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1

JEE Main 2021 (Online) 27th August Morning Shift

MCQ (Single Correct Answer)
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

MCQ (Single Correct Answer)
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

MCQ (Single Correct Answer)
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

MCQ (Single Correct Answer)
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

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

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

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

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

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

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

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

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

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

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

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

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

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

Questions Asked from Properties of Triangle

On those following papers in MCQ (Single Correct Answer)
Number in Brackets after Paper Indicates No. of Questions
JEE Main 2021 (Online) 31st August Morning Shift (1)
JEE Main 2021 (Online) 27th August Evening Shift (1)
JEE Main 2021 (Online) 27th August Morning Shift (1)
JEE Main 2021 (Online) 26th August Evening Shift (1)
JEE Main 2021 (Online) 25th July Morning Shift (1)
JEE Main 2021 (Online) 20th July Evening Shift (1)
JEE Main 2021 (Online) 20th July Morning Shift (1)
JEE Main 2021 (Online) 18th March Evening Shift (1)
JEE Main 2021 (Online) 26th February Evening Shift (1)
JEE Main 2021 (Online) 25th February Morning Shift (1)
JEE Main 2021 (Online) 24th February Evening Shift (1)
JEE Main 2021 (Online) 24th February Morning Shift (1)
JEE Main 2020 (Online) 6th September Evening Slot (1)
JEE Main 2020 (Online) 4th September Evening Slot (1)
JEE Main 2020 (Online) 4th September Morning Slot (1)
JEE Main 2019 (Online) 12th April Evening Slot (1)
JEE Main 2019 (Online) 10th April Evening Slot (1)
JEE Main 2019 (Online) 10th April Morning Slot (1)
JEE Main 2019 (Online) 9th April Evening Slot (1)
JEE Main 2019 (Online) 8th April Evening Slot (2)
JEE Main 2019 (Online) 12th January Evening Slot (1)
JEE Main 2019 (Online) 11th January Evening Slot (1)
JEE Main 2019 (Online) 11th January Morning Slot (1)
JEE Main 2019 (Online) 10th January Evening Slot (1)
JEE Main 2019 (Online) 10th January Morning Slot (1)
JEE Main 2018 (Online) 16th April Morning Slot (1)
JEE Main 2018 (Online) 15th April Evening Slot (1)
JEE Main 2018 (Online) 15th April Morning Slot (1)
JEE Main 2016 (Online) 10th April Morning Slot (1)
JEE Main 2015 (Offline) (1)
JEE Main 2014 (Offline) (1)
AIEEE 2010 (1)
AIEEE 2008 (1)
AIEEE 2007 (1)
AIEEE 2005 (2)
AIEEE 2004 (2)
AIEEE 2003 (3)
AIEEE 2002 (2)

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