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Question: Find the acute angle \[\theta \] through which coordinate axes should be rotated for the point \[A(2...

Find the acute angle θ\theta through which coordinate axes should be rotated for the point A(2,4)A(2,4) to attend new abscissa 44.
A). tan⁡θ=34\tan \theta = \dfrac{3}{4}
B). tan⁡θ=56\tan \theta = \dfrac{5}{6}
C). tan⁡θ=78\tan \theta = \dfrac{7}{8}
D). None of these

Explanation

Solution

We need to find the acute angle θ\theta through which coordinate axes should be rotated for the point A(2,4)A(2,4) to attend new abscissa 44. To find the new position we need to use the identity given below in the hint section.
Formula used:
The acute angleθ\theta through which coordinate axes should be rotated for the point A(x,y)A(x,y) to attend new abscissa XX is given by,
X=xcos⁡θ+ysin⁡θX = x\cos \theta + y\sin \theta

Complete step-by-step solution:
We have to find the acute angle θ\theta through which coordinate axes should be rotated for the point A(2,4)A(2,4) to attend new abscissa 44.
Let us note the given information,
A(x,y)=A(2,4)A(x,y) = A(2,4)
Let us use the below identity,
The acute angle θ\theta through which coordinate axes should be rotated for the point A(x,y)A(x,y) to attend new abscissa XX is given by,
X=xcos⁡θ+ysin⁡θX = x\cos \theta + y\sin \theta
On putting values (x,y)=(2,4)(x,y) = (2,4) in above equation we get,
4=2cos⁡θ+4sin⁡θ4 = 2\cos \theta + 4\sin \theta
On dividing the equation by 22 on both sides we get,
2=cos⁡θ+2sin⁡θ2 = \cos \theta + 2\sin \theta
On rearranging the terms on both sides of the equation we get,
cos⁡θ=2−2sin⁡θ\cos \theta = 2 - 2\sin \theta
On squaring both side we get,
cos⁡2θ=(2−2sin⁡θ)2{\cos ^2}\theta = {\left( {2 - 2\sin \theta } \right)^2}
On performing square of the bracket using the formula (a−b)2=a2−2ab+b2{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2} on R.H.S. we get,
cos⁡2θ=4−8sin⁡θ+4sin⁡2θ{\cos ^2}\theta = 4 - 8\sin \theta + 4{\sin ^2}\theta
On putting value cos⁡2θ=1−sin⁡2θ{\cos ^2}\theta = 1 - {\sin ^2}\theta in above equation we get,
1−sin⁡2θ=4−8sin⁡θ+4sin⁡2θ1 - {\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,
5sin⁡2θ−8sin⁡θ+3=05{\sin ^2}\theta - 8\sin \theta + 3 = 0
On splitting the middle term to factorize the equation we get,
5sin⁡2θ−5sin⁡θ−3sin⁡θ+3=05{\sin ^2}\theta - 5\sin \theta - 3\sin \theta + 3 = 0
On taking common terms out we get,
5sin⁡θ(1−sin⁡θ)−3(1−sin⁡θ)=05\sin \theta (1 - \sin \theta ) - 3(1 - \sin \theta ) = 0
On taking common term out we get,
(sin⁡θ−1)(5sin⁡θ−3)=0\left( {\sin \theta - 1} \right)\left( {5\sin \theta - 3} \right) = 0
On equating both factors to zero we get,
sin⁡θ−1=0,5sin⁡θ−3=0\sin \theta - 1 = 0,5\sin \theta - 3 = 0
On rearranging the terms of the equation we get,
sin⁡θ=1,sin⁡θ=35\sin \theta = 1,\sin \theta = \dfrac{3}{5}
On considering first value,
sin⁡θ=1\sin \theta = 1
From above value we can write that,
tan⁡θ=∞\tan \theta = \infty
Thus this value sin⁡θ=1\sin \theta = 1is invalid.
On considering second value,
sin⁡θ=35\sin \theta = \dfrac{3}{5}
From above value we can write that,
cos⁡θ=45\cos \theta = \dfrac{4}{5}
On taking ratios of both values we can write that,
tan⁡θ=sin⁡θcos⁡θ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}
On putting both values we can write that,
tan⁡θ=3545\tan \theta = \dfrac{{\dfrac{3}{5}}}{{\dfrac{4}{5}}}
On cancelling the common denominator and performing the operation we get,
tan⁡θ=34\tan \theta = \dfrac{3}{4}
Hence option A)tan⁡θ=34\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.