If v lies in the first quadrant and makes an angle π/3 with the positive x-axis and ivi =4, find v in component form

V ( Vcos 60° , Vsin60°)  or   V ( 2, 2√3)

Step-by-step explanation:

A vector and its components form a right triangle, then we need to get trigonometric functions  ( sin and cos) of the angle of the module with x-axis

π/3  = 180⁰/3  = 60⁰  Angle between vector and positive x-axis

and  sin 60⁰ = √3 /2       cos 60⁰ = 1/2

Let call V(x)  component of vector v in x-axis,   and  V(y) component of vector v in y-axis then

V(x)  =  |v|*cos 60⁰         ⇒  V(x)  =  |v|*1/2 ⇒  V(x) = 4*1/2  ⇒  V(x)  = 2

And

V(y)  =   |v|*sin 60⁰        ⇒   V(y)  =  |v|*√3/2  ⇒   V(y)  =  4*√3/2  V(y) = 2*√3

V = (Vx,Vy)     V ( Vcos 60° , Vsin60°)   V ( 2, 2√3)

addition

step-by-step explanation:

x² + 11x + 18 < 0  

[-11 ± sqrt(121 - 72)]/2 = (-11 ± 7)/2  

solution: (-9, -2)

step-by-step explanation: