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1

### AIEEE 2007

A tower stands at the centre of a circular park. $$A$$ and $$B$$ are two points on the boundary of the park such that $$AB(=a)$$ subtends an angle of $${60^ \circ }$$ at the foot of the tower, and the angle of elevation of the top of the tower from $$A$$ or $$B$$ is $${30^ \circ }$$. The height of the tower is
A
$$a/\sqrt 3$$
B
$$a\sqrt 3$$
C
$$2a/\sqrt 3$$
D
$$2a\sqrt 3$$

## Explanation

In the $$\Delta AOB,\,\,\angle AOB = {60^ \circ },$$ and

$$\angle OBA = \angle OAB$$

(since $$OA=OB=AB$$ radius of same circle).

$$\therefore$$ $$\Delta AOB$$ is a equilateral triangle.

Let the height of tower is $$h$$ $$m.$$ Given distance between two points $$A$$ & $$B$$ lie on boundary of

circular park, subtends an angle of $${60^ \circ }$$ at the foot of the tower

is $$AB$$ i.e. $$AB$$$$=a.$$ A tower $$OC$$ stands at the center of a circular

park. Angle of elevation of the top of the tower from $$A$$ and $$B$$ is $${30^ \circ }$$ .

In $$\Delta OAC\,\,\tan {30^ \circ } = {h \over a}$$

$$\Rightarrow {1 \over {\sqrt 3 }} = {h \over a} \Rightarrow h = {a \over {\sqrt 3 }}$$
2

### AIEEE 2005

In a triangle $$ABC$$, let $$\angle C = {\pi \over 2}$$. If $$r$$ is the inradius and $$R$$ is the circumradius of the triangle $$ABC$$, then $$2(r+R)$$ equals
A
$$b+c$$
B
$$a+b$$
C
$$a+b+c$$
D
$$c+a$$

## Explanation

We know by sin c rule

$${c \over {\sin C}} = 2R \Rightarrow c = 2R\sin C$$

$$\Rightarrow c = 2R$$

$$\left( \, \right.$$as $$\,\,\,\angle C = {90^ \circ }$$ $$\left. \, \right)$$

Also $$\tan {C \over 2} = {r \over {s - c}}$$

$$\Rightarrow \tan {\pi \over 4} = {r \over {s - c}}$$

$$\left( \, \right.$$as $$\,\,\,\angle C = {90^ \circ }$$ $$\left. \, \right)$$

$$\Rightarrow r = s - c = {{a + b - c} \over 2}$$

$$\Rightarrow 2r + c = a + b$$

$$\Rightarrow 2r + 2R = a + b$$

(using $$c=2R$$)
3

### AIEEE 2005

If in a $$\Delta ABC$$, the altitudes from the vertices $$A, B, C$$ on opposite sides are in H.P, then $$\sin A,\sin B,\sin C$$ are in
A
G. P.
B
A. P.
C
A.P-G.P.
D
H. P

## Explanation

$$\Delta = {1 \over 2}{p_1}a = {1 \over 2}{p_2}b = {1 \over 2}{p_3}b$$

$${p_1},{p_2},{p_3},$$ are in $$H.P.$$

$$\Rightarrow {{2\Delta } \over a},{{2\Delta } \over b},{{2\Delta } \over c}$$ are in $$H.P.$$

$$\Rightarrow {1 \over a},{1 \over b},{1 \over c},$$ are in $$H.P.$$

$$\Rightarrow a,b,c$$ are in $$A.P.$$

$$\Rightarrow$$ $$K\sin A,K\sin B,K\sin C$$ are in $$A.P.$$

$$\Rightarrow$$ $$\sin A,\sin B,\sin C$$ are in $$A.P.$$
4

### 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 $$C$$

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

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

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

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

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