**1133**
Sourish Ghosh
**·**2014-03-02 02:47:58
sinisinr = 1.414

i = 90-r2

Use wolframalpha to solve the first equation.

http://www.wolframalpha.com/input/?i=sin%28%2890-x%29%2F2%29+-+1.414*sin%28x%29+%3D+0

Area: Ï€[2tanr + cot(90+r2)]^{2}

**2305**
Shaswata Roy
**·**2014-03-07 21:48:44
I don't have that solution anymore.

Neither do I have that question anymore.

But anyways the solution below is somewhat similar.

sin(i)sin(r)=Î¼....(1)

The triangles FEI and FCI are congruent.

â†’ <FCI =<FEI = i

â†’<HFC = <FCI = i(Alternate angles)

Since FI and FG are tangents drawn from the same point (F) to the circle.

<IFC = <CFG = i+r

<IFH = <IFC +<CFH = i+r + i=2i+r=90Â°....(2)

Tangents drawn from a point to a circle are of equal lengths.

Using this property,

FI = FJ and GD = GJ

Therefore,

FG = FJ+GJ = FI+GD = FI + GH +HD =2FI+GH (Because HD = FI)

â†’2sec(r) = 2tan(r) + 2tan(i)

â†’ 1=sin(r)+sin(i)cos(r)cos(i)

â†’1=sin(i)Î¼+sin(i)cos(90-2i)cos(i)

â†’1 = sin(i)Î¼+2sin^{2}(i)

This is a quadratic in sin(i).

Out of the 2 roots we need to take the positive root.

Than we can find sin(r) from equation(1).

The length GD = GH+HD=2tan(r)+tan(i)

And area = Ï€(GD)^{2}

·0·Reply·2014-03-06 05:32:52