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# 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: 14th Jul 2024
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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.