1. 8.4 Angles of Elevation and Depression SOL: G8 Objectives: The Student Will … Solve problems involving angles of elevation Solve problems involving angles of depression

2. Angles of Elevation Is the angle between the line of sight and the horizontal when an observer looks upward.

3. Example 1: At the circus, a person in the audience at ground level watches the high-wire routine. A 5-foot-6-inches tall acrobat is standing on a platform that is 25 feet off the ground. How far is the audience member from the base of the platform, if the angle of elevation from the audience member’s line of sight to the top of the acrobat is 27  25 ft 27  d tan = opp tan 27  = 30.5 d (d)tan 27  = (d) 30.5 d (d)tan 27 = 30.5 (d)tan 27  30.5 tan 27  tan 27  = d = 59.8  5.5 ft 30.5 tan 27  d = opposite 30.5 adjacent adj ≈ 60 

4. Angle of Depression Is the angle between the line of sight when an observer looks downward, and the horizontal.

5. Example 2: A wheelchair ramp is 3 meters long and inclines at 6 . Find the height of the ramp to the nearest tenth centimeter. 66 3 m. h opp sin = sin 6  = h 3 (3)sin 6  = (3) h 3 (3)sin 6  = h.3135 m = h What trig function uses opposite and hypotenuse opposite hypotenuse hyp 31.4 cm =

6. Common Mistake The angle of depression is often not an angle of the triangle, but the complement to an angle of the triangle.

7. Example 3: Vernon is on the top dock of a cruise ship and observes two dolphins following each other directly away from the ship in a straight line. Vernon’s position is 154 meters above sea level, and the angles of depression to the two dolphins are 35 and 36. Find the distance between the two dolphins to the nearest meter. 154 m 35  36  35  36  d n m d = m – n

8. Example 3: continued d = m – n 154 tan 35  = 154 m (m)tan 35  = (m) (m)tan 35  = 154 (m)tan 35  154 tan 35  tan 35 = m = 219.93 tan 36  = 154 n (n)tan 36  = (n) 154 n (n)tan 36  = 154 (n)tan 36  154 tan 36  tan 36  = n = 211.96 Then, d = m – n 154 35  m adjacent opposite m 154 36  n adjacent opposite = 219.93 – 211.96= 7.97≈ 8 m 154 tan 35 m = 154 tan 35 m =

9. Example 4: The top of a signal tower 120 meters above sea level. The angle of depression from the top of tower to a passing ship is 25 . How many meters from the foot of the tower is the ship? d 25  120 tan 25  = 120 d (d)tan 25 = (d) (d)tan 25 = 120 (d)tan 25 120 tan 25 tan 25 = d = 257.3 d

10. 11000 Example 5: After flying at an altitude of 500 meters, a helicopter starts to descend when its ground distance from the landing pad is 11 kilometers. What is the angle of depression for this part of the flight? xx xx 500 meters 11 km = 11,000 meters 500 tan x  = 11000 500 tan -1 ( ) = x  2.602562205  = x 2.6  = x