New coordinates after clockwise rotation of triangle? The figure below represents a triangle $PQR$ with initial coordinates of the vertices as $P(1,3)$,
$Q(4,5)$ and $R(5,3.5)$. The triangle is rotated in the $X-Y$ plane about the vertex $P$ by angle $\theta$ in
clockwise direction. If $sin(\theta) = 0.6$ and $cos(\theta) = 0.8$, the new coordinates of the vertex $Q$ are



*

*$ (4.6, 2.8)$

*$ (3.2, 4.6)$

*$ (7.9, 5.5) $

*$ (5.5, 7.9) $



We calculated the distance between $P$ and $Q$ using distance formula(i.e., $d=\sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2}$ ). It came to be root $13$ and same for only option $(1)$.
 A: HINT :
If $P=(x_1,y_1)$ and
$Q=(x_2,y_2)$
the new point $= (x_3,y_3)$;
$y_3-y_1 = - sin(\theta) \times (x_2-x_1) + cos(\theta) \times (y_2-y_1)$ 
$x_3-x_1 = cos(\theta) \times (x_2-x_1) + sin(\theta) \times (y_2-y_1)$
A: That's a nice approach to the problem -- well done!
To express it in slightly fancy terms: you found an invariant of the transformation (distance between $Q$ or $Q'$ and the fixed point $P$) and then checked which of the potential values for $Q'$ satisfied the invariant. 
That's often a good technique for solving such problems, but also for solving real problems (i.e., not multiple-choice-problem on an exam or in a textbook). Sometimes it's fairly easy to prove that a result lies in some finite set (e.g., "the integers mod 2"), at which point there are only a few possible answers...and if you can find the right invariant, you can detect the right answer without being able to compute it directly. 
You can slightly simplify, in your case, by noting that $\sqrt{13} < 4$; since answers 3 and 4 have $x$-distance greater than 4 from $P$, neither can be the correct solution. 
A: Draw right triangle $TPQ$ so that $T$ has coordinates $(4,3)$ and let's say that angle $TPQ$ is $\alpha$. Then
$$\sin\alpha=\frac{2}{\sqrt{13}},\cos\alpha=\frac{3}{\sqrt{13}}.$$
Considering triangle $T'PQ'$ where angle $T'PQ'$ is $\alpha-\theta$, use the formulas for $\sin(\alpha-\theta)$ and $\cos(\alpha-\theta)$ with the values given in the question for $\theta$. This gives 
$$\sin(\alpha-\theta) =\frac{-1}{5\sqrt{13}},\quad\cos(\alpha-\theta)=\frac{18}{5\sqrt{13}}.$$
From here the coordinates of $Q'$ are $(1+18/5,3-1/5)=(4.6,2.8)$.
