Asked by Trig Picture
maybe i drew it wrong...
h t t p : / / i 3 2 . t i n y p i c . c o m / i p m k o 5 . j p g
the way my teacher taught me the angle of depression was the anlge that is made with a horizontal line or something like that
i believed i drew it right
oh and by the way this is the diagram i drew for that problem i asked earlier which I\'m not sure to do i though it drew it right
h t t p : / / i 3 2 . t i n y p i c . c o m / i p m k o 5 . j p g
the way my teacher taught me the angle of depression was the anlge that is made with a horizontal line or something like that
i believed i drew it right
oh and by the way this is the diagram i drew for that problem i asked earlier which I\'m not sure to do i though it drew it right
Answers
Answered by
Ms. Sue
Here's the link to the diagram.
http://i32.tinypic.com/ipmko5.jpg
http://i32.tinypic.com/ipmko5.jpg
Answered by
MathMate
The angle PAQ is the angle of elevation viewed from the ship, which is the same as the angle of depression viewed from the airplane (8°20'). The same for ship B.
This is because the angles of elevation and depression are alternate angles between two parallel lines.
I agree with the two calculations above using H/tan(θ) for the horizontal distance, and the answer of 13.143 km.
See original post:
http://www.jiskha.com/display.cgi?id=1252701406
This is because the angles of elevation and depression are alternate angles between two parallel lines.
I agree with the two calculations above using H/tan(θ) for the horizontal distance, and the answer of 13.143 km.
See original post:
http://www.jiskha.com/display.cgi?id=1252701406
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