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Q.(a) Prove that sin⁡600°cos⁡390°+cos⁡480°sin⁡150°=−1\sin 600° \cos 390° + \cos 480° \sin 150° = -1.

(OR)
(b) Solve the following linear programming problem by graphical method : Minimize Z=20x1+40x2Z = 20x_1 + 40x_2 subject to the constraints 36x1+6x2≥10836x_1 + 6x_2 \geq 108, 3x1+12x2≥363x_1 + 12x_2 \geq 36, 20x1+10x2≥10020x_1 + 10x_2 \geq 100 and x1,x2≥0x_1, x_2 \geq 0.
Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Commerce Board 2020Subjective· 5mImportance★★★★★
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  1. Reduce each trig ratio to a standard angle and add: result −1-1. (b) Graph the three constraints; the corner points give minimum Z=160Z=160 at (4,2)(4,2). A 5-mark either/or from the Trigonometry / Operations Research units of the Tamil Nadu HSC Class-11 Business Mathematics syllabus. Both alternatives are solved. (a) Prove sin⁡600∘cos⁡390∘+cos⁡480∘sin⁡150∘=−1\sin600^\circ\cos390^\circ+\cos480^\circ\sin150^\circ=-1. Step 1 — Reduce each angle. sin⁡600∘=sin⁡(600∘−360∘)=sin⁡240∘=−sin⁡60∘=−32,\sin600^\circ=\sin(600^\circ-360^\circ)=\sin240^\circ=-\sin60^\circ=-\tfrac{\sqrt3}{2}, cos⁡390∘=cos⁡(390∘−360∘)=cos⁡30∘=32,\cos390^\circ=\cos(390^\circ-360^\circ)=\cos30^\circ=\tfrac{\sqrt3}{2}, cos⁡480∘=cos⁡(480∘−360∘)=cos⁡120∘=−12,\cos480^\circ=\cos(480^\circ-360^\circ)=\cos120^\circ=-\tfrac12, sin⁡150∘=sin⁡(180∘−30∘)=sin⁡30∘=12.\sin150^\circ=\sin(180^\circ-30^\circ)=\sin30^\circ=\tfrac12. Step 2 — Substitute. sin⁡600∘cos⁡390∘=(−32)(32)=−34,\sin600^\circ\cos390^\circ=\left(-\tfrac{\sqrt3}{2}\right)\left(\tfrac{\sqrt3}{2}\right)=-\tfrac34, cos⁡480∘sin⁡150∘=(−12)(12)=−14.\cos480^\circ\sin150^\circ=\left(-\tfrac12\right)\left(\tfrac12\right)=-\tfrac14. Step 3 — Add. −34−14=−1.-\tfrac34-\tfrac14=-1. Hence proved.
  2. LPP by graphical method. Minimize Z=20x1+40x2Z=20x_1+40x_2 subject to 36x1+6x2≥108, 3x1+12x2≥36, 20x1+10x2≥100, x1,x2≥0.36x_1+6x_2\ge108,\ 3x_1+12x_2\ge36,\ 20x_1+10x_2\ge100,\ x_1,x_2\ge0. Step 1 — Simplify the constraints. 6x1+x2≥18,x1+4x2≥12,2x1+x2≥10.6x_1+x_2\ge18,\qquad x_1+4x_2\ge12,\qquad 2x_1+x_2\ge10. Step 2 — Boundary lines and intercepts. 6x1+x2=186x_1+x_2=18 meets axes at (3,0),(0,18)(3,0),(0,18); x1+4x2=12x_1+4x_2=12 at (12,0),(0,3)(12,0),(0,3); 2x1+x2=102x_1+x_2=10 at (5,0),(0,10)(5,0),(0,10). The feasible region (all ≥\ge) is unbounded, lying above every line in the first quadrant. …
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