Angle BAC and vertical angle P-Q (2066)

(a) Horizontal angle BAC (instrument at A)

Mean direction of each line = [FL + (FR - 180°)]/2.

$$\begin{aligned} \text{Mean direction AB} &= \frac{0°00′00″ + 0°00′10″}{2} = 0°00′05″\\ \text{Mean direction AC} &= \frac{80°40′00″ + 80°40′20″}{2} = 80°40′10″\\ \angle BAC &= 80°40′10″ - 0°00′05″ = 80°40′05″ \end{aligned}$$

(b) Vertical angle P to Q (instrument at P)

Readings: FL = 68°30′30″ and FR = 291°29′00″. The vertical circle reads zenith angles, so FL + FR should be 360°.

$$\begin{aligned} FL + FR &= 359°59′30″ \;(\text{index error present})\\ \text{Index error}\ i &= \frac{(FL + FR) - 360°}{2} = -0°00′15″\\ \text{Mean zenith distance}\ Z &= \frac{FL - FR + 360°}{2} = \frac{68°30′30″ - 291°29′00″ + 360°}{2} = 68°30′45″\\ \text{Vertical angle} &= 90° - Z = 21°29′15″ \;(\text{angle of elevation}) \end{aligned}$$

Answer: horizontal angle BAC = 80°40′05″; vertical angle of Q from P = 21°29′15″ (elevation), index error -0°00′15″.

Note: the vertical circle is taken as a zenith-angle circle (reading 0° at the zenith); this is confirmed by FL + FR being almost exactly 360°.