Split up each force into horizontal and vertical components.
• 300 N at N30°E :
(300 N) (cos(30°) i + sin(30°) j)
• 400 N at N60°E :
(400 N) (cos(60°) i + sin(60°) j)
• 500 N at N80°E :
(500 N) (cos(80°) i + sin(80°) j)
The resultant force is the sum of these forces,
∑ F = (300 cos(30°) + 400 cos(60°) + 500 cos(80°)) i
… … … + (300 sin(30°) + 400 sin(60°) + 500 sin(80°)) j N
∑ F ≈ (546.632 i + 988.814 j) N
so ∑ F has a magnitude of approximately 1129.85 N and points in the direction of approximately N61.0655°E.
#1
Clear shown in graph
#2
No
At. no where graph passes through origin so y intercept is not 0 at anywhere
#3
Domain is x values
#4
Range is y values set
Answer:
a = 11.71 ; b = 15.56
Step-by-step explanation:
For this problem, we need two things. The law of sines, and the sum of the interior angles of a triangle.
The law of sines is simply:
sin(A)/a = sin(B)/b = sin(C)/c
And the sum of interior angles of a triangle is 180.
45 + 110 + <C = 180
<C = 25
We can find the sides by simply applying the law of sines.
length b
7/sin(25) = b/sin(110)
b = 7sin(110)/sin(25)
b = 15.56
length a
7/sin(25) = a/sin(45)
a = 7sin(45)/sin(25)
a = 11.71
