Only two horizontal forces act on a 3.0 kg body that can move over a frictionless floor. One force is 9.0 N, acting due east, and the other is 8.0 N, acting 62 degrees north of west. What is the magnitude of the body's acceleration?

I found an answer like:

total components along East
= 3 - 9 * sin 62
= - 4.95 N (along west)

total component along North
= 8* cos 62
= 3.76 N

Resultant force
= square root [ 4.95^2 + 3.76^2 ]
= 6.21N

Acc
= 6.21 / 3
= 2.07 m/s^2

But I can't understand here,3 - 9 * sin 62 and
here 8* cos 62
how they come from? please somebody help

Question

Only two horizontal forces act on a 3.0 kg body that can move over a frictionless floor. One force is 9.0 N, acting due east, and the other is 8.0 N, acting 62 degrees north of west. What is the magnitude of the body's acceleration?


I found an answer like:

total components along East
= 3 - 9 * sin 62
= - 4.95 N (along west)

total component along North
= 8* cos 62
= 3.76 N

Resultant force
= square root [ 4.95^2 + 3.76^2 ]
= 6.21N

Acc
= 6.21 / 3
= 2.07 m/s^2 

But I can't understand here,3 - 9 * sin 62 and
here 8* cos 62
how they come from? please somebody help

ghazi · Answer

|dw:1347736127914:dw| find resultant of 9N and 8 N with an angle of 162/2=81