Question

To determine the gradient between two points A and B, a tacheometer was set up at another station C and the following observations were taken, keeping the staff vertical:
Assume k = 100, C = 0. If the horizontal angle ACB is ${35}^{\circ }{20}^{\prime }$, then the average gradient between A and B is

Staff at	Vertical angle	Stadia readings
A	$+4^{\circ }{20}^{\prime }{0}^{\prime \prime }$	1.300, 1.610, 1.920
B	$+0^{\circ }{10}^{\prime }{40}^{\prime \prime }$	1.100, 1.410, 1.720

• A. 1 in 10.62
• B. 1 in 11.38
• C. 1 in 8.77
• D. 1 in 7.23

Answer 1

The correct option is C 1 in 8.77


Staff of A,

Let the distance,
$CA=D_1$

$\therefore D_1=ksco{s}^2\theta$

$\Rightarrow D_1=100\times 0.620\times co{s}^2\left(4^{\circ }{20}^{\prime }{0}^{\prime \prime }\right)=61.646m$

$V_1=\frac{1}{2}kssin2\theta =\frac{1}{2}\times 100\times 0.62sin{8}^{\circ }{40}^{\prime }=4.671m$

$\therefore R.LofA=R.Loflineofcollimation+V_1-r_1$

Let R.L. of line of collimation be 100.00 m

R.L of A = 100.00 + 4.671 - 1.61 = 103.061 m

Staff at B,

Let the distance,
$CB=D_2$

$\therefore D_2=100\times 0.62co{s}^2\left(0^{\circ }{10}^{\prime }{40}^{\prime \prime }\right)=61.999m$

$V_2=\frac{1}{2}\times 100\times 0.62sin\left({21}^{\prime }{20}^{\prime \prime }\right)=0.192m$

$\therefore$ R.L. of B = 100.0 + 0.192 - 1.410 = 98.782 m

$\therefore$ Difference of R.L. of A and B = 103.061 - 98.782 = 4.279 m

The distance AB can be calculated from cosine law as,

$cos\left({35}^{\circ }{20}^{\prime }\right)=\frac{(61.646{)}^2+(61.999{)}^2-(AB{)}^2}{2\times 61.646\times 61.999}$

$\Rightarrow AB=37.525m$

Gradient of AB = $\frac{4.279}{37.525}=0.114=1in8.77$