janessia
Isaac is looking through binoculars on a whale watching trip when he notices a sea otter in the distance. If he is 20 feet above sea level in the boat, and the angle of depression is 30 degrees, how far away from the boat is the otter (to the nearest foot)?
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mark_o.
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hello are you there? do u think you can draw this first before you can solve?
mark_o.
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@ janessia
janessia
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hmm idk
janessia
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how would i draw it?
mark_o.
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do you know the angle of depression?
mark_o.
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|dw:1351530728427:dw|
mark_o.
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the height y=20 , the x= distance
angle theta = angle of depression
janessia
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so how do i solve this?
mark_o.
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|dw:1351530990457:dw|
mark_o.
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you can try tangent
janessia
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opposite hypotenuse?
janessia
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opposite adjacent
janessia
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?
mark_o.
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sorry my pc got frozen,,, yes try that tangent
mark_o.
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let me know what you getting
mark_o.
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\[\tan \theta =\frac{ 20 }{ x } ,......\]
then what is x=?
mark_o.
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memorized
sin A= opposite/ hypotenuse
cos A= adjacent/hypotenuse
tan A= opposite/adjacent
janessia
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ohk but hmmmm??? im so confused
janessia
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just forget it its fine
mark_o.
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\[\tan \theta =\frac{ 20 }{ x }\]
so here we want x, therefore
multiply both sides by x, to get
\[xtan \theta =\frac{ 20 }{ x }.x\]
\[x \tan \theta =20\] divide both sides by tan o gives
\[\frac{ x \tan \theta }{ \tan \theta }=\frac{ 20 }{ \tan \theta } \]
w/c gives
\[x=\frac{ 20 }{ \tan \theta }....\]
x=?
mark_o.
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did you get it?
angle of depression
\[\theta =30\]
mark_o.
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so
x= 20/ tan 30 what did you get?