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
•
Total views: 203.4k
•
Views today: 2.83k
Answer
203.4k+ views
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.
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.
Recently Updated Pages
Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE
