please solve this example class 10 cbse
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---if 3A=cos(A-26),where 3A=acute angle
cos(90-3A)=Cos(A-26)
90-3A=A-26
-3A-A=-26-90
-4A=-106
A=-106+4
A=-102
cos(90-3A)=Cos(A-26)
90-3A=A-26
-3A-A=-26-90
-4A=-106
A=-106+4
A=-102
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