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1

### JEE Main 2021 (Online) 27th August Morning Shift

Let $${{\sin A} \over {\sin B}} = {{\sin (A - C)} \over {\sin (C - B)}}$$, where A, B, C are angles of triangle ABC. If the lengths of the sides opposite these angles are a, b, c respectively, then :
A
b2 $$-$$ a2 = a2 + c2
B
b2, c2, a2 are in A.P.
C
c2, a2, b2 are in A.P.
D
a2, b2, c2 are in A.P.

## Explanation

$${{\sin A} \over {\sin B}} = {{\sin (A - C)} \over {\sin (C - B)}}$$

As A, B, C are angles of triangle.

A + B + C = $$\pi$$

A = $$\pi$$ $$-$$ (B + C) ...... (1)

Similarly sinB = sin(A + C) ..... (2)

From (1) and (2)

$${{\sin (B + C)} \over {\sin (A + C)}} = {{\sin (A - C)} \over {\sin (C - B)}}$$

$$\sin (C + B).\sin (C - B) = \sin (A - C)\sin (A + C)$$

$${\sin ^2}C - {\sin ^2}B = {\sin ^2}A - {\sin ^2}C$$

$$\because$$ $$\{ \sin (x + y)\sin (x - y) = {\sin ^2}x - {\sin ^2}y\}$$

$$2{\sin ^2}C = {\sin ^2}A + {\sin ^2}B$$

By sine rule

$$2{c^2} = {a^2} + {b^2}$$

$$\Rightarrow$$ b2, c2 and a2 are in A.P.
2

### JEE Main 2021 (Online) 26th August Evening Shift

A 10 inches long pencil AB with mid point C and a small eraser P are placed on the horizontal top of a table such that PC = $$\sqrt 5$$ inches and $$\angle$$PCB = tan-1(2). The acute angle through which the pencil must be rotated about C so that the perpendicular distance between eraser and pencil becomes exactly 1 inch is :

A
$${\tan ^{ - 1}}\left( {{3 \over 4}} \right)$$
B
tan$$-$$1(1)
C
$${\tan ^{ - 1}}\left( {{4 \over 3}} \right)$$
D
$${\tan ^{ - 1}}\left( {{1 \over 2}} \right)$$

## Explanation

From figure,

$$\sin \beta = {1 \over {\sqrt 5 }}$$

$$\therefore$$ $$\tan \beta = {1 \over 2}$$

$$\tan (\alpha + \beta ) = 2$$

$${{\tan \alpha + \tan \beta )} \over {1 - \tan \alpha .\tan \beta }} = 2$$

$${{\tan \alpha + {1 \over 2}} \over {1 - \tan \alpha \left( {{1 \over 2}} \right)}} = 2$$

$$\tan \alpha = {3 \over 4}$$

$$\alpha = {\tan ^{ - 1}}\left( {{3 \over 4}} \right)$$
3

### JEE Main 2021 (Online) 25th July Morning Shift

A spherical gas balloon of radius 16 meter subtends an angle 60$$^\circ$$ at the eye of the observer A while the angle of elevation of its center from the eye of A is 75$$^\circ$$. Then the height (in meter) of the top most point of the balloon from the level of the observer's eye is :
A
$$8(2 + 2\sqrt 3 + \sqrt 2 )$$
B
$$8(\sqrt 6 + \sqrt 2 + 2)$$
C
$$8(\sqrt 2 + 2 + \sqrt 3 )$$
D
$$8(\sqrt 6 - \sqrt 2 + 2)$$

## Explanation

O $$\to$$ centre of sphere

P, Q $$\to$$ point of contact of tangents from A

Let T be top most point of balloon & R be foot of perpendicular from O to ground.

From triangle OAP, OA = 16cosec30$$^\circ$$ = 32

From triangle ABO, OR = OA sin75$$^\circ$$ = $$32{{\left( {\sqrt 3 + 1} \right)} \over {2\sqrt 2 }}$$

So level of top most point = OR + OT

$$= 8\left( {\sqrt 6 + \sqrt 2 + 2} \right)$$
4

### JEE Main 2021 (Online) 20th July Evening Shift

Let in a right angled triangle, the smallest angle be $$\theta$$. If a triangle formed by taking the reciprocal of its sides is also a right angled triangle, then sin$$\theta$$ is equal to :
A
$${{\sqrt 5 + 1} \over 4}$$
B
$${{\sqrt 5 - 1} \over 2}$$
C
$${{\sqrt 2 - 1} \over 2}$$
D
$${{\sqrt 5 - 1} \over 4}$$

## Explanation

Let a $$\Delta$$ABC having C = 90$$^\circ$$ and A = $$\theta$$

$${{\sin \theta } \over a} = {{\cos \theta } \over b} = {1 \over c}$$ ..... (i)

Also for triangle of reciprocals

$$\cos A = {{{{\left( {{1 \over c}} \right)}^2} + {{\left( {{1 \over b}} \right)}^2} - {{\left( {{1 \over a}} \right)}^2}} \over {2\left( {{1 \over c}} \right)\left( {{1 \over b}} \right)}}$$

$${1 \over {{c^2}}} + {1 \over {{{(c\cos \theta )}^2}}} = {1 \over {{{(c\sin \theta )}^2}}}$$

$$\Rightarrow 1 + {\sec ^2}\theta = \cos e{c^2}\theta$$

$$\Rightarrow {1 \over 4} = {{{{\cos }^2}\theta } \over {4{{\sin }^2}\theta {{\cos }^2}\theta }}$$

$$\Rightarrow {1 \over 4} = {{{{\cos }^2}\theta } \over {{{\sin }^2}2\theta }}$$

$$\Rightarrow 1 - {\cos ^2}2\theta = 4\cos 2\theta$$

$${\cos ^2}2\theta + 4\cos 2\theta - 1 = 0$$

$$\cos 2\theta = {{ - 4 \pm \sqrt {16 + 4} } \over 2}$$

$$\cos 2\theta = - 2 \pm \sqrt 5$$

$$\cos 2\theta = \sqrt 5 - 2 = 1 - 2{\sin ^2}\theta$$

$$\Rightarrow 2{\sin ^2}\theta = 3 - \sqrt 5$$

$$\Rightarrow {\sin ^2}\theta = {{3 - \sqrt 5 } \over 2}$$

$$\Rightarrow \sin \theta = {{\sqrt 5 - 1} \over 2}$$

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