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**Illustration 1: A straight line through the vertex P of a triangle PQR intersects the side QR at the point S and the circum circle of the triangle PQR at the point T. If S is not the center of the circum circle then which of the following relation is true? (2008)****1. 1/PS + 1/ST < 2/âˆšQS.SR****2. 1/PS + 1/ST > 2/âˆšQS.SR****3. 1/PS + 1/ST < 4/QR****4. 1/PS + 1/ST > 4/QR****Solution:** Let us suppose that a straight line through the vertex P of a given ?PQR intersects the side QR at the point S and the circum circle of the triangle PQR at the point T. Now, the points P, Q, R and T are concyclic and hence PS.ST = QS.SR

Now we know that A.M > G.M

So, we have, (PS + ST)/ 2 > âˆšPS.ST

1/ PS + 1/ ST > 2/ âˆšPS.ST = 2/ âˆšQS.SR

Also, (SQ + QR)/ 2 > âˆšQS.SR

So, QR/2 > âˆšQS.SR

1/âˆšQS.SR > 2/QR

Hence, 2/ âˆšQS.SR > 4/ QR

So, we obtain 1/ PS + 1/ ST > 2/ âˆšQS.SR > 4/ QR.

**Illustration 2: If in a triangle ABC, ****(2 cos A)/a + (cos B)/b + (2 cos C)/c = a/bc + b/ca. ****Find the value of angle A. (1993)****Solution:** The given condition is (2 cos A)/a + (cos B)/b + (2 cos C)/c = a/bc + b/ca. â€¦.. (1)

We know that in ?ABC, by cosine rule we have

cos A = b^{2} + c^{2} - a^{2 }/2bc

and so, cos B = c^{2 }+ a^{2} - b^{2} /2ac

and cos C = a^{2 }+ b^{2 }- c^{2} /2ab

We now substitute these values in equation (1) and hence obtain,

[2 (b^{2 }+ c^{2} - a^{2})/ 2abc] + [(c^{2} + a^{2} - b^{2}) /2abc] + [2(a^{2} + b^{2 }- c^{2})/ 2abc] = a/bc + b/ca

This gives [2(b^{2 }+ c^{2} - a^{2}) + (c^{2} + a^{2 }- b^{2}) + 2(a^{2} + b^{2} - c^{2})] / abc

= (a^{2} + b^{2})/abc

This means 3b^{2 }+ c^{2} + a^{2} = 2a^{2} + 2b^{2}

Hence, b^{2 }+ c^{2} = a^{2}.

Hence we get the measure of angle A is 90Â°.

**Illustration 3: In is the area of n-sided regular polygon inscribed in a circle of unit radius and On be the area of the polygon circumscribing the given circle, then prove that****I _{n} = O_{n}/2 (1 + âˆš{1 â€“ (2I_{n}/n)^{2}}) (2003)**

We know that I

Hence, since r = 1, so 2I

Also, O

Hence, O

Therefore 2I

Hence, I

= (1 + cos 2Ï€/n)/2

Therefore, from equation (1) we obtain,

I

Hence, I

Hence proved.

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