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1

### AIEEE 2008

$$AB$$ is a vertical pole with $$B$$ at the ground level and $$A$$ at the top. $$A$$ man finds that the angle of elevation of the point $$A$$ from a certain point $$C$$ on the ground is $${60^ \circ }$$. He moves away from the pole along the line $$BC$$ to a point $$D$$ such that $$CD=7$$ m. From $$D$$ the angle of elevation of the point $$A$$ is $${45^ \circ }$$. Then the height of the pole is
A
$${{7\sqrt 3 } \over 2} {1 \over {\sqrt {3 - 1} }}m$$
B
$${{7\sqrt 3 } \over 2}\left( {\sqrt {3 + 1} } \right)m$$
C
$${{7\sqrt 3 } \over 2}\left( {\sqrt {3 - 1} } \right)m$$
D
$${{7\sqrt 3 } \over 2} {1 \over {\sqrt {3 + 1} }}m$$

## Explanation

In $$\Delta ABC$$

$${h \over x} = \tan {60^ \circ } = \sqrt 3$$

$$\Rightarrow x = {h \over {\sqrt 3 }}$$

In $$\Delta ABD{h \over {x + 7}}$$

$$= \tan {45^ \circ } = 1$$

$$\Rightarrow h = x + 7 \Rightarrow h - {h \over {\sqrt 3 }} = 7$$

$$\Rightarrow h = {{7\sqrt 3 } \over {\sqrt 3 - 1}} \times {{\sqrt 3 + 1} \over {\sqrt 3 + 1}}$$

$$\Rightarrow h = {{7\sqrt 3 } \over 2}\left( {\sqrt 3 + 1\,m} \right)$$
2

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

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

### 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 $$c=2R$$)
4

### 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 $$A.P.$$

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