Question 1182585: It is desired to determine the distance between two
inaccessible objects A and B. Both A and B are visible
from two other points C and D. Given the following
measurements, determine the distance AB: DC = 153.4 m,
ADB = 58°22’, ADC = 81° 17’, BCA = 65° 43’ and BCD =
100° 08’
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Here's how to determine the distance AB using the given measurements:
**1. Calculate angles BDC and ABD:**
* **Angle BDC:**
* BDC = BCD - BCA
* BDC = 100° 08’ - 65° 43’
* BDC = 34° 25’
* **Angle ABD:**
* In triangle ADB, the sum of angles is 180°
* ABD = 180° - ADB - BDC
* ABD = 180° - 58° 22’ - 34° 25’
* ABD = 87° 13’
**2. Calculate angles ACD and BAC:**
* **Angle ACD:**
* In triangle BCD, the sum of angles is 180°
* ACD = 180° - ADC - BCA
* ACD = 180° - 81° 17’ - 65° 43’
* ACD = 33°
* **Angle BAC:**
* In triangle ABC, the sum of angles is 180°
* BAC = 180° - BCA - ACB
* We know BCA = 65° 43’ and ACB = BCD - ACD = 100° 08' - 33° = 67° 08'
* BAC = 180° - 65° 43’ - 67° 08’
* BAC = 47° 09’
**3. Use the Law of Sines to find AD and AC:**
* **Side AD:**
* AD / sin(BDC) = DC / sin(ADB)
* AD = (DC * sin(BDC)) / sin(ADB)
* AD = (153.4 * sin(34° 25’)) / sin(58° 22’)
* AD ≈ 98.2 m
* **Side AC:**
* AC / sin(BCA) = DC / sin(ADC)
* AC = (DC * sin(BCA)) / sin(ADC)
* AC = (153.4 * sin(65° 43’)) / sin(81° 17’)
* AC ≈ 142.6 m
**4. Use the Law of Cosines to find AB:**
* In triangle ABC:
* AB² = AC² + BC² - 2 * AC * BC * cos(ACB)
* We already know AC and we can find BC using the Law of Sines in triangle BCD.
* BC / sin(BDC) = DC / sin(DBC)
* We need to find angle DBC = 180 - BDC - BCD = 180 - 34° 25' - 100° 08' = 45° 27'
* BC = (153.4 * sin(34° 25')) / sin(45° 27') = 119.8 m
* AB² = 142.6² + 119.8² - 2 * 142.6 * 119.8 * cos(67° 08’)
* AB ≈ 122.1 m
**Therefore, the distance between A and B is approximately 122.1 meters.**
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