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Mathematics
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OpenStudy (anonymous):
help me
12 years ago
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OpenStudy (anonymous):
The length of the shadow of a building is 90 meters, as shown below:
What is the height of the building?
180 m
51.96 m
155.88 m
60 m
i know i have to find the adjacent i believe but i just need someone to guide me
12 years ago
OpenStudy (anonymous):
@jim_thompson5910
12 years ago
OpenStudy (dls):
\[\Huge \tan \theta = \frac{perpendicular}{base}\]
12 years ago
OpenStudy (dls):
|dw:1404247333746:dw|
12 years ago
OpenStudy (dls):
and all you need to know is that \[\Huge \tan 30 = \frac{1}{\sqrt 3}\]
12 years ago
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OpenStudy (anonymous):
i did something similar with someone earlier so i think it should be easy for me to understand
12 years ago
OpenStudy (anonymous):
okay let me solve it
12 years ago
OpenStudy (dls):
\[\Huge \frac{1}{\sqrt 3 } = \frac{H}{90}\]
Solve for H.
12 years ago
OpenStudy (anonymous):
cross multiply?
12 years ago
OpenStudy (dls):
yep then divide both sides by root 3
12 years ago
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OpenStudy (anonymous):
52?
12 years ago
OpenStudy (dls):
what did you do? H should be
\[\Huge H=\frac{90}{ \sqrt 3}\]
12 years ago
OpenStudy (dls):
to change the form,
\[\LARGE H = \frac{90}{\sqrt 3} \times \frac{\sqrt 3}{\sqrt 3} = 30 \sqrt 3\]
12 years ago
OpenStudy (anonymous):
i know then i get 52?
12 years ago
OpenStudy (anonymous):
i got 51.96 and i rounded
12 years ago
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OpenStudy (dls):
oh..u wrote an exact answer so I thought u did smt wrong XD its okay
12 years ago
OpenStudy (anonymous):
i submitted it and got 100% thanks for all the help
12 years ago
OpenStudy (dls):
uw
12 years ago
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