---
title: "Angle ABC and vertical angle P-Q (2062)"
subject: "Surveying I"
chapter: 7
key: "semester_3/surveying_i:horizontal-and-vertical-angle-figures-abc-61-30"
source: https://www.bctengineer.com/q/semester_3/surveying_i/horizontal-and-vertical-angle-figures-abc-61-30
site: BCT Engineering
---

# Angle ABC and vertical angle P-Q (2062)

Surveying I · Chapter 7: Transit and Theodolite

- Asked in: 2062 Baisakh, 8 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-abc-61-30
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-7

## Question

Find the horizontal angle ABC from the following field observations and also the vertical angle from P to Q. [Figure: at B, line to A observed FR 179°59′20″ and FL 00°00′00″; line to C observed FL 160°20′30″ and FR 340°20′00″. Find horizontal angle ABC.] [Figure: instrument at P sighting to Q, observed FL 61°30′00″ and FR 298°30′30″. Find vertical angle from P to Q.]

## Short answer

Horizontal angle ABC: BA mean -0°00′20″, BC mean 160°20′15″, difference 160°20′35″. Vertical: FL = 61°30′00″, FR = 298°30′30″, index error +15″, zenith distance 61°29′45″, vertical angle = 28°30′15″ elevation.

## Answer

#### (a) Horizontal angle ABC (instrument at B)
Mean direction = [FL + (FR - 180°)]/2.

$$
\begin{aligned}
\text{BA}: &\; FL = 0°00′00″,\; FR = 179°59′20″ \Rightarrow \text{mean} = 359°59′40″\ (= -0°00′20″)\\
\text{BC}: &\; FL = 160°20′30″,\; FR = 340°20′00″ \Rightarrow \text{mean} = \frac{160°20′30″ + 160°20′00″}{2} = 160°20′15″\\
\angle ABC &= 160°20′15″ - (-0°00′20″) = 160°20′35″
\end{aligned}
$$

#### (b) Vertical angle P to Q (instrument at P)
FL = 61°30′00″, FR = 298°30′30″ (zenith-angle circle).

$$
\begin{aligned}
\text{Index error} &= \frac{(FL+FR) - 360°}{2} = \frac{360°00′30″ - 360°}{2} = 0°00′15″\\
Z &= \frac{FL - FR + 360°}{2} = \frac{61°30′00″ - 298°30′30″ + 360°}{2} = 61°29′45″\\
\text{Vertical angle} &= 90° - Z = 28°30′15″
\end{aligned}
$$

**Answer:** horizontal angle ABC = **160°20′35″**; vertical angle from P to Q = **28°30′15″** (elevation).

*Note:* a zenith-type vertical circle is assumed.

## Check yourself

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

1. At B, FL to A is 0°00′00″ and FR is 179°59′20″. The mean direction of BA is:
   A. -0°00′20″ (359°59′40″)
   B. +0°00′20″
   C. +0°00′40″
   D. 179°59′40″
   Answer: A. FR - 180° = -40″; the mean of 0 and -40″ is -20″. (Key point: Direction BA near zero)

2. With BC = 160°20′15″, angle ABC is:
   A. 160°19′55″
   B. 160°20′35″
   C. 160°20′15″
   D. 160°20′25″
   Answer: B. 160°20′15″ - (-0°00′20″) = 160°20′35″. (Key point: Horizontal angle)

3. For FL = 61°30′00″ and FR = 298°30′30″, the index error is:
   A. -15″
   B. +30″
   C. +15″
   D. 0″
   Answer: C. (FL + FR - 360°)/2 = (360°00′30″ - 360°)/2 = +15″. (Key point: Index error)

4. The vertical angle from P to Q is:
   A. 28°29′45″ elevation
   B. 61°29′45″ elevation
   C. 28°30′15″ depression
   D. 28°30′15″ elevation
   Answer: D. Z = (FL - FR + 360°)/2 = 61°29′45″; VA = 90° - Z = 28°30′15″. (Key point: Vertical angle)

## Related in Surveying I

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