Asked by Anonymous
find the acute anlge A, to the nearest seconds when ;
A.log cos A= 9.12575
you might say that The cosine is >1, there's no real number solution.
but there is . I think they mean
log cos A= 9.12575 -10 .
okay now i don't know how to get rid of the log but i "think" i can convert it to degrees ,minutes, seconds .
I also have a similar one which is .
B. log sin A = 9.91655
please i need all the help i can get trust me i tryed already and i already did the rest of them. for example this one
log tan A = 0.06323
tan(A) = 1.1567...
A = ~ 49.156º
A =~ 49º 9' 23"
A.log cos A= 9.12575
you might say that The cosine is >1, there's no real number solution.
but there is . I think they mean
log cos A= 9.12575 -10 .
okay now i don't know how to get rid of the log but i "think" i can convert it to degrees ,minutes, seconds .
I also have a similar one which is .
B. log sin A = 9.91655
please i need all the help i can get trust me i tryed already and i already did the rest of them. for example this one
log tan A = 0.06323
tan(A) = 1.1567...
A = ~ 49.156º
A =~ 49º 9' 23"
Answers
Answered by
Reiny
Your example with log tan A ....
is done correctly.
However log cos A = 9.12575 definitely does NOT have a solution, since you would get
cos A = 10^9.1257 which of course is not possible.
If you think they meant
log cos A = 9.12575 - 10 , then
log cos A = -.87425
( 2nd F log)
cos A = .1335826
angle A = 82.32333° or 82° 19' 24"
question b) would be the same situation.
is done correctly.
However log cos A = 9.12575 definitely does NOT have a solution, since you would get
cos A = 10^9.1257 which of course is not possible.
If you think they meant
log cos A = 9.12575 - 10 , then
log cos A = -.87425
( 2nd F log)
cos A = .1335826
angle A = 82.32333° or 82° 19' 24"
question b) would be the same situation.
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