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Mathematics 10 Online
OpenStudy (anonymous):

help me

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

OpenStudy (anonymous):

@jim_thompson5910

OpenStudy (dls):

\[\Huge \tan \theta = \frac{perpendicular}{base}\]

OpenStudy (dls):

|dw:1404247333746:dw|

OpenStudy (dls):

and all you need to know is that \[\Huge \tan 30 = \frac{1}{\sqrt 3}\]

OpenStudy (anonymous):

i did something similar with someone earlier so i think it should be easy for me to understand

OpenStudy (anonymous):

okay let me solve it

OpenStudy (dls):

\[\Huge \frac{1}{\sqrt 3 } = \frac{H}{90}\] Solve for H.

OpenStudy (anonymous):

cross multiply?

OpenStudy (dls):

yep then divide both sides by root 3

OpenStudy (anonymous):

52?

OpenStudy (dls):

what did you do? H should be \[\Huge H=\frac{90}{ \sqrt 3}\]

OpenStudy (dls):

to change the form, \[\LARGE H = \frac{90}{\sqrt 3} \times \frac{\sqrt 3}{\sqrt 3} = 30 \sqrt 3\]

OpenStudy (anonymous):

i know then i get 52?

OpenStudy (anonymous):

i got 51.96 and i rounded

OpenStudy (anonymous):

http://www.wolframalpha.com/input/?i=30%E2%88%9A3

OpenStudy (dls):

oh..u wrote an exact answer so I thought u did smt wrong XD its okay

OpenStudy (anonymous):

i submitted it and got 100% thanks for all the help

OpenStudy (dls):

uw

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