check my work
Two boats start their journey from the same point A and travel along directions AC and AD, as shown below What is the distance, CD, between the boats? 284.3 ft 115.5 ft 230.9 ft >173.2 ft
tan (30) = BD /100 tan (60) = BC/ 100
BD-BC= CD
100 SQT 3 = 173.205
attention please !!! tan 30 = 100/BD and the second is invers too
the tan law say opposite divide adjacent
@TheSmartOne please
ok. ? correct it please
(100 sqrt 3) -(100/sqrt3) = 116.2
i think this would be correct
i mean 115.47
@agent0smith
tan30=100/BD tan60 = 100/BC BD = 100/tan30 BC = 100/tan60 CD = BD -BC = 100/tan30 - 100/tan60 do you can solve it this now please ?
tan30 = (sqrt3)/3 tan60 = sqrt3
thats what i did . tan 30 = \[\sqrt{3/(3)} = \sqrt{3}\] tan 60 = \[\sqrt{3}\]
so then \[100 \sqrt{3}\] \[100/ \sqrt{3}\]
173.205- 57.735 = 115.47
300/sqrt3 -100/sqrt3 = 200/sqrt3 = 200*sqrt3 /3 = 200*1,73 /3 = 346/3 = 115,3 so CD = 115,3 approximately like my result so i think is right now ok. ?
Smith do you agree this ?
please
ty
i don't really understand what you did but we got the same answer. although mines was a bit more accurate
100/(sqrt3 /3) = 300/sqrt3 yes ?
were did you get the 300 from?
100/sqrt3 /3 = (100*3)/sqrt3 yes ?
a/(b/c) = ac/b
ok
do you understand it now ? just i ve wrote there directly
nope , i only get it the way i did it. however i'm going to go with 115.47. i think this is right.
thank you for the help . it was correct
np was my pleasure good luck bye bye
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