8.4 Angles of Elevation and Depression Solve problems involving angles of elevation and angles of depression.
Line of sight Angle of Elevation –angle between the line of sight and the horizontal when looking upward. Find horizontal line (eye level) and then angle (raise arm). Make a drawing.
Line of sight Angle of Depression –angle between the horizontal and the line of sight when an observer looks downward. Find horizontal line (eye level) and then angle (lower arm). Make a drawing.
An angle of elevation is the angle formed by a horizontal line and a line of sight to a point above the line. In the diagram, 1 is the angle of elevation from the warrior to the dragon. An angle of depression is the angle formed by a horizontal line and a line of sight to a point below the line. 2 is the angle of depression from the dragon to the warrior. Angle of depression 2 1 Angle of elevation
to the angle of depression from the other point. Since horizontal lines are parallel, 1 2 by the Alternate Interior Angles Theorem. Therefore the angle of elevation from one point is congruent to the angle of depression from the other point. Angle of depression 2 1 Angle of elevation
Classifying Angles of Elevation and Depression Classify each angle as an angle of elevation or an angle of depression. 1 1 is formed by a horizontal line and a line of sight to a point below the line. It is an angle of depression.
Classifying Angles of Elevation and Depression Classify each angle as an angle of elevation or an angle of depression. 4 4 is formed by a horizontal line and a line of sight to a point above the line. It is an angle of elevation.
Check It Out! Example 3 Use the diagram above to classify each angle as an angle of elevation or angle of depression. 3a. 5 5 is formed by a horizontal line and a line of sight to a point below the line. It is an angle of depression. 3b. 6 6 is formed by a horizontal line and a line of sight to a point above the line. It is an angle of elevation.
Example 4: Finding Distance by Using Angle of Elevation The Black Tower casts a 67-meter shadow. If the angle of elevation from the tip of the shadow to the top of the tower is 70º, how tall is the Black Tower? Round to the nearest meter. Draw a sketch to represent the given information. Let A represent the tip of the shadow, and let B represent the top of the tower. Let y be the height of the tower.
Example 4 Continued You are given the side adjacent to A, and y is the side opposite A. So write a tangent ratio. y = 67 tan 70° Multiply both sides by 67. y 184 m Simplify the expression.
Multiply both sides by x and divide by tan 29°. 3500 ft 29° What if…? Suppose a dragon is at an altitude of 3500 ft and the angle of elevation from the dragon to you is 29°. What is the horizontal distance between the dragon and you? Round to the nearest foot. You are given the side opposite A, and x is the side adjacent to A. So write a tangent ratio. Multiply both sides by x and divide by tan 29°. x 6314 ft Simplify the expression.
Finding Distance by Using Angle of Depression A Healer stands at the edge of a crevasse that is 115 ft wide. The angle of depression from the edge where she stands to the bottom of the opposite side is 52º. How deep is the crevasse at this point? Round to the nearest foot.
Draw a sketch to represent the given information Draw a sketch to represent the given information. Let C represent the ice climber and let B represent the bottom of the opposite side of the crevasse. Let y be the depth of the crevasse.
By the Alternate Interior Angles Theorem, mB = 52°. Write a tangent ratio. y = 115 tan 52° Multiply both sides by 115. y 147 ft Simplify the expression.
Finding Distance between Two Objects A dragon flying at an altitude of 12,000 ft sights two cannons directly in front of him. The angle of depression to one cannon is 78°, and the angle of depression to the second cannon is 19°. What is the distance between the two cannons? Round to the nearest foot.
Step 1 Draw a sketch. Let P represent the Dragon and let A and B represent the two cannons. Let x be the distance between the two cannons. 78° 19° 12,000 ft
Step 2 Find y. By the Alternate Interior Angles Theorem, mCAP = 78°. In ∆APC, So
Step 3 Find z. By the Alternate Interior Angles Theorem, mCBP = 19°. In ∆BPC, So
Step 4 Find x. x = z – y x 34,851 – 2551 32,300 ft So the two airports are about 32,300 ft apart.
Practice problems 1. A slide has an angle of elevation of 25˚. It is 60 feet from the end of the slide to the stairway beneath the top of the slide. How long is the slide? 25˚ 60 ft x 2. A warrior is standing 150 feet away from a tower. He measures the angle of elevation from where he is to the top of the tower to be 30˚. How tall is the tower? x 30˚ 150 ft
3. A moving sidewalk takes zoo visitors up a hill 3. A moving sidewalk takes zoo visitors up a hill. The vertical distance of the sidewalk is 48 feet. Its angle of elevation is 15˚. About how long is the sidewalk? x 48 ft 15˚ 4. A dragon flying 4000 feet above ground begins a 2˚ descent (angle of depression) to land on a clifftop. How many miles from the cliff is the dragon when it starts its descent? (Hint: 1 mile = 5280 feet) 2˚ 4000 ft. x 2˚
5. A 100-foot-tall lighthouse stands at the top of a 150-foot-tall cliff. The angle of depression from the top of the lighthouse to a ship is 27˚. About how far from the cliff is the ship? 100 ft 250 ft 150 ft 27˚ x 6. A building is 50 feet high. At a distance away from the building, an observer notices that the angle of elevation to the top of the building is 41º. How far is the observer from the base of the building? 50 ft 41˚ x
7. An airplane is flying at a height of 2 miles above the ground 7. An airplane is flying at a height of 2 miles above the ground. The distance along the ground from the airplane to the airport is 5 miles. What is the angle of depression from the airplane to the airport? 2 mi x˚ 5 mi 8. A campsite is 9.41 miles from a point directly below the mountain top. If the angle of elevation is 12° from the camp to the top of the mountain, how high is the mountain? x 12˚ 9.41 mi