In which is following equation for find out horizontal distance in trigonometric leveling when base of object

Question

In which is following equation for find out horizontal distance in trigonometric leveling when base of object is inaccessible and nearest station is high to away station.

Examoogle · Accepted Answer

The correct answer is Option D: Both B and C. When the base of an object is inaccessible and the instrument station nearer to the object is at a higher elevation than the farther station, the horizontal distance $D$ is given by $D = \frac{(b - h_d \cot \alpha_2) 	an \alpha_2}{	an \alpha_1 - 	an \alpha_2}$. Expanding this expression yields $D = \frac{b 	an \alpha_2 - h_d}{	an \alpha_1 - 	an \alpha_2}$, making both Option B and Option C mathematically equivalent and correct.

Step-by-Step Solution & Concepts:
When the base of an object is inaccessible and the instrument station nearer to the object is at a higher elevation than the farther station, the horizontal distance $D$ is given by $D = \frac{(b - h_d \cot \alpha_2) 	an \alpha_2}{	an \alpha_1 - 	an \alpha_2}$. Expanding this expression yields $D = \frac{b 	an \alpha_2 - h_d}{	an \alpha_1 - 	an \alpha_2}$, making both Option B and Option C mathematically equivalent and correct.

Working Principle: In trigonometrical levelling, when instrument stations are at different levels, the height difference $h_d$ between the two instrument axes affects the geometry. By setting up simultaneous equations using the vertical angles $\alpha_1$ and $\alpha_2$ measured from the near and far stations respectively, the horizontal distance $D$ is solved by eliminating the vertical height of the object above the instrument axis.

Key Formulas: $D = \frac{(b - h_d \cot \alpha_2) 	an \alpha_2}{	an \alpha_1 - 	an \alpha_2}$ — Horizontal distance formula when near station is higher; $D = \frac{b 	an \alpha_2 - h_d}{	an \alpha_1 - 	an \alpha_2}$ — Simplified equivalent expression for horizontal distance

Key Points: $\alpha_1$ is the angle of elevation measured from the near station A.; $\alpha_2$ is the angle of elevation measured from the far station B.; $b$ is the horizontal distance between instrument stations A and B.; $h_d$ is the difference in height between the instrument axes at stations A and B.

Advantages: Allows accurate distance measurement without direct access to the object base; Accounts for level differences between instrument stations

Disadvantages: Requires precise measurement of vertical angles and inter-station distance; Sensitive to atmospheric refraction errors over long distances

Applications: Determining heights of inaccessible structures like towers and chimneys; Topographical surveying in hilly or uneven terrain

Additional Info: Option A represents the case where the nearer instrument station is at a LOWER elevation than the farther station.. Multiplying $(b - h_d \cot \alpha_2)$ by $	an \alpha_2$ gives $b 	an \alpha_2 - h_d \cot \alpha_2 	an \alpha_2 = b 	an \alpha_2 - h_d$, showing why B and C are identical.

Answer: D is correct — Both Option B and Option C represent the exact same mathematical formula for horizontal distance when the near station is higher than the far station.

Exam Tip: Remember that if the near station is LOWER, the sign of $h_d$ changes to positive ($b + h_d \cot \alpha_2$). If HIGHER, the sign is negative ($b - h_d \cot \alpha_2$).

Examoogle · Answer

Option A: D=b+hdcotα2tanα2tanα1-tanα2

Examoogle · Answer

Option B: D=b-hdcotα2tanα2tanα1-tanα2

Examoogle · Answer

Option C: D=btanα2-hdtanα1-tanα2

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