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# 8-5 Angles of Elevation and Depression - PowerPoint PPT Presentation

8-5 Angles of Elevation and Depression. You used similar triangles to measure distances indirectly. . Solve problems involving angles of elevation and depression. Use angles of elevation and depression to find the distance between two objects. Vocabulary.

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You used similar triangles to measure distances indirectly.

• Solve problems involving angles of elevation and depression.

• Use angles of elevation and depression to find the distance between two objects.

Angle of elevation – the angle formed by a horizontal line and an observer’s line of sight to an object above the horizontal line.

Angle of depression – the angle formed by a horizontal line and an observer’s line of sight to an object below the horizontal line

CIRCUS ACTS At the circus, a person in the audience at ground level watches the high-wire routine. A 5-foot-6-inch 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°?

Make a drawing.

Since QR is 25 feet and RS is 5 feet 6 inches or 5.5 feet, QS is 30.5 feet. Let x represent PQ.

Multiply both sides by x.

Simplify.

Answer:The audience member is about 60 feet from the base of the platform.

DIVING At a diving competition, a 6-foot-tall diver stands atop the 32-foot platform. The front edge of the platform projects 5 feet beyond the ends of the pool. The pool itself is 50 feet in length. A camera is set up at the opposite end of the pool even with the pool’s edge. If the camera is angled so that its line of sight extends to the top of the diver’s head, what is the camera’s angle of elevation to the nearest degree?

A. 37°

B. 35°

C. 40°

D. 50°

Since are parallel, mBAC = mACD by the Alternate Interior Angles Theorem.

DISTANCE Maria is at the top of a cliff and sees a seal in the water. If the cliff is 40 feet above the water and the angle of depression is 52°, what is the horizontal distance from the seal to the cliff, to the nearest foot?

Make a sketch of the situation.

Let x represent the horizontal distance from the seal to the cliff, DC.

C = 52°; AD = 40, and DC= x

Multiply each side by x.

Divide each side by tan 52°.

Luisa is in a hot air balloon 30 feet above the ground. She sees the landing spot at an angle of depression of 34. What is the horizontal distance between the hot air balloon and the landing spot to the nearest foot?

A. 19 ft

B. 20 ft

C. 44 ft

D. 58 ft

Two angles of Elevation or Depression sees the landing spot at an angle of depression of 34

• Angles of elevation or depression to two different objects can be used to estimate the distance between those objects.

• Angles from two different positions of observation to the same object can be used to estimate the object’s height.

DISTANCE sees the landing spot at an angle of depression of 34 Vernon is on the top deck 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.

Understand ΔMLK andΔMLJare right triangles. The distance between the dolphins isJK or JL – KL. Use the right triangles to find these two lengths.

Plan sees the landing spot at an angle of depression of 34 Because are horizontal lines, they are parallel. Thus, and because they are alternate interior angles. This means that

Divide each side by tan

Solve

Multiply each side by JL.

Use a calculator.

Divide each side by tan sees the landing spot at an angle of depression of 34

Multiply each side by KL.

Use a calculator.

Answer:The distance between the dolphins is JK – KL. JL – KL≈ 219.93 – 211.96, or about 8 meters.

8-5 Assignment sees the landing spot at an angle of depression of 34

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