Ask
your own question, for FREE!
Ask question now!
Mathematics
10 Online
OpenStudy (anonymous):
help me
11 years ago
Join the QuestionCove community and study together with friends!
Sign Up
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
11 years ago
OpenStudy (anonymous):
@jim_thompson5910
11 years ago
OpenStudy (dls):
\[\Huge \tan \theta = \frac{perpendicular}{base}\]
11 years ago
OpenStudy (dls):
|dw:1404247333746:dw|
11 years ago
OpenStudy (dls):
and all you need to know is that \[\Huge \tan 30 = \frac{1}{\sqrt 3}\]
11 years ago
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
i did something similar with someone earlier so i think it should be easy for me to understand
11 years ago
OpenStudy (anonymous):
okay let me solve it
11 years ago
OpenStudy (dls):
\[\Huge \frac{1}{\sqrt 3 } = \frac{H}{90}\]
Solve for H.
11 years ago
OpenStudy (anonymous):
cross multiply?
11 years ago
OpenStudy (dls):
yep then divide both sides by root 3
11 years ago
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
52?
11 years ago
OpenStudy (dls):
what did you do? H should be
\[\Huge H=\frac{90}{ \sqrt 3}\]
11 years ago
OpenStudy (dls):
to change the form,
\[\LARGE H = \frac{90}{\sqrt 3} \times \frac{\sqrt 3}{\sqrt 3} = 30 \sqrt 3\]
11 years ago
OpenStudy (anonymous):
i know then i get 52?
11 years ago
OpenStudy (anonymous):
i got 51.96 and i rounded
11 years ago
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (dls):
oh..u wrote an exact answer so I thought u did smt wrong XD its okay
11 years ago
OpenStudy (anonymous):
i submitted it and got 100% thanks for all the help
11 years ago
OpenStudy (dls):
uw
11 years ago
Can't find your answer?
Make a FREE account and ask your own questions, OR help others and earn volunteer hours! Join our real-time social learning platform and learn together with your friends!
Sign Up
Ask Question