Question

▶️In a clock the minute hand has radius 49 cm, so find distance and displacement if minute on 20 minutes​

Answer 1

Answer:

Algebraic equation, statement of the equality of two expressions formulated by applying to a set of variables the algebraic operations, namely, addition, subtraction, multiplication, division, raising to a power, and extraction of a root. Examples are x3 + 1 and (y4x2 + 2xy – y)/(x – 1) = 12.

Answer 2

$\bf\huge{\underline{\pink{ given }}}$

Angle made by minute hand in 60 min= 360

So angle made in 20 min = 360/3 = 120

as 60/3 = 20 So its 1/3 rd of 360

Or

Or360× 20/60 = 6× 20 = 120

Arc length ( ASB)( Distance ) = theta × radius

= 120 × pie/180 ( radian converted) × 49

= 120 × 22 × 49/(7× 180)

= 120 × 22 × 49/(7× 180)= 120× 22 × 7 / 180

= 120 × 22 × 49/(7× 180)= 120× 22 × 7 / 180= 2× 22× 7/ 3 = 308/3= 102.66

Displacement ( AB)

Draw perpendicular from O to AB

Draw perpendicular from O to ABAs its isosceles triangle

Draw perpendicular from O to ABAs its isosceles triangleso it would be perpendicular bisector

Draw perpendicular from O to ABAs its isosceles triangleso it would be perpendicular bisectoralso angle bisect

Draw perpendicular from O to ABAs its isosceles triangleso it would be perpendicular bisectoralso angle bisectAOF = 60

Draw perpendicular from O to ABAs its isosceles triangleso it would be perpendicular bisectoralso angle bisectAOF = 60In OAF

Draw perpendicular from O to ABAs its isosceles triangleso it would be perpendicular bisectoralso angle bisectAOF = 60In OAFAF = sin60 × AO = 49 × √3/2

Draw perpendicular from O to ABAs its isosceles triangleso it would be perpendicular bisectoralso angle bisectAOF = 60In OAFAF = sin60 × AO = 49 × √3/2As AF = FB

Draw perpendicular from O to ABAs its isosceles triangleso it would be perpendicular bisectoralso angle bisectAOF = 60In OAFAF = sin60 × AO = 49 × √3/2As AF = FBSo AB = 2× 49√3/2 = 49√3 which is displacement