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Angles of Elevation 8-4 and Depression Warm Up Lesson Presentation

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1 Angles of Elevation 8-4 and Depression Warm Up Lesson Presentation
Lesson Quiz Holt Geometry

2 1. Identify the pairs of alternate interior angles.
Warm Up 1. Identify the pairs of alternate interior angles. 2. Use your calculator to find tan 30° to the nearest hundredth. 3. Solve Round to the nearest hundredth. 2 and 7; 3 and 6 0.58

3 Objective Solve problems involving angles of elevation and angles of depression.

4 Vocabulary angle of elevation angle of depression

5 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 tower T to the plane P. 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 plane to the tower.

6 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.

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

8 Example 1A: 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.

9 Example 1B: 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.

10 Check It Out! Example 1 Use the diagram above to classify each angle as an angle of elevation or angle of depression. 1a. 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. 1b. 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.

11 Example 2: Finding Distance by Using Angle of Elevation
The Seattle Space Needle casts a 67-meter shadow. If the angle of elevation from the tip of the shadow to the top of the Space Needle is 70º, how tall is the Space Needle? 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 Space Needle. Let y be the height of the Space Needle.

12 Example 2 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.

13 Multiply both sides by x and divide by tan 29°.
Check It Out! Example 2 What if…? Suppose the plane is at an altitude of 3500 ft and the angle of elevation from the airport to the plane is 29°. What is the horizontal distance between the plane and the airport? 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°. 3500 ft 29° x  6314 ft Simplify the expression.

14 Example 3: Finding Distance by Using Angle of Depression
An ice climber 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.

15 Example 3 Continued 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.

16 Example 3 Continued By the Alternate Interior Angles Theorem, mB = 52°. Write a tangent ratio. y = 115 tan 52° Multiply both sides by 115. y  147 ft Simplify the expression.

17 By the Alternate Interior Angles Theorem, mF = 3°.
Check It Out! Example 3 What if…? Suppose the ranger sees another fire and the angle of depression to the fire is 3°. What is the horizontal distance to this fire? Round to the nearest foot. By the Alternate Interior Angles Theorem, mF = 3°. Write a tangent ratio. Multiply both sides by x and divide by tan 3°. x  1717 ft Simplify the expression.

18 Example 4: Shipping Application
An observer in a lighthouse is 69 ft above the water. He sights two boats in the water directly in front of him. The angle of depression to the nearest boat is 48º. The angle of depression to the other boat is 22º. What is the distance between the two boats? Round to the nearest foot.

19 Example 4 Application Step 1 Draw a sketch. Let L represent the observer in the lighthouse and let A and B represent the two boats. Let x be the distance between the two boats.

20 By the Alternate Interior Angles Theorem, mCAL = 58°.
Example 4 Continued Step 2 Find y. By the Alternate Interior Angles Theorem, mCAL = 58°. In ∆ALC, So .

21 Example 4 Continued Step 3 Find z. By the Alternate Interior Angles Theorem, mCBL = 22°. In ∆BLC, So

22 Example 4 Continued Step 4 Find x. x = z – y x  – 62.1  109 ft So the two boats are about 109 ft apart.

23 Check It Out! Example 4 A pilot flying at an altitude of 12,000 ft sights two airports directly in front of him. The angle of depression to one airport is 78°, and the angle of depression to the second airport is 19°. What is the distance between the two airports? Round to the nearest foot.

24 Check It Out! Example 4 Continued
Step 1 Draw a sketch. Let P represent the pilot and let A and B represent the two airports. Let x be the distance between the two airports. 78° 19° 12,000 ft

25 Check It Out! Example 4 Continued
Step 2 Find y. By the Alternate Interior Angles Theorem, mCAP = 78°. In ∆APC, So

26 Check It Out! Example 4 Continued
Step 3 Find z. By the Alternate Interior Angles Theorem, mCBP = 19°. In ∆BPC, So

27 Check It Out! Example 4 Continued
Step 4 Find x. x = z – y x  34,851 – 2551  32,300 ft So the two airports are about 32,300 ft apart.

28 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 30.5 d tan 27 = opposite 5.5 ft (d)tan 27 = (d) 30.5 d 25 ft 30.5 (d)tan 27 = 30.5 27 d adjacent (d)tan 27 tan 27 tan 27 = opp tan = 30.5 tan 27 d = adj d = 59.8 ≈60

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

30 Example 3: d = m – n 35 36 d 154 m n 36 35 m
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 Find the distance between the two dolphins to the nearest meter. 154 m 35 36 d d = m – n n 36 35 m

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

32 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? 120 tan 25 = d 120 d (d)tan 25 = (d) 25 (d)tan 25 = 120 d (d)tan tan tan 25 = d = 257.3

33 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? 500 tan x = 11000 11000 500 tan-1 ( ) = x x 500 meters  = x x 2.6 = x 11 km = 11,000 meters

34 Lesson Quiz: Part I Classify each angle as an angle of elevation or angle of depression. 1. 6 2. 9 angle of depression angle of elevation

35 Lesson Quiz: Part II 3. A plane is flying at an altitude of 14,500 ft. The angle of depression from the plane to a control tower is 15°. What is the horizontal distance from the plane to the tower? Round to the nearest foot. 4. A woman is standing 12 ft from a sculpture. The angle of elevation from her eye to the top of the sculpture is 30°, and the angle of depression to its base is 22°. How tall is the sculpture to the nearest foot? 54,115 ft 12 ft


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