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1

### AIEEE 2004

The sides of a triangle are $$\sin \alpha ,\,\cos \alpha$$ and $$\sqrt {1 + \sin \alpha \cos \alpha }$$ for some $$0 < \alpha < {\pi \over 2}$$. Then the greatest angle of the triangle is
A
$${150^ \circ }$$
B
$${90^ \circ }$$
C
$${120^ \circ }$$
D
$${60^ \circ }$$

## Explanation

Let $$a = \sin \alpha ,b = \cos \alpha$$

and $$c = \sqrt {1 + \sin \alpha \cos \alpha }$$

Clearly $$a$$ and $$b < 1$$ but $$c > 1$$

as $$\,\,\,\sin \alpha > 0$$ and $$\cos \alpha > 0$$

$$\therefore$$ $$c$$ is the greatest side and greatest angle is $$C$$

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

$$= {{{{\sin }^2}\alpha + {{\cos }^2}\alpha - 1 - \sin \alpha \cos \alpha } \over {2\,\sin \alpha \cos \alpha }}$$

$$= - {1 \over 2}$$

$$\therefore$$ $$C = {120^ \circ }$$
2

### AIEEE 2004

A person standing on the bank of a river observes that the angle of elevation of the top of a tree on the opposite bank of the river is $${60^ \circ }$$ and when he retires $$40$$ meters away from the tree the angle of elevation becomes $${30^ \circ }$$. The breadth of the river is
A
$$60\,\,m$$
B
$$30\,\,m$$
C
$$40\,\,m$$
D
$$20\,\,m$$

## Explanation

From the figure

$$\tan {60^ \circ } = {y \over x}$$

$$\Rightarrow y = \sqrt {3x} .......\left( 1 \right)$$

$$\tan {30^ \circ } = {y \over {x + 40}}$$

$$\Rightarrow y = {{x + 40} \over {\sqrt 3 }}........\left( 2 \right)$$

From $$(1)$$ and $$(2),$$

$$\sqrt 3 x = {{x + 40} \over {\sqrt 3 }} \Rightarrow x = 20m$$
3

### AIEEE 2003

If in a $$\Delta ABC$$ $$a{\cos ^2}\left( {{C \over 2}} \right) + {\cos ^2}\left( {{A \over 2}} \right) = {{3b} \over 2},$$ then the sides $$a, b$$ and $$c$$
A
satisfy $$a+b=c$$
B
are in A.P
C
are in G.P
D
are in H.P

## Explanation

If $$a\,{\cos ^2}\left( {{C \over 2}} \right) + c\,{\cos ^2}\left( {{A \over 2}} \right) = {{3b} \over 2}$$

$$a\left[ {\cos C + 1} \right] + c\left[ {\cos A + 1} \right] = 3b$$

$$\left( {a + c} \right) + \left( {a\cos C + c\cos \,B} \right) = 3b$$

$$a + c + b = 3b$$ or $$a + c = 2b$$

or $$a,b,c$$ are in $$A.P.$$
4

### AIEEE 2003

In a triangle $$ABC$$, medians $$AD$$ and $$BE$$ are drawn. If $$AD=4$$,
$$\angle DAB = {\pi \over 6}$$ and $$\angle ABE = {\pi \over 3}$$, then the area of the $$\angle \Delta ABC$$ is
A
$${{64} \over 3}$$
B
$${8 \over 3}$$
C
$${{16} \over 3}$$
D
$${{32} \over {3\sqrt 3 }}$$

## Explanation

$$AP = {2 \over 3}AD = {8 \over 3};\,\,PD = {4 \over 3};\,\,$$

Let $$PB=x$$

$$\tan {60^ \circ } = {{8/3} \over x}$$

or $$x = {8 \over {3\sqrt 3 }}$$

Area of $$\Delta ABD$$

$$= {1 \over 2} \times 4 \times {8 \over {3\sqrt 3 }} = {{16} \over {3\sqrt 3 }}$$

$$\therefore$$ Area of $$\Delta ABC$$

$$= 2 \times {{16} \over {3\sqrt 3 }} = {{32} \over {3\sqrt 3 }}$$

$$\left[ \, \right.$$ As median of a $$\Delta$$ divides it into two $$\Delta 's$$ of equal area. $$\left. \, \right]$$

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