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October 30, 2014

October 30, 2014

Posted by **Trig Picture** on Friday, September 11, 2009 at 6:16pm.

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

- Trig Picture -
**Ms. Sue**, Friday, September 11, 2009 at 6:32pmHere's the link to the diagram.

http://i32.tinypic.com/ipmko5.jpg

- Trig Picture -
**MathMate**, Friday, September 11, 2009 at 7:35pmThe 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

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