Components of Vectors — Question 2

Question 2

Bearing angles exchange the sine and cosine

A survey line is 20 m long on a bearing of $35^\circ$ east of north. Use east and north as the positive coordinate directions. The bearing is measured from north toward east.

1. Find the east and north components and the standard angle from positive $x$.

2. A report gives $(16.38,11.47)$ m as the east and north components. What bearing does that pair describe?

3. Check your formulas at bearings $0^\circ$ and $90^\circ$. Explain why these checks help.

Question 2 - Solution

Bearing angles exchange the sine and cosine

(a) Use the angle next to the north leg.

$$\boxed{(A_x,A_y)=20(\sin35^\circ,\cos35^\circ)=(11.472,16.383)\,\mathrm m},\quad\theta=55^\circ$$ $A_x=20\sin35^\circ\approx11.47\ \mathrm m$ and $A_y=20\cos35^\circ\approx16.38\ \mathrm m$. The standard angle is $90^\circ-35^\circ=55^\circ$.

(b) Infer the angle from the ratio.

$$\beta_{\mathrm{report}}=\tan^{-1}(16.38/11.47)\approx55.0^\circ\ne35^\circ$$ The reported pair has $\tan\beta=A_x/A_y\approx16.38/11.47$, giving $\beta\approx55.0^\circ$ east of north. Its length is approximately correct but its direction is wrong. A magnitude check alone cannot detect exchanged components.

(c) Test the limiting directions.

$$\beta=0^\circ\Rightarrow(0,20)\,\mathrm m,\qquad\beta=90^\circ\Rightarrow(20,0)\,\mathrm m$$ At $\beta=0^\circ$, $(A_x,A_y)=(0,20)$ m, due north. At $90^\circ$ the pair is $(20,0)$ m, due east. These limits identify which component should contain sine; they also preserve the named reference axis.