Question 1182585
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.**