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mr dom is standing on a 40ft ocean bluff.he has two dogs below If his line of sight is...

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r4rob92 | (Level 1) Valedictorian

Posted May 12, 2013 at 9:58 PM via web

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mr dom is standing on a 40ft ocean bluff.he has two dogs below If his line of sight is 6ft above the ground and angles of depression to dogs are 34degrees and 48 degrees how far apart are the dogs?

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embizze | High School Teacher | (Level 1) Educator Emeritus

Posted May 13, 2013 at 1:20 AM (Answer #1)

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We have two right triangles that share a leg. The length of the leg is 46 feet. (The height of the bluff plus the man's eye height.)

In the first triangle, the angle opposite the 46 foot leg is `48^@` .(The angle of depression from the man to the dog is the same as the angle of elevation from the dog to the man.)

Then if this dog's distance from the bluff is `d_1` then:

`tan 48^@=46/d_1`  ==> `d_1=46/(tan 48^@)~~41.4"ft"`

In the second triangle the angle opposite the 46 foot leg is `34^@` . And if the distance to the second dog is `d_2` we have:

`tan 34^@=46/d_2` ==> `d_2=46/(tan 34)~~68.2"ft"`

The distance between the dogs is `d_2-d_1=68.2-41.4=26.8"ft"`

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