# Find the acute angle \[\theta \] through which coordinate axes should be rotated for the point \[A(2,4)\] to attend new abscissa \[4\].

A). \[\tan \theta = \dfrac{3}{4}\]

B). \[\tan \theta = \dfrac{5}{6}\]

C). \[\tan \theta = \dfrac{7}{8}\]

D). None of these

Last updated date: 13th Mar 2023

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Answer

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**Hint:**We need to find the acute angle \[\theta \] through which coordinate axes should be rotated for the point \[A(2,4)\] to attend new abscissa \[4\]. To find the new position we need to use the identity given below in the hint section.

**Formula used:**

The acute angle\[\theta \] through which coordinate axes should be rotated for the point \[A(x,y)\] to attend new abscissa \[X\] is given by,

\[X = x\cos \theta + y\sin \theta \]

**Complete step-by-step solution:**

We have to find the acute angle \[\theta \] through which coordinate axes should be rotated for the point \[A(2,4)\] to attend new abscissa \[4\].

Let us note the given information,

\[A(x,y) = A(2,4)\]

Let us use the below identity,

The acute angle \[\theta \] through which coordinate axes should be rotated for the point \[A(x,y)\] to attend new abscissa \[X\] is given by,

\[X = x\cos \theta + y\sin \theta \]

On putting values \[(x,y) = (2,4)\] in above equation we get,

\[4 = 2\cos \theta + 4\sin \theta \]

On dividing the equation by $2$ on both sides we get,

\[2 = \cos \theta + 2\sin \theta \]

On rearranging the terms on both sides of the equation we get,

\[\cos \theta = 2 - 2\sin \theta \]

On squaring both side we get,

\[{\cos ^2}\theta = {\left( {2 - 2\sin \theta } \right)^2}\]

On performing square of the bracket using the formula \[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\] on R.H.S. we get,

\[{\cos ^2}\theta = 4 - 8\sin \theta + 4{\sin ^2}\theta \]

On putting value \[{\cos ^2}\theta = 1 - {\sin ^2}\theta \] in above equation we get,

\[1 - {\sin ^2}\theta = 4 - 8\sin \theta + 4{\sin ^2}\theta \]

On arranging all the terms on L.H.S. and performing the addition we get,

\[5{\sin ^2}\theta - 8\sin \theta + 3 = 0\]

On splitting the middle term to factorize the equation we get,

\[5{\sin ^2}\theta - 5\sin \theta - 3\sin \theta + 3 = 0\]

On taking common terms out we get,

\[5\sin \theta (1 - \sin \theta ) - 3(1 - \sin \theta ) = 0\]

On taking common term out we get,

\[\left( {\sin \theta - 1} \right)\left( {5\sin \theta - 3} \right) = 0\]

On equating both factors to zero we get,

\[\sin \theta - 1 = 0,5\sin \theta - 3 = 0\]

On rearranging the terms of the equation we get,

\[\sin \theta = 1,\sin \theta = \dfrac{3}{5}\]

On considering first value,

\[\sin \theta = 1\]

From above value we can write that,

\[\tan \theta = \infty \]

Thus this value \[\sin \theta = 1\]is invalid.

On considering second value,

\[\sin \theta = \dfrac{3}{5}\]

From above value we can write that,

\[\cos \theta = \dfrac{4}{5}\]

On taking ratios of both values we can write that,

\[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]

On putting both values we can write that,

\[\tan \theta = \dfrac{{\dfrac{3}{5}}}{{\dfrac{4}{5}}}\]

On cancelling the common denominator and performing the operation we get,

\[\tan \theta = \dfrac{3}{4}\]

**Hence option A)$\tan \theta = \dfrac{3}{4}$ is correct.**

**Note:**We need to calculate the new angle using the identity and by performing operations. We need to choose a valid value and we need to discard the value which gives the answer which is out of the range.

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