A vector is inclined at θ to the horizontal and its vertical component is Y Then magnitude of - Physics - Physical World

Question

A   v e c t o r   i s   i n c l i n e d   a t
  θ   t o   t h e   h o r i z o n t a l
  a n d   i t s   v e r t i c a l   c o m p o n
e n t   i s   Y   T h e n     m a g n i t
u d e   o f   t h a t   v e c t o r   i s 1 )
  Y sin θ 2 )   Y cos e c θ 3 )   Y cos
θ 4 )   Y s e c θ

Pintu B. · Accepted Answer

Dear Student ,
The magnitude of the vector is = Y sin( 90 - θ)= Y
cosθ

Regards

A v e c t o r i s i n c l i n e d a t θ t o t h e h o r i z o n t a l a n d i t s v e r t i c a l c o m p o n e n t i s Y . T h e n m a g n i t u d e o f t h a t v e c t o r i s 1 ) Y sin θ 2 ) Y cos e c θ 3 ) Y cos θ 4 ) Y s e c θ

$A v e c t o r i s i n c l i n e d a t θ t o t h e h o r i z o n t a l a n d i t s v e r t i c a l c o m p o n e n t i s Y . T h e n m a g n i t u d e o f t h a t v e c t o r i s 1) Y \sin θ 2) Y \cos e c θ 3) Y \cos θ 4) Y s e c θ$