Trigonometry question?

Question

Find the acute angle A,to the nearest second when:

a. log cos A=9.12575
b.log sin A=9.91655

if you could explain step by step this would be helpful,thank you.

Answer 1

These problems are written properly if you allow for imaginary angles. Imaginary angles can produce trigonometric values exceeding 1... or less than -1

Answer 2

Are you sure you wrote those down correctly? Cosine and sine vary from -1 to 1 and in acute angles from 0 to 1, however log(cos A)=9.12575 implies that 10^9.12575= Cos A, assuming your logarithms are in base ten, and this menas Cos A equals a number much greater than one which is imposible.
If you meant log(cos A)= -9.12575 then this implies Cos A= 10^-9.12575 =0.0000000000746 which means taht A=89.9999999573 degrees.
To the nearest second A= 90 degrees 0 minutes 0 seconds
For the second, Sin A=0.000000000121 so A=0.000000006943 degrees.
And to the closeste second A=0 degrees with 0 minutes and 0 seconds

Answer 3

Angles to the nearest second may exceed the accuracy of the inputs. First, you need to convert the log sin and log cos to sin and cos. There should be a (-10) added on to each log value, because cos and sin are at most equal to 1, whose log is 0.

In other words, the cos A= (0.1)*10^(.12575) and sin A=0.1*10^(.91655). BTW: these are TWO different angles, no single angle meets both values.

Answer 4

First off, the problem should have said log cos A = 9.12575 - 10. Then, just type it into a calculator, as 2nd log (9.12575 - 10) which gives cos A then do 2nd cos answer in correct mode (degree/radian). log with no base is assumed to be common log.

Answer 5

I really really don't mean to be rude, but the expression log MUST have a base specified, so you know what logarithm to extract. Others might know what base you're referring to, but personally, i'm stumped until you mention that. Sorry