Question
Let $$A = \left( {1,\,0} \right)$$ and $$B = \left( {2,\,1} \right).$$ The line $$AB$$ turns about $$A$$ through an angle $$\frac{\pi }{6}$$ in the clockwise sense, and the new position of $$B$$ is $$B'.$$ Then $$B'$$ has the coordinates :
A.
$$\left( {\frac{{3 + \sqrt 3 }}{2},\,\frac{{\sqrt 3 - 1}}{2}} \right)$$
B.
$$\left( {\frac{{3 - \sqrt 3 }}{2},\,\frac{{\sqrt 3 + 1}}{2}} \right)$$
C.
$$\left( {\frac{{1 - \sqrt 3 }}{2},\,\frac{{1 + \sqrt 3 }}{2}} \right)$$
D.
none of these
Answer :
$$\left( {\frac{{3 + \sqrt 3 }}{2},\,\frac{{\sqrt 3 - 1}}{2}} \right)$$
Solution :
$$\eqalign{
& {\text{Given, points are }}A = \left( {1,\,0} \right){\text{ and }}B = \left( {2,\,1} \right) \cr
& {\text{Slope of }}AB = \frac{{1 - 0}}{{2 - 1}} = 1 \cr
& {\text{Then angle of }}AB{\text{ with }}x{\text{ - axis is}} \cr
& \angle BAX = {45^ \circ } \cr
& {\text{Hence, }}\angle B'AX = {45^ \circ } - {30^ \circ } = {15^ \circ } \cr
& {\text{Therefore for }}B'\left( {h,\,k} \right) \cr
& h = 1 + \sqrt 2 \,\cos \,{15^ \circ },\,k = \sqrt 2 \,\sin \,{15^ \circ } \cr
& {\text{We have, }}\sin \left( {{{15}^ \circ }} \right) = \frac{{\sqrt 6 - \sqrt 2 }}{4} \cr
& \Rightarrow \cos \left( {{{15}^ \circ }} \right) = \frac{{\sqrt 6 + \sqrt 2 }}{4} \cr
& \Rightarrow h = \frac{{3 + \sqrt 3 }}{2},\,k = \frac{{\sqrt 3 - 1}}{2} \cr} $$