# Question #aff5e

Mar 18, 2016

$\sin \left(\angle A B C\right) = \frac{6 \sin \left({65}^{o}\right)}{\sqrt{52 - 48 \cos \left({65}^{o}\right)}}$
$\angle B C D = {180}^{o} - \angle A B C$
$A B = 22$
$B C = 2 \cdot \sqrt{52 - 48 \cos \left({65}^{o}\right)}$

#### Explanation:

$D C | | X Y | | A B$
$\angle D X Y = {65}^{o}$
$D X = X A = 6$
$D C = 14$
$X Y = 18$

Draw a line through point $D$ parallel to line $B C$. It intersects $X Y$ at point $Z$ and it intersects $A B$ at point $E$.

Quadrilateral $C D Z Y$ parallelogram since opposite sides are parallel. Therefore, $Y Z = C D = 14$ and $X Z = X Y - Y Z = 18 - 14 = 4$.

Consider $\Delta D X Z$. We know its two sides $D X = 6$ and $X Z = 4$ and an angle between them $\angle D X Z = {65}^{o}$. Therefore, we can determine any element of this triangle.

Side $D Z$ can be determined using the Law of Cosines:
$D {Z}^{2} = D {X}^{2} + X {Z}^{2} - 2 \cdot D X \cdot X Z \cdot \cos \left(\angle D X Z\right)$
$D {Z}^{2} = {6}^{2} + {4}^{2} - 2 \cdot 6 \cdot 4 \cdot \cos \left({65}^{o}\right) = 52 - 48 \cos \left({65}^{o}\right)$
$D Z = \sqrt{52 - 48 \cos \left({65}^{o}\right)}$

Angle $\angle D Z X$ can be determined using the Law of Sines:
$\frac{D Z}{\sin} \left(\angle D X Z\right) = \frac{D X}{\sin} \left(\angle D Z X\right)$

or, since we know all elements of this triangle,
$\frac{\sqrt{52 - 48 \cos \left({65}^{o}\right)}}{\sin} \left({65}^{o}\right) = \frac{6}{\sin} \left(\angle D Z X\right)$

and, therefore,
$\sin \left(\angle D Z X\right) = \frac{6 \sin \left({65}^{o}\right)}{\sqrt{52 - 48 \cos \left({65}^{o}\right)}}$

Since $D Z | | B C$, $\angle D Z X = \angle C Y X = \angle A B C$, so
$\sin \left(\angle A B C\right) = \frac{6 \sin \left({65}^{o}\right)}{\sqrt{52 - 48 \cos \left({65}^{o}\right)}}$

Angle $\angle B C D$ is, obviously, ${180}^{o} - \angle A B C$

Triangles $\Delta D X Z$ and $\Delta D A E$ are similar since $X Z | | A E$. Also note that the factor of scaling between them is $2$ since $A D = 2 \cdot X D$. Therefore, $A E = 2 \cdot X Z = 2 \cdot 4 = 8$, which makes
$A B = A E + E B = 8 + 14 = 22$

Since $D Z = C Y = \frac{B C}{2}$,
$B C = 2 \cdot D Z = 2 \cdot \sqrt{52 - 48 \cos \left({65}^{o}\right)}$

I do recommend you to check all my calculations.