Chainage of R across a river

A chain line crosses a river, so QR cannot be chained; it is found by trigonometry from the perpendicular QS set out at Q (the bearings at S are measured with a theodolite or compass).

Given: QS = 90 m (perpendicular to PQR), PQ = 45 m, chainage of Q = 650 m, bearing of SR = 87°30′20″, bearing of SP = 177°30′20″.

              S
            / | \
          /   |    \
        /     | 90   \
      /       |        \
 ---P---------Q-----------R---
       45 m        ?    (river between Q and R)

Step 1: Angle PSR

$$\angle PSR = 177^\circ 30'20'' - 87^\circ 30'20'' = 90^\circ 00'00''$$

Step 2: Angle QSP from the right triangle SQP

$$\tan \angle QSP = \frac{PQ}{QS} = \frac{45}{90} = 0.5 \;\Rightarrow\; \angle QSP = 26^\circ 33'54''$$

Step 3: Angle QSR

Q lies between P and R, so

$$\angle QSR = \angle PSR - \angle QSP = 90^\circ - 26^\circ 33'54'' = 63^\circ 26'06''$$

Step 4: Length QR from the right triangle SQR

$$QR = QS \tan \angle QSR = 90 \times \tan 63^\circ 26'06'' = 90 \times 2.000 = 180.00\ \text{m}$$

(Check: in the right-angled triangle PSR, $QS^{2} = PQ\times QR$, so $QR = 8100/45 = 180$ m.)

Step 5: Chainage of R

$$\text{Chainage of R} = 650 + 180 = 830\ \text{m}$$

(Chainage of P = 650 − 45 = 605 m.)

Answer: width of river QR = 180 m; chainage of R = 830 m.