Question

find the angle of depression from point a to point c.
6 mi
angle of depression = ?°

Explanation:

Step1: Recall angle of depression property

The angle of depression is equal to the angle of elevation (alternate interior angles, since the horizontal line from A and the horizontal line at B are parallel). In the right triangle \( \triangle ABC \), we know one angle is \( 58^\circ \), and the triangle is right - angled at B. The sum of angles in a triangle is \( 180^\circ \), so the non - right angles sum to \( 90^\circ \). Let the angle at A (other than \( 58^\circ \)) be \( x \), then \( 58^\circ+x = 90^\circ \), so \( x=90^\circ - 58^\circ \).

Step2: Calculate the angle of depression

Since the angle of depression from A to C is equal to the angle at C (angle of elevation) or can be found by using the fact that the horizontal line from A and the vertical line AB form a right angle. The angle of depression is equal to \( 90^\circ-58^\circ = 32^\circ \)? Wait, no. Wait, the angle of depression: the horizontal line from A is parallel to BC (since AB is vertical and BC is horizontal). So the angle of depression (between the horizontal line from A and the line AC) is equal to the angle at C (angle of elevation). In triangle \( ABC \), angle at B is \( 90^\circ \), angle at A between AB and AC is \( 58^\circ \), so angle at C is \( 90^\circ - 58^\circ=32^\circ \)? Wait, no, wait. Wait, the angle between AB (vertical) and AC is \( 58^\circ \), so the angle between the horizontal (from A) and AC is \( 90^\circ - 58^\circ = 32^\circ \)? Wait, no, let's think again. The horizontal line from A is parallel to BC (because AB is perpendicular to BC, and the horizontal line from A is perpendicular to AB as well). So the angle of depression (let's call it \( \theta \)) and the angle at C are alternate interior angles. In triangle \( ABC \), angle at B is \( 90^\circ \), angle at A (between AB and AC) is \( 58^\circ \), so angle at C is \( 90 - 58=32^\circ \)? Wait, no, that's not right. Wait, AB is 6 miles, vertical, BC is horizontal. The angle between AB and AC is \( 58^\circ \), so the angle between the horizontal (from A) and AC is \( 90 - 58 = 32^\circ \)? Wait, no, actually, the angle of depression is equal to the angle whose tangent is \( \frac{BC}{AB} \), but maybe a simpler way: the sum of angles in the right triangle: angle at A (between horizontal and AC) + angle at A (between AB and AC) = \( 90^\circ \). So angle of depression \( = 90^\circ-58^\circ = 32^\circ \)? Wait, no, I think I made a mistake. Wait, the angle between AB (vertical) and AC is \( 58^\circ \), so the angle between the horizontal (from A) and AC is \( 90 - 58 = 32^\circ \). Alternatively, since the triangle is right - angled at B, the two non - right angles are complementary. So if one angle (between AB and AC) is \( 58^\circ \), the other angle (between horizontal and AC, which is the angle of depression) is \( 90 - 58 = 32^\circ \). Wait, no, wait. Let's draw this mentally. Point A is the airplane, AB is vertical down to B (on the ground), BC is horizontal from B to C. The horizontal line from A is parallel to BC. The angle of depression from A to C is the angle between the horizontal line from A and the line AC. This angle is equal to the angle at C (angle of elevation) because of alternate interior angles. In triangle ABC, angle at B is \( 90^\circ \), angle at A (between AB and AC) is \( 58^\circ \), so angle at C is \( 90 - 58 = 32^\circ \). So the angle of depression is \( 32^\circ \). Wait, but let's check again. The angle between AB (vertical) and AC is \( 58^\circ \), so the angle between horizontal (from A) and AC is \( 90 - 58 = 32^\circ \). Yes, because the horizontal and vertical are perpendicular ( \( 90^\circ \) between them). So the angle of depression is \( 32^\circ \).

Answer:

\( 32 \)