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A theodolite was set up at a station P. The angle of depression to a vane 2 m above the foot of a staff held at another station Q was 45°. The horizontal distance between stations P and Q is 20 m. The staff reading at a benchmark S of RL 433.050 m is 2.905 m. Neglecting the errors due to curvature and refraction, the RL of the station Q (in m), is​
  • a)
    431.050
  • b)
    435.955
  • c)
    413.050
  • d)
    413.955
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
A theodolite was set up at a station P. The angle of depression to a v...

x/20 = tan 45°
x = 20 m
RL of Q = 433.05 + 2.905 – x – 2
= 433.05 + 2.905 – 20 – 2
= 413.955 m
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Most Upvoted Answer
A theodolite was set up at a station P. The angle of depression to a v...
Degrees. The distance between the stations P and Q is 10 m. Find the height of the theodolite above the ground at station P.

To solve this problem, we can use trigonometry. Let's denote the height of the theodolite above the ground at station P as h.

From the information given, we can create a right triangle with the following dimensions:
- The side opposite to the angle of depression is 2 m (the height of the vane).
- The side adjacent to the angle of depression is 10 m (the distance between the stations P and Q).
- The hypotenuse is the distance from the theodolite to the vane.

Using the tangent function, we can write the equation:

tan(45 degrees) = 2 m / h

Next, let's solve for h.

tan(45 degrees) = 1

1 = 2 m / h

h = 2 m / 1

h = 2 m

Therefore, the height of the theodolite above the ground at station P is 2 meters.
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A theodolite was set up at a station P. The angle of depression to a vane 2 m above the foot of a staff held at another station Q was 45°. The horizontal distance between stations P and Q is 20 m. The staff reading at a benchmark S of RL 433.050 m is 2.905 m. Neglecting the errors due to curvature and refraction, the RL of the station Q (in m), isa)431.050b)435.955c)413.050d)413.955Correct answer is option 'D'. Can you explain this answer?
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A theodolite was set up at a station P. The angle of depression to a vane 2 m above the foot of a staff held at another station Q was 45°. The horizontal distance between stations P and Q is 20 m. The staff reading at a benchmark S of RL 433.050 m is 2.905 m. Neglecting the errors due to curvature and refraction, the RL of the station Q (in m), isa)431.050b)435.955c)413.050d)413.955Correct answer is option 'D'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A theodolite was set up at a station P. The angle of depression to a vane 2 m above the foot of a staff held at another station Q was 45°. The horizontal distance between stations P and Q is 20 m. The staff reading at a benchmark S of RL 433.050 m is 2.905 m. Neglecting the errors due to curvature and refraction, the RL of the station Q (in m), isa)431.050b)435.955c)413.050d)413.955Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A theodolite was set up at a station P. The angle of depression to a vane 2 m above the foot of a staff held at another station Q was 45°. The horizontal distance between stations P and Q is 20 m. The staff reading at a benchmark S of RL 433.050 m is 2.905 m. Neglecting the errors due to curvature and refraction, the RL of the station Q (in m), isa)431.050b)435.955c)413.050d)413.955Correct answer is option 'D'. Can you explain this answer?.
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