Find magnitude of `vecB` and direction of `vecA` . If `vecB` makes angle `37^@` and `vecC` makes `53^@` with x axis and `vecA` has magnitude equal to 10 and `vecc` has 5. `("given "vecA + vecB + vecC = 0)`

Explanation

`-vecA = vecC + vecB`
`vecA = A_(x) hati + A_(y) hatj`
`implies -vecA = - A_(x)hati + A_(y)hatj`
`implies A_(x) = -(B cos 37^@ + C cos 53^@)`
`A_(y) = -(B sin 37^@ + C cos 53^@)`
`|vecA|^(2) = A_(x)^(2) + A_(y)^(2)`
`A^(2) = (B xx 4/5 + C xx 3/5)^(2) + (B xx 3/5 + C xx 4/5)^(2)`
`10^2 + ((4B)/5 + 3)^(2) + ((3B)/5 + 4)^(2)`
`implies 100 = 16/25 B^2 + 9/25 B^2 + 25 + 2 ((3 xx 4)/(5) + (4 xx 3)/(5))B`
`implies B^2 + 48/5B - 75 = 0`
B = 5 (magnitude can not be negative)
& Angle made by A
`implies A_(x) = -(20/5 + 3) = -12`
`A_(y) = -(15/5 + 4) = -7`
`tan theta = (A_y)/(A_x) = (-7)/(-12)`
`theta = 180^(@) + 25^@ = 205^@`.