<p>In a △ABC; inscribed circle with centre I touches sides AB, AC and BC at D, E, F respectively. Let area of quadrilateral ADIE is 5 units and area of quadrilateral BFID is 10 units. Find the value of cos(C/2)/sin((A-B)/2).</p>
Step-by-Step Solution
Key Concept: Use the property that tangent segments from a vertex to the incircle are equal (AD = AE, BD = BF, CE = CF), and express quadrilateral areas in terms of inradius and tangent lengths to set up equations relating the sides.
<p><strong>Step 1: Set up tangent length notation</strong></p><p>Let the incircle have radius r and centre I. By tangent properties: AD = AE = x, BD = BF = y, CE = CF = z.</p><p><strong>Step 2: Express quadrilateral areas</strong></p><p>Area of quadrilateral ADIE = Area(△ADI) + Area(△AEI) = ½·AD·r + ½·AE·r = ½·x·r + ½·x·r = xr = 5</p><p>Area of quadrilateral BFID = Area(△BFI) + Area(△BDI) = ½·BF·r + ½·BD·r = ½·y·r + ½·y·r = yr = 10</p><p><strong>Step 3: Find x and y in terms of r</strong></p><p>From xr = 5: x = 5/r</p><p>From yr = 10: y = 10/r</p><p><strong>Step 4: Express sides in terms of tangent lengths</strong></p><p>AB = x + y = 5/r + 10/r = 15/r, so c = 15/r</p><p>Let the third quadrilateral area CEIF = zr. Then: a = y + z, b = x + z</p><p><strong>Step 5: Use the target expression</strong></p><p>We need: cos(C/2)/sin((A-B)/2)</p><p>Using cos(C/2) = √[s(s-c)/(ab)] where s is semi-perimeter, and sin((A-B)/2) = (a-b)cos(C/2)/(2c) (from standard identities).</p><p><strong>Step 6: Simplify the ratio</strong></p><p>cos(C/2)/sin((A-B)/2) = cos(C/2)/[(a-b)cos(C/2)/(2c)] = 2c/(a-b)</p><p>Since a = y + z = 10/r + z and b = x + z = 5/r + z:</p><p>a - b = (10/r + z) - (5/r + z) = 5/r</p><p>Therefore: 2c/(a-b) = 2(15/r)/(5/r) = 30/5 = 6</p><p><strong>∴ Answer: 6</strong></p>
Correct Answer: 6