The vertical angle from B to the vane at A is an elevation of 4°15'. What does the answer conclude?

A is higher than B, so B (RL 1280) is the farther station and A the nearer

How do the corrections change the RL of the temple top?

They are added to the heights, so the RL rises slightly: 1376.828 m against 1376.820 m

They are subtracted, so the RL is lower

Why is x found by a few iterations when corrections are applied?

Because the correction C depends on the distance x, which is itself unknown

Because the distance D is unknown

2071 Magh · 8 marks

RL of temple top with refraction correction

Short answer

Without correction: axis B = 1281.420 m, axis A = 1289.713 m, x = 259.290 m, RL P = 1376.820 m. Applying C = 0.0673 d^2 gives x = 259.300 m and RL P = 1376.828 m.

The top of temple was sighted from two stations A and B at very different level. The observed vertical angle from A and B to top of temple 'P' were 30°36' and 20°12' respectively. The vertical angle B to A to a vane at 1.5 m above the foot of the vane was 4°15'. The height of instrument at A and B were 1.47 m and 1.42 m. The distance between two instrument station was 112 m. RL of B was 1280.00 m. Find the RL of the top of the temple. Also apply the correction for refraction and curvature.

Let B (RL 1280.00 m, lower station) be the farther station and A the nearer one: the sight from B to A is an angle of elevation, so A is higher. D = 112 m is taken as the horizontal distance AB. Curvature and refraction correction: $C = 0.0673\,d^2$ m with $d$ in km (added to the height of a sighted point).

(a) Without correction

Axis of B $= 1280.000 + 1.42 = 1281.420$ m
Sight B to the vane at A (1.5 m above foot): $\alpha = 4^\circ15'$ . Vane level $= 1281.420 + 112\tan 4^\circ15' = 1289.743$ m, ground A $= 1288.243$ m
Axis of A $= 1288.243 + 1.47 = 1289.713$ m

Let $x$ = horizontal distance from B to the temple.

x = \frac{(1289.713 - 1281.420) - 112\tan 30^\circ36'}{\tan 20^\circ12' - \tan 30^\circ36'} = \frac{8.293 - 66.237}{0.36793 - 0.59140} = 259.290\ \text{m}

RL_P = 1281.420 + 259.290\tan 20^\circ12' = 1376.820\ \text{m}

(b) With curvature and refraction

Each calculated height is increased by $0.0673\,(d/1000)^2$ . For the sight B to A, $d = 112$ m: $C = 0.0008$ m, so the axis of A becomes 1289.714 m.

The height of P must be the same from both stations:

1281.420 + x\tan 20^\circ12' + C(x) = 1289.714 + (x - 112)\tan 30^\circ36' + C(x-112)

Solving (the correction terms are small, so a few iterations are enough):

x = 259.300\ \text{m}, \qquad C(x) = 0.0045\ \text{m}

RL_P = 1281.420 + 259.300\tan 20^\circ12' + 0.005 = 1376.828\ \text{m}

Answer: RL of the top of the temple = 1376.820 m without correction; 1376.828 m after applying curvature and refraction.

✓ Answer checked by Claude Opus 5.5 (high effort)

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RL of temple top with refraction correction

(a) Without correction

(b) With curvature and refraction

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