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 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 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 = $$\theta$$