Physics, let a=(3m 20deg south of esast, (B=2m north and (C=5m 70deg south of west?

Write a b and c in component form, using unit vectors. Find the magnitude of D=A+B+C

Physics, let a=(3m 20deg south of esast, (B=2m north and (C=5m 70deg south of west?
A(x) = 3cos20; A(y) = - 3sin20


B(x) = 2 ; B(y) = 0


C(x) = -5cos70 ; C(y) = -5sin70





D(x) = 3cos20 + 2 - 5cos70 = 3.10897715


D(y) = - 3sin20 - 5sin70 = -5.72452353


(D)^2 = (3.10897715)^2 + ( -5.72452353)^2


(D)^2 = 42.4359086


D = sqrt(42.4359086) = 6.51428496





arctan(-5.72452353 / 3.10897715)=


arctan(-1.84128839) = -61.4937045 degrees





So:


D = 6.51 meters at an angle of 61.5 degrees South of East








P.S. I used Google Calculator to find the numerical values of the cos and sin and arctangent and sqrt, etc, etc.
Reply:Components of Vectors in the X-axis and the Y axis are as follows:


For A


3* cosine 20 = 2.79 east


3*sine 20 = 0.34 *3 =1.02 south


For B


2 pointing towards the north


C has 5*cosine 70 =5* 0.342=1.71 west


and 5*sine 70=5* 0.939 = 4.695 south





Total magnitude of the vectors in the Y-axis = 2-1.02-4.695 =


-3.62 to the south


And 2.79-1.71= 1.08 to the east


D = A+B+C = sqrt{(1.08)^2 + (3.62)^2}


=3.77 at an angle = arctan theta = 3.62/1.08=73.38 degrees


Answer is D = 3.77 pointing at an angle = 73 .38 south of the east