The vector u has a magnitude ∥u ∥= 12 ft and direction with generalized angle α = 135°. Find the magnitude of its vertical and horizontal components

The magnitude of the vertical component of vector u is  

The magnitude of the horizontal component of vector u is  

Step-by-step explanation:

A vector quantity has, a magnitude and a direction

The horizontal component of vector is ║u║ cos α

The vertical component of vector is ║u║ sin α

∵ The magnitude of vector u is 12 ft

∵ The measure of angle α is 135°

- Substitute the magnitude and the angle in the rule of

   each component

∴ = 12 cos(135)

∴ = 12 sin(135)

∵ cos(135) =

∴ = 12 ( )

∴ =  

- The magnitude of a number means the number without its sign

∴ The magnitude of the horizontal component of vector u is  

∵ sin(135) =

∴ = 12 ( )

∴ =  

∴ The magnitude of the vertical component of vector u is  

-60.75

step-by-step explanation:

just add the numbers then put the negative sign on it

answer is (b.

164 ft2)

a base  

18

b base  

7

h height  

8