Prolonging line past a high building

Data

BC = 121.92 m, perpendicular to AB at B. At C, $\angle BCD = 30^\circ$ and $\angle BCE = 40^\circ$. D and E are to lie on the prolongation of AB beyond the building. Chainage of B $= 95.10$ m.

 A ------ B ---[ building ]--- D ----- E ---->
          |                  /       /
 121.92   |              /      /
          |          /     /
          C ----------------
          (30 deg at C to D, 40 deg at C to E)

Triangles CBD and CBE are right-angled at B.

Length CD

$$CD = \frac{BC}{\cos 30^\circ} = \frac{121.92}{0.8660} = 140.78\ \text{m}$$

Length CE

$$CE = \frac{BC}{\cos 40^\circ} = \frac{121.92}{0.7660} = 159.16\ \text{m}$$

Distances BD and BE

$$BD = BC\tan 30^\circ = 121.92\times 0.5774 = 70.39\ \text{m}$$ $$BE = BC\tan 40^\circ = 121.92\times 0.8391 = 102.30\ \text{m}$$

Chainage of D

$$\text{Chainage of D} = 95.10 + 70.39 = 165.49\ \text{m}$$

(Chainage of E $= 95.10 + 102.30 = 197.40$ m.)

Answer: CD = 140.78 m, CE = 159.16 m; chainage of D = 165.49 m.