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1

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 $$c=2R$$)
2

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 $$A.P.$$
3

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 $$C$$

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

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

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

$$\therefore$$ $$C = {120^ \circ }$$
4

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 $$(2),$$

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

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