) − ( 3 − 1 ) = 2 ( 4 ) − 2 = 2(4) - 2 = 2 ( 4 ) − 2 = 8 − 2 = 6 = 8 - 2 = 6 = 8 − 2 = 6
∴ c = 6 \therefore c = \sqrt{6} ∴ c = 6
উত্তর: c = 6 c = \sqrt{6} c = 6
খ) AI-তৈরি নমুনা উত্তর
উদ্দীপকের চিত্রানুসারে, △ A B C \triangle ABC △ A B C B C BC B C D D D ∠ A C D = 2 θ \angle ACD = 2\theta ∠ A C D = 2 θ ∴ ∠ A C D = A + B \therefore \angle ACD = A + B ∴ ∠ A C D = A + B 2 θ = A + B 2\theta = A + B 2 θ = A + B
দেওয়া আছে, A + B = 120 ∘ A + B = 120^\circ A + B = 12 0 ∘ ∴ 2 θ = 120 ∘ ⟹ θ = 60 ∘ \therefore 2\theta = 120^\circ \implies \theta = 60^\circ ∴ 2 θ = 12 0 ∘ ⟹ θ = 6 0 ∘
বামপক্ষ = sin 2 ( θ + α ) + sin 2 ( θ − α ) − cos 2 α = \sin^2(\theta + \alpha) + \sin^2(\theta - \alpha) - \cos^2 \alpha = sin 2 ( θ + α ) + sin 2 ( θ − α ) − cos 2 α = sin 2 ( 60 ∘ + α ) + sin 2 ( 60 ∘ − α ) − cos 2 α = \sin^2(60^\circ + \alpha) + \sin^2(60^\circ - \alpha) - \cos^2 \alpha = sin 2 ( 6 0 ∘ + α ) + sin 2 ( 6 0 ∘ − α ) − cos 2 α = 1 2 [ 2 sin 2 ( 60 ∘ + α ) + 2 sin 2 ( 60 ∘ − α ) ] − cos 2 α = \frac{1}{2} [2\sin^2(60^\circ + \alpha) + 2\sin^2(60^\circ - \alpha)] - \cos^2 \alpha = 2 1 [ 2 sin 2 ( 6 0 ∘ + α ) + 2 sin 2 ( 6 0 ∘ − α )] − cos 2 α = 1 2 [ 1 − cos ( 120 ∘ + 2 α ) + 1 − cos ( 120 ∘ − 2 α ) ] − cos 2 α = \frac{1}{2} [1 - \cos(120^\circ + 2\alpha) + 1 - \cos(120^\circ - 2\alpha)] - \cos^2 \alpha = 2 1 [ 1 − cos ( 12 0 ∘ + 2 α ) + 1 − cos ( 12 0 ∘ − 2 α )] − cos 2 α = 1 2 [ 2 − { cos ( 120 ∘ + 2 α ) + cos ( 120 ∘ − 2 α ) } ] − cos 2 α = \frac{1}{2} [2 - \{\cos(120^\circ + 2\alpha) + \cos(120^\circ - 2\alpha)\}] - \cos^2 \alpha = 2 1 [ 2 − { cos ( 12 0 ∘ + 2 α ) + cos ( 12 0 ∘ − 2 α )}] − cos 2 α = 1 2 [ 2 − 2 cos 120 ∘ cos 2 α ] − cos 2 α = \frac{1}{2} [2 - 2\cos 120^\circ \cos 2\alpha] - \cos^2 \alpha = 2 1 [ 2 − 2 cos 12 0 ∘ cos 2 α ] − cos 2 α = 1 2 [ 2 − 2 ( − 1 2 ) cos 2 α ] − cos 2 α = \frac{1}{2} [2 - 2\left(-\frac{1}{2}\right) \cos 2\alpha] - \cos^2 \alpha = 2 1 [ 2 − 2 ( − 2 1 ) cos 2 α ] − cos 2 α = 1 2 [ 2 + cos 2 α ] − cos 2 α = \frac{1}{2} [2 + \cos 2\alpha] - \cos^2 \alpha = 2 1 [ 2 + cos 2 α ] − cos 2 α = 1 + 1 2 cos 2 α − cos 2 α = 1 + \frac{1}{2}\cos 2\alpha - \cos^2 \alpha = 1 + 2 1 cos 2 α − cos 2 α = 1 + 1 2 ( 2 cos 2 α − 1 ) − cos 2 α = 1 + \frac{1}{2}(2\cos^2 \alpha - 1) - \cos^2 \alpha = 1 + 2 1 ( 2 cos 2 α − 1 ) − cos 2 α = 1 + cos 2 α − 1 2 − cos 2 α = 1 + \cos^2 \alpha - \frac{1}{2} - \cos^2 \alpha = 1 + cos 2 α − 2 1 − cos 2 α = 1 − 1 2 = 1 2 = = 1 - \frac{1}{2} = \frac{1}{2} = = 1 − 2 1 = 2 1 =
গ) AI-তৈরি নমুনা উত্তর
উদ্দীপকের চিত্রানুযায়ী, B C D BCD B C D ∴ C + 2 θ = 180 ∘ \therefore C + 2\theta = 180^\circ ∴ C + 2 θ = 18 0 ∘ C = 180 ∘ − 2 θ C = 180^\circ - 2\theta C = 18 0 ∘ − 2 θ A B C ABC A B C A + B + C = 180 ∘ A + B + C = 180^\circ A + B + C = 18 0 ∘ A + B = 180 ∘ − C = 180 ∘ − ( 180 ∘ − 2 θ ) = 2 θ A + B = 180^\circ - C = 180^\circ - (180^\circ - 2\theta) = 2\theta A + B = 18 0 ∘ − C = 18 0 ∘ − ( 18 0 ∘ − 2 θ ) = 2 θ
বামপক্ষ = sin 2 A − sin 2 B + sin 2 C = \sin 2A - \sin 2B + \sin 2C = sin 2 A − sin 2 B + sin 2 C = 2 cos ( A + B ) sin ( A − B ) + 2 sin C cos C = 2 \cos(A + B) \sin(A - B) + 2 \sin C \cos C = 2 cos ( A + B ) sin ( A − B ) + 2 sin C cos C = 2 cos ( 180 ∘ − C ) sin ( A − B ) + 2 sin C cos C = 2 \cos(180^\circ - C) \sin(A - B) + 2 \sin C \cos C = 2 cos ( 18 0 ∘ − C ) sin ( A − B ) + 2 sin C cos C = − 2 cos C sin ( A − B ) + 2 sin C cos C = -2 \cos C \sin(A - B) + 2 \sin C \cos C = − 2 cos C sin ( A − B ) + 2 sin C cos C = − 2 cos C [ sin ( A − B ) − sin C ] = -2 \cos C [\sin(A - B) - \sin C] = − 2 cos C [ sin ( A − B ) − sin C ] = − 2 cos C [ sin ( A − B ) − sin ( 180 ∘ − ( A + B ) ) ] = -2 \cos C [\sin(A - B) - \sin(180^\circ - (A + B))] = − 2 cos C [ sin ( A − B ) − sin ( 18 0 ∘ − ( A + B ))] = − 2 cos C [ sin ( A − B ) − sin ( A + B ) ] = -2 \cos C [\sin(A - B) - \sin(A + B)] = − 2 cos C [ sin ( A − B ) − sin ( A + B )] = -2 \cos C [-\(\sin(A + B) - \sin(A - B)\)] = 2 cos C [ 2 cos A sin B ] = 2 \cos C [2 \cos A \sin B] = 2 cos C [ 2 cos A sin B ] = 4 cos A sin B cos C = 4 \cos A \sin B \cos C = 4 cos A sin B cos C
যেহেতু C = 180 ∘ − 2 θ C = 180^\circ - 2\theta C = 18 0 ∘ − 2 θ cos C = cos ( 180 ∘ − 2 θ ) = − cos 2 θ \cos C = \cos(180^\circ - 2\theta) = -\cos 2\theta cos C = cos ( 18 0 ∘ − 2 θ ) = − cos 2 θ
∴ 4 cos A sin B ( − cos 2 θ ) \therefore 4 \cos A \sin B (-\cos 2\theta) ∴ 4 cos A sin B ( − cos 2 θ ) = − 4 cos A sin B cos 2 θ = = -4 \cos A \sin B \cos 2\theta = = − 4 cos A sin B cos 2 θ =
