ক) AI-তৈরি নমুনা উত্তর
দেওয়া আছে, tan θ = y x \tan \theta = \frac{y}{x} tan θ = x y
বামপক্ষ = x cos 2 θ + y sin 2 θ = x \cos 2\theta + y \sin 2\theta = x cos 2 θ + y sin 2 θ = x ( 1 − tan 2 θ 1 + tan 2 θ ) + y ( 2 tan θ 1 + tan 2 θ ) = x \left( \frac{1 - \tan^2\theta}{1 + \tan^2\theta} \right) + y \left( \frac{2\tan\theta}{1 + \tan^2\theta} \right) = x ( 1 + t a n 2 θ 1 − t a n 2 θ ) + y ( 1 + t a n 2 θ 2 t a n θ ) = x ( 1 − y 2 x 2 1 + y 2 x 2 ) + y ( 2 ⋅ y x 1 + y 2 x 2 ) = x \left( \frac{1 - \frac{y^2}{x^2}}{1 + \frac{y^2}{x^2}} \right) + y \left( \frac{2\cdot \frac{y}{x}}{1 + \frac{y^2}{x^2}} \right) = x ( 1 + x 2 y 2 1 − x 2 y 2 ) + y ( 1 + x 2 y 2 2 ⋅ x y ) = x ( x 2 − y 2 x 2 x 2 + y 2 x 2 ) + y ( 2 y x x 2 + y 2 x 2 ) = x \left( \frac{\frac{x^2 - y^2}{x^2}}{\frac{x^2 + y^2}{x^2}} \right) + y \left( \frac{\frac{2y}{x}}{\frac{x^2 + y^2}{x^2}} \right) = x ( x 2 x 2 + y 2 x 2 x 2 − y 2 ) + y ( x 2 x 2 + y 2 x 2 y ) = x ⋅ x 2 − y 2 x 2 + y 2 + y ⋅ 2 x y x 2 + y 2 = x \cdot \frac{x^2 - y^2}{x^2 + y^2} + y \cdot \frac{2xy}{x^2 + y^2} = x ⋅ x 2 + y 2 x 2 − y 2 + y ⋅ x 2 + y 2 2 x y = x ( x 2 − y 2 ) + 2 x y 2 x 2 + y 2 = \frac{x(x^2 - y^2) + 2xy^2}{x^2 + y^2} = x 2 + y 2 x ( x 2 − y 2 ) + 2 x y 2 = x 3 − x y 2 + 2 x y 2 x 2 + y 2 = \frac{x^3 - xy^2 + 2xy^2}{x^2 + y^2} = x 2 + y 2 x 3 − x y 2 + 2 x y 2 = x 3 + x y 2 x 2 + y 2 = \frac{x^3 + xy^2}{x^2 + y^2} = x 2 + y 2 x 3 + x y 2 = x ( x 2 + y 2 ) x 2 + y 2 = x = ডানপক্ষ = \frac{x(x^2 + y^2)}{x^2 + y^2} = x = \text{ডানপক্ষ} = x 2 + y 2 x ( x 2 + y 2 ) = x = ডানপক্ষ
∴ x cos 2 θ + y sin 2 θ = x \therefore x \cos 2\theta + y \sin 2\theta = x ∴ x cos 2 θ + y sin 2 θ = x
খ) AI-তৈরি নমুনা উত্তর
দৃশ্যকল্প-১ হতে,O A = 12 km OA = 12\text{ km} O A = 12 km O B = 9 km OB = 9\text{ km} O B = 9 km O N ∥ B N ′ ON \parallel BN' O N ∥ B N ′ O B OB O B 180 ∘ 180^\circ 18 0 ∘ O B OB O B 45 ∘ 45^\circ 4 5 ∘ ∠ O B N ′ = 45 ∘ \angle OBN' = 45^\circ ∠ O B N ′ = 4 5 ∘ ∠ A O B = 180 ∘ − 45 ∘ = 135 ∘ \angle AOB = 180^\circ - 45^\circ = 135^\circ ∠ A O B = 18 0 ∘ − 4 5 ∘ = 13 5 ∘
এখন, △ A O B \triangle AOB △ A O B A B 2 = O A 2 + O B 2 − 2 ⋅ O A ⋅ O B ⋅ cos ( ∠ A O B ) AB^2 = OA^2 + OB^2 - 2 \cdot OA \cdot OB \cdot \cos(\angle AOB) A B 2 = O A 2 + O B 2 − 2 ⋅ O A ⋅ O B ⋅ cos ( ∠ A O B ) ⇒ A B 2 = 12 2 + 9 2 − 2 × 12 × 9 × cos ( 135 ∘ ) \Rightarrow AB^2 = 12^2 + 9^2 - 2 \times 12 \times 9 \times \cos(135^\circ) ⇒ A B 2 = 1 2 2 + 9 2 − 2 × 12 × 9 × cos ( 13 5 ∘ ) ⇒ A B 2 = 144 + 81 − 216 × ( − 1 2 ) \Rightarrow AB^2 = 144 + 81 - 216 \times \left(-\frac{1}{\sqrt{2}}\right) ⇒ A B 2 = 144 + 81 − 216 × ( − 2 1 ) ⇒ A B 2 = 225 + 108 2 \Rightarrow AB^2 = 225 + 108\sqrt{2} ⇒ A B 2 = 225 + 108 2 ⇒ A B 2 ≈ 225 + 152.735 = 377.735 \Rightarrow AB^2 \approx 225 + 152.735 = 377.735 ⇒ A B 2 ≈ 225 + 152.735 = 377.735 ⇒ A B = 225 + 108 2 ≈ 19.44 km \Rightarrow AB = \sqrt{225 + 108\sqrt{2}} \approx 19.44\text{ km} ⇒ A B = 225 + 108 2 ≈ 19.44 km
অতএব, A B AB A B 19.44 km 19.44\text{ km} 19.44 km 225 + 108 2 km \sqrt{225 + 108\sqrt{2}}\text{ km} 225 + 108 2 km
গ) AI-তৈরি নমুনা উত্তর
দৃশ্যকল্প-২ হতে, △ P Q R \triangle PQR △ P QR Q R = l QR = l QR = l P R = m PR = m P R = m P Q = n PQ = n P Q = n T = l + m + n T = l + m + n T = l + m + n
দেওয়া আছে,1 T − m + 1 T − l = 3 T \frac{1}{T-m} + \frac{1}{T-l} = \frac{3}{T} T − m 1 + T − l 1 = T 3 ⇒ ( T − l ) + ( T − m ) ( T − m ) ( T − l ) = 3 T \Rightarrow \frac{(T-l) + (T-m)}{(T-m)(T-l)} = \frac{3}{T} ⇒ ( T − m ) ( T − l ) ( T − l ) + ( T − m ) = T 3 ⇒ 2 T − ( l + m ) T 2 − ( l + m ) T + l m = 3 T \Rightarrow \frac{2T - (l+m)}{T^2 - (l+m)T + lm} = \frac{3}{T} ⇒ T 2 − ( l + m ) T + l m 2 T − ( l + m ) = T 3
যেহেতু T = l + m + n T = l + m + n T = l + m + n l + m = T − n l + m = T - n l + m = T − n ⇒ 2 T − ( T − n ) T 2 − ( T − n ) T + l m = 3 T \Rightarrow \frac{2T - (T - n)}{T^2 - (T - n)T + lm} = \frac{3}{T} ⇒ T 2 − ( T − n ) T + l m 2 T − ( T − n ) = T 3 ⇒ T + n T 2 − T 2 + n T + l m = 3 T \Rightarrow \frac{T + n}{T^2 - T^2 + nT + lm} = \frac{3}{T} ⇒ T 2 − T 2 + n T + l m T + n = T 3 ⇒ T + n n T + l m = 3 T \Rightarrow \frac{T + n}{nT + lm} = \frac{3}{T} ⇒ n T + l m T + n = T 3 ⇒ T ( T + n ) = 3 ( n T + l m ) \Rightarrow T(T + n) = 3(nT + lm) ⇒ T ( T + n ) = 3 ( n T + l m ) ⇒ T 2 + n T = 3 n T + 3 l m \Rightarrow T^2 + nT = 3nT + 3lm ⇒ T 2 + n T = 3 n T + 3 l m ⇒ T 2 − 2 n T − 3 l m = 0 \Rightarrow T^2 - 2nT - 3lm = 0 ⇒ T 2 − 2 n T − 3 l m = 0 ⇒ ( l + m + n ) 2 − 2 n ( l + m + n ) − 3 l m = 0 \Rightarrow (l + m + n)^2 - 2n(l + m + n) - 3lm = 0 ⇒ ( l + m + n ) 2 − 2 n ( l + m + n ) − 3 l m = 0 ⇒ ( l + m ) 2 + 2 ( l + m ) n + n 2 − 2 n ( l + m ) − 2 n 2 − 3 l m = 0 \Rightarrow (l + m)^2 + 2(l + m)n + n^2 - 2n(l + m) - 2n^2 - 3lm = 0 ⇒ ( l + m ) 2 + 2 ( l + m ) n + n 2 − 2 n ( l + m ) − 2 n 2 − 3 l m = 0 ⇒ ( l + m ) 2 − n 2 − 3 l m = 0 \Rightarrow (l + m)^2 - n^2 - 3lm = 0 ⇒ ( l + m ) 2 − n 2 − 3 l m = 0 ⇒ l 2 + 2 l m + m 2 − n 2 − 3 l m = 0 \Rightarrow l^2 + 2lm + m^2 - n^2 - 3lm = 0 ⇒ l 2 + 2 l m + m 2 − n 2 − 3 l m = 0 ⇒ l 2 + m 2 − n 2 − l m = 0 \Rightarrow l^2 + m^2 - n^2 - lm = 0 ⇒ l 2 + m 2 − n 2 − l m = 0 ⇒ l 2 + m 2 − n 2 = l m \Rightarrow l^2 + m^2 - n^2 = lm ⇒ l 2 + m 2 − n 2 = l m
আমরা জানি, কোসাইন সূত্রানুসারে:cos R = l 2 + m 2 − n 2 2 l m \cos R = \frac{l^2 + m^2 - n^2}{2lm} cos R = 2 l m l 2 + m 2 − n 2 ⇒ cos R = l m 2 l m = 1 2 \Rightarrow \cos R = \frac{lm}{2lm} = \frac{1}{2} ⇒ cos R = 2 l m l m = 2 1 ⇒ cos R = cos 60 ∘ \Rightarrow \cos R = \cos 60^\circ ⇒ cos R = cos 6 0 ∘ ⇒ R = 60 ∘ \Rightarrow R = 60^\circ ⇒ R = 6 0 ∘ π 3 \frac{\pi}{3} 3 π
অতএব, R R R 60 ∘ 60^\circ 6 0 ∘ π 3 \frac{\pi}{3} 3 π
