Question 1171450
Let's break down this problem step-by-step using trigonometry.

**1. Visualize the Situation**

Imagine the CN Tower as a vertical line. The woman is at the top, and the two landmarks are on the ground. We have a triangle formed by the woman and the two landmarks.

**2. Define Variables**

* Let the height of the observation deck be `h = 1150 ft`.
* Let the angle between the lines of sight to the landmarks be `θ = 43°`.
* Let the angle between the vertical and the line of sight to the first landmark be `α = 62°`.
* Let the angle between the vertical and the line of sight to the second landmark be `β = 54°`.
* Let the distance from the base of the tower to the first landmark be `d1`.
* Let the distance from the base of the tower to the second landmark be `d2`.
* Let the distance between the two landmarks be `D`.

**3. Calculate Distances to Landmarks (d1 and d2)**

We can use the tangent function to relate the height and distances:

* `tan(α) = d1 / h`
    * `d1 = h * tan(α) = 1150 * tan(62°) ≈ 1150 * 1.8807 ≈ 2162.8 ft`
* `tan(β) = d2 / h`
    * `d2 = h * tan(β) = 1150 * tan(54°) ≈ 1150 * 1.3764 ≈ 1582.9 ft`

**4. Apply the Law of Cosines**

Now, we have a triangle with sides `d1` and `d2`, and the angle between them is `θ = 43°`. We want to find the distance `D` between the landmarks. We can use the Law of Cosines:

* `D² = d1² + d2² - 2 * d1 * d2 * cos(θ)`
* `D² = (2162.8)² + (1582.9)² - 2 * 2162.8 * 1582.9 * cos(43°) `
* `D² ≈ 4677703.84 + 2505572.41 - 2 * 2162.8 * 1582.9 * 0.73135`
* `D² ≈ 7183276.25 - 5005813.1`
* `D² ≈ 2177463.15`
* `D ≈ √2177463.15 ≈ 1475.6 ft`

**Therefore, the distance between the two landmarks is approximately 1475.6 feet.**