# If $ \sin 4A = \cos (A - {20^ \circ }) $ , where 4A is an acute angle, Find the value of A in degrees.

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**Hint**: This sum is from the chapter trigonometric functions. Though this sum may seem complicated with the use of words like acute angle , $ \sin 4A $ ,it is very easy if you know how to break down these trigonometric functions. This sum involves only one step where you need to convert $ \sin \theta $ to $ \cos \theta $ . For this it is very important to know the relationship between $ \sin \theta $ & $ \cos \theta $ . Also it is necessary to know all the identities pertaining to trigonometric functions.

**:**

__Complete step-by-step answer__Given that $ 4A $ is an acute angle, an acute angle is an angle which measures less than $ {90^ \circ } $ .

Since it is given that $ 4A $ is acute angle, that means $ A $ is also an acute angle.

We will have to use $ \sin \theta = \cos (90 - \theta ) $ for this particular problem.

$ \Rightarrow \sin 4A = \cos (90 - 4A).......(1) $

We can now substitute value of $ Equation1 $ in the given problem in order to find the value of $ A $

$ \Rightarrow \cos ({90^ \circ } - 4A) = \cos (A - {20^0}).......(2) $

Since both the LHS & RHS are terms having $ \cos () $ we can equate the values inside the bracket directly to find the value of $ A $ .

\[ \Rightarrow {90^ \circ } - 4A = A - {20^0}.......(3)\]

\[ \Rightarrow 5A = {110^0}.......(4)\]

$ \Rightarrow A = {22^ \circ } $

Thus the value of $ A $ is $ {22^ \circ } $ . We can also say this answer is correct as the value of both $ 4A $ is less than $ {90^ \circ } $ and hence it is an acute angle.

**So, the correct answer is “ $ A = {22^ \circ } $ ”.**

**Note**: Though this sum looks easy, it wouldn’t have been possible to solve it, if the student was not aware about the relationship between $ \cos \theta $ & $ \sin \theta $ . For all the sums it is advisable to note down the identities and memories of them so that students are not stuck in any of the sums. All the sums involving trigonometric functions are just application of the identities and nothing else. So once the student is thorough with the identities he/she can easily solve the sum.

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