Find the value of #/_A#, where #0^@<=A<=90^@#?

#Sin(90^@-3A).Csc42^@=1#

1 Answer
Nghi N
Mar 28, 2018

#A = 15^@98 #

Explanation:

sin (90 - 3A) = cos 3A
#cos 3A = 1/(csc 42)= sin 42^@ = 0.67#
cos 3A = 0.67
Calculator and unit circle give 2 solutions:
#3A = +- 47^@93 + k360^@#^@
#A = +- 15^@98 + k120^@#
Inside the interval (0, 90) the answer is:
#A = 15^@98#

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