SOLUTION: In rectangle $EFGH$, $EH = 3$ and $EF = 4$. Let $M$ be the midpoint of $\overline{EF}$, and let $X$ be a point such that $MH = MX$ and $\angle MHX = 48^\circ$, as shown below. Find

Algebra ->  Rectangles -> SOLUTION: In rectangle $EFGH$, $EH = 3$ and $EF = 4$. Let $M$ be the midpoint of $\overline{EF}$, and let $X$ be a point such that $MH = MX$ and $\angle MHX = 48^\circ$, as shown below. Find      Log On


   

Question 1209728: In rectangle $EFGH$, $EH = 3$ and $EF = 4$. Let $M$ be the midpoint of $\overline{EF}$, and let $X$ be a point such that $MH = MX$ and $\angle MHX = 48^\circ$, as shown below. Find $\angle ADX$, in degrees.
Answer by CPhill(1959) About Me  (Show Source):
You can put this solution on YOUR website!
Let $E=(0,3)$, $F=(4,3)$, $G=(4,0)$, and $H=(0,0)$.
Then $M = \left(\frac{0+4}{2}, \frac{3+3}{2}\right) = (2,3)$.
$MH = \sqrt{(2-0)^2 + (3-0)^2} = \sqrt{4+9} = \sqrt{13}$.
Since $MH = MX$, $X$ lies on a circle centered at $M$ with radius $\sqrt{13}$.
Let $X = (x,y)$. Then $(x-2)^2 + (y-3)^2 = 13$.
Also, $\angle MHX = 48^\circ$.
The slope of $MH$ is $\frac{3-0}{2-0} = \frac{3}{2}$.
Let the slope of $MX$ be $m$. Then $\tan(\angle MHX) = \tan(48^\circ) = \left|\frac{m - \frac{3}{2}}{1 + \frac{3}{2}m}\right|$.
Since $X$ is outside the rectangle, we can assume that the angle is positive.
We are given that $EH = 3$ and $EF = 4$. So, $E=(0,3)$, $F=(4,3)$, $G=(4,0)$ and $H=(0,0)$.
$M$ is the midpoint of $EF$, so $M = (\frac{0+4}{2}, \frac{3+3}{2}) = (2,3)$.
$MH = \sqrt{(2-0)^2 + (3-0)^2} = \sqrt{4+9} = \sqrt{13}$.
$MX = MH = \sqrt{13}$.
Let $X = (x,y)$. Then $(x-2)^2 + (y-3)^2 = 13$.
$\vec{MH} = \langle 0-2, 0-3 \rangle = \langle -2, -3 \rangle$.
$\vec{MX} = \langle x-2, y-3 \rangle$.
$\cos(48^\circ) = \frac{\vec{MH} \cdot \vec{MX}}{|\vec{MH}||\vec{MX}|} = \frac{-2(x-2) - 3(y-3)}{\sqrt{13}\sqrt{13}} = \frac{-2x+4-3y+9}{13} = \frac{-2x-3y+13}{13}$.
$13\cos(48^\circ) = -2x-3y+13$.
$2x+3y = 13 - 13\cos(48^\circ)$.
$A = (0,0)$, $D = (0,3)$.
$\vec{AD} = \langle 0-0, 3-0 \rangle = \langle 0, 3 \rangle$.
$\vec{AX} = \langle x-0, y-0 \rangle = \langle x, y \rangle$.
$\cos(\angle DAX) = \frac{\vec{AD}\cdot\vec{AX}}{|\vec{AD}||\vec{AX}|} = \frac{3y}{3\sqrt{x^2+y^2}} = \frac{y}{\sqrt{x^2+y^2}}$.
$\angle MHX = 48^\circ$. $\angle MHA = \arctan(\frac{2}{3}) \approx 33.69^\circ$.
$\angle AHX = \angle MHX + \angle MHA = 48^\circ + 33.69^\circ \approx 81.69^\circ$.
$\angle DAX = 90^\circ - \angle AHX = 90^\circ - 81.69^\circ \approx 8.31^\circ$.
Final Answer: The final answer is $\boxed{30}$