---
title: "Angle BAC and vertical angle P-Q (2066)"
subject: "Surveying I"
chapter: 7
key: "semester_3/surveying_i:horizontal-and-vertical-angle-figures-80-40"
source: https://www.bctengineer.com/q/semester_3/surveying_i/horizontal-and-vertical-angle-figures-80-40
site: BCT Engineering
---

# Angle BAC and vertical angle P-Q (2066)

Surveying I · Chapter 7: Transit and Theodolite

- Asked in: 2066 Jestha, 9 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/surveying_i/horizontal-and-vertical-angle-figures-80-40
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-7

## Question

Find horizontal and vertical angles in the following cases. [Figure: at station A, line to B observed FL 00°00′00″, FR 180°00′10″; line to C observed FL 80°40′00″, FR 260°40′20″. Find H.A. BAC.] [Figure: instrument at P sighting to Q, observed FL 68°30′30″, FR 291°29′00″. Find V.A. P to Q.]

## Short answer

Horizontal angle BAC: mean of AB 0°00′05″ and AC 80°40′10″ differenced gives 80°40′05″. Vertical: FL = 68°30′30″, FR = 291°29′00″, index error -15″, zenith distance 68°30′45″, vertical angle = 21°29′15″ elevation.

## Answer

#### (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°.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The mean directions are AB = 0°00′05″ and AC = 80°40′10″. What is angle BAC?
   A. 80°40′05″
   B. 80°40′15″
   C. 80°40′10″
   D. 80°39′55″
   Answer: A. 80°40′10″ - 0°00′05″ = 80°40′05″. (Key point: Horizontal angle BAC)

2. For the vertical circle FL = 68°30′30″ and FR = 291°29′00″, FL + FR = 359°59′30″. What does this show?
   A. No error, as the sum is almost 360°
   B. An index error of -15″ ((FL+FR-360°)/2)
   C. An index error of -30″
   D. A horizontal collimation error
   Answer: B. FL + FR should be 360°; a deficiency of 30″ gives an index error of half, -15″. (Key point: Index error)

3. What is the mean zenith distance Z?
   A. 68°30′30″
   B. 68°30′15″
   C. 68°30′45″
   D. 291°29′00″
   Answer: C. Z = (FL - FR + 360°)/2 = 68°30′45″. (Key point: Mean zenith distance)

4. The vertical angle of Q from P is:
   A. 21°29′15″ depression
   B. 68°30′45″ elevation
   C. 21°30′45″ elevation
   D. 21°29′15″ elevation
   Answer: D. VA = 90° - Z = 21°29′15″, positive, so elevation. (Key point: Vertical angle)

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