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1

### JEE Advanced 2014 Paper 2 Offline

In a triangle the sum of two sides is $$x$$ and the product of the same sides is $$y$$. If $${x^2} - {c^2} = y$$, where $$c$$ is the third side of the triangle, then the ratio of the in radius to the circum-radius of the triangle is
A
$${{3y} \over {2x\left( {x + c} \right)}}$$
B
$${{3y} \over {2c\left( {x + c} \right)}}$$
C
$${{3y} \over {4x\left( {x + c} \right)}}$$
D
$${{3y} \over {4c\left( {x + c} \right)}}$$
2

### IIT-JEE 2012 Paper 2 Offline

Let $$PQR$$ be a triangle of area $$\Delta$$ with $$a=2$$, $$b = {7 \over 2}$$ and $$c = {5 \over 2}$$; where $$a, b,$$ and $$c$$ are the lengths of the sides of the triangle opposite to the angles at $$P.Q$$ and $$R$$ respectively. Then $${{2\sin P - \sin 2P} \over {2\sin P + \sin 2P}}$$ equals.
A
$${3 \over {4\Delta }}$$
B
$${45 \over {4\Delta }}$$
C
$${\left( {{3 \over {4\Delta }}} \right)^2}$$
D
$${\left( {{45 \over {4\Delta }}} \right)^2}$$
3

### IIT-JEE 2010 Paper 1 Offline

Let $$ABC$$ be a triangle such that $$\angle ACB = {\pi \over 6}$$ and let $$a, b$$ and $$c$$ denote the lengths of the sides opposite to $$A$$, $$B$$ and $$C$$ respectively. The value(s) of $$x$$ for which $$a = {x^2} + x + 1,\,\,\,b = {x^2} - 1\,\,\,$$ and $$c = 2x + 1$$ is (are)
A
$$- \left( {2 + \sqrt 3 } \right)$$
B
$${1 + \sqrt 3 }$$
C
$${2 + \sqrt 3 }$$
D
$${4 \sqrt 3 }$$

## Explanation

Using, $$\cos C = {{{a^2} + {b^2} - {c^2}} \over {2ab}}$$

$$\Rightarrow {{\sqrt 3 } \over 2} = {{{{({x^2} + x + 1)}^2} + {{({x^2} - 1)}^2} - {{(2x + 1)}^2}} \over {2({x^2} + x + 1)({x^2} - 1)}}$$

$$\Rightarrow (x + 2)(x + 1)(x - 1)x + {({x^2} - 1)^2} = \sqrt 3 ({x^2} + x + 1)({x^2} - 1)$$

$$\Rightarrow {x^2} + 2x + ({x^2} - 1) = \sqrt 3 ({x^2} + x + 1)$$

$$\Rightarrow (2 - \sqrt 3 ){x^2} + (2 - \sqrt 3 )x - (\sqrt 3 + 1) = 0$$

$$\Rightarrow x = - (2 + \sqrt 3 )$$ and $$x = 1 + \sqrt 3$$

But, $$x = - (2 + \sqrt 3 ) \Rightarrow c$$ is negative.

$$\therefore$$ $$x = 1 + \sqrt 3$$ is the only solution.

Hence, (b) is the correct option.

4

### IIT-JEE 2010 Paper 1 Offline

If the angles $$A, B$$ and $$C$$ of a triangle are in an arithmetic progression and if $$a, b$$ and $$c$$ denote the lengths of the sides opposite to $$A, B$$ and $$C$$ respectively, then the value of the expression $${a \over c}\sin 2C + {c \over a}\sin 2A$$ is
A
$${1 \over 2}$$
B
$${{\sqrt 3 } \over 2}$$
C
$$1$$
D
$${\sqrt 3 }$$

## Explanation

Since, A, B, C are in AP

$$\Rightarrow$$ 2B = A + C i.e., $$\angle$$B = 60$$^\circ$$

$$\therefore$$ $${a \over c}$$(2 sin C cos C) + $${c \over a}$$ (2 sin A cos A)

= 2k (a cos C + c cos A)

[using, $${a \over {\sin A}} = {b \over {\sin B}} = {c \over {\sin C}} = {1 \over k}$$]

= 2k (b)

= 2 sin B

[using, b = a cos C + c cos A]

= $$\sqrt3$$

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