If sin 3A = cos (A - $10^{o}$ ) and 3A is acute then $∠$ A = ?

(a) $35^{o}$ (b) $25^{o}$ (c) $20^{o}$ (d) $45^{o}$

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Solution

We know that, $s i n 3 A = c o s (90 - 3 A)$

Now, $c o s (90 - 3 A) = c o s (A - 10) \Rightarrow 90 - 3 A = A - 10 \Rightarrow 100 = 4 A ∴ A = 25^{o}$

So, the correct choice is (b).

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