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The image of a star is projected onto a planetarium wall by
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13 Sep 2020, 08:26
13 Sep 2020, 08:26
Expert Reply
00:00
Question Stats:
61% (01:21) correct
38% (01:53) wrong based on 39 sessions
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The image of a star is projected onto a planetarium wall by a projector that sits atop a vertical 4-foot stand. If the projector is directed 30 degrees above the horizontal, and the image appears 16 feet above the level floor of the planetarium, then, in feet, how far is the projector from the wall?
A. \(12 \sqrt{2}\)
B. \(12 \sqrt{3}\)
C. \(16 \sqrt{2}\)
D. \(16 \sqrt{3}\)
E. \(24\)
A. \(12 \sqrt{2}\)
B. \(12 \sqrt{3}\)
C. \(16 \sqrt{2}\)
D. \(16 \sqrt{3}\)
E. \(24\)
Re: The image of a star is projected onto a planetarium wall by
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29 Sep 2020, 10:32
29 Sep 2020, 10:32
Expert Reply
Post A Detailed Correct Solution For The Above Questions And Get A Kudos.
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
For more theory and practice see our GRE - Math Book
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
For more theory and practice see our GRE - Math Book
Re: The image of a star is projected onto a planetarium wall by
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29 Sep 2020, 22:04
29 Sep 2020, 22:04
2
we need to find BC from the figure (here P-projector
W-wall)
from triangle PBC
tan30=(4/AB)
AB=4√3
from triangle AWC
tan30=(16/AC)
AC=16√3
now BC=AC-AB=16√3-4√3=12√3
option B is correct
W-wall)
from triangle PBC
tan30=(4/AB)
AB=4√3
from triangle AWC
tan30=(16/AC)
AC=16√3
now BC=AC-AB=16√3-4√3=12√3
option B is correct
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