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| 1. |
If sin 3A=cos(A-16) where 3A |
| Answer» Qustion:\xa0If sin 3A = cos (A – 26°), where 3A is an acute angle, find the value of A.Solution:Sin 3A = cos (A-26 )cos (90 - 3A) = cos (A-26) [ sinA = cos (90- A )By comparison90-3A = A-2690+26=A+3A116=4AA= 29 | |