Question: A boat is heading towards a lighthouse, where Jeremiah is watching from a vertical distance of $126$ feet above the water. Jeremiah measures an angle of depression to the boat at point A to be $11^\circ$. At some later time, Jeremiah takes another measurement and finds the angle of depression to the boat (now at point B) to be $78^\circ$. Find the distance from point A to point B. Round your answer to the nearest tenth of a foot if necessary.
1. **State the problem:**
Jeremiah is $126$ feet above water in a lighthouse. He measures angles of depression to a boat at two points, A and B, as $11^\circ$ and $78^\circ$ respectively. We need to find the horizontal distance between points A and B.
2. **Set up the scenario:**
Let the horizontal distances from the lighthouse base to points A and B be $x_A$ and $x_B$ respectively.
3. **Use the angle of depression and right triangle properties:**
The vertical height is $126$ feet, and the angles of depression correspond to angles of elevation from the boat to Jeremiah.
4. **Apply the tangent function:**
$$\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{126}{x}$$
For point A:
$$\tan(11^\circ) = \frac{126}{x_A} \implies x_A = \frac{126}{\tan(11^\circ)}$$
For point B:
$$\tan(78^\circ) = \frac{126}{x_B} \implies x_B = \frac{126}{\tan(78^\circ)}$$
5. **Calculate $x_A$ and $x_B$:**
$$x_A = \frac{126}{\tan(11^\circ)}$$
$$x_B = \frac{126}{\tan(78^\circ)}$$
6. **Find the distance between points A and B:**
$$d = x_A - x_B = \frac{126}{\tan(11^\circ)} - \frac{126}{\tan(78^\circ)}$$
7. **Calculate values:**
$$\tan(11^\circ) \approx 0.1944$$
$$\tan(78^\circ) \approx 4.7046$$
So,
$$x_A = \frac{126}{0.1944} \approx 647.9$$
$$x_B = \frac{126}{4.7046} \approx 26.8$$
8. **Distance between A and B:**
$$d = 647.9 - 26.8 = 621.1$$
**Final answer:**
The distance from point A to point B is approximately **621.1 feet**.