A spelunker is surveying a cave. She follows a passage 140 m straight west, then 250 m in a direction 45∘ east of south, and then 280 m at 30∘ east of north. After a fourth unmeasured displacement, she finds herself back where she started.
for the last displacement x=188.61 and y = 20.39 therefore the vector is a straight 190m line with an angle of ?
measured from the positive x-axis,
tanθ = y/x = 20.39/188.61
so, θ = 6.17°
That's E 6.17° N
or, as a course heading, 83.83°
total east = -140 + 250 sin 45 + 280 sin 30 = 176.78
total north = 0 -250 cos 45 +280 cos 30 = 65.71
I do not understand where you get that last x and y
in my opinion you must go west 176.78 and south 65.71
tan^-1 65.71/176.78 = 20.4 deg south of west
i got confused while in a part but i already checked it was 20.4 deg
To find the angle of the last displacement vector, we can use trigonometry. The x and y components of the vector are given as x = 188.61 and y = 20.39, respectively.
First, we can calculate the magnitude of the vector using the Pythagorean theorem:
magnitude = sqrt( x^2 + y^2 )
= sqrt( 188.61^2 + 20.39^2 )
= sqrt( 38226.5721 + 415.7281 )
= sqrt( 38642.30 )
≈ 190.00
Next, we can find the angle by using inverse trigonometric functions. The angle θ can be determined using the equation:
θ = arctan( y / x )
= arctan( 20.39 /188.61 )
≈ arctan( 0.108 )
Calculating this angle in degrees yields approximately:
θ ≈ 0.108°
Therefore, the angle of the last displacement vector is approximately 0.108° from the positive x-axis.