Find the two other sides in a 15-30-135 triangle A triangle has angle measures of 15, 30, and 135 degrees. The side opposite the 15 angle is x feet, the side opposite the 30 angle is y feet, and the side opposite the 135 angle is 2 feet.
Find x and y without the law of sine.
 A: Apply sine rule in given triangle as follows 
$$\frac{x}{\sin 15^\circ}=\frac{y}{\sin 30^\circ}=\frac{2}{\sin 135^\circ}$$
considering first & third, 
$$\frac{x}{\sin 15^\circ}=\frac{2}{\sin 135^\circ}$$
$$x=\frac{2\sin 15^\circ}{\sin 135^\circ}=\frac{2\sin (45^\circ-30^\circ)}{\sin 45^\circ}=\frac{2\left(\frac{1}{\sqrt 2}\frac{\sqrt 3}{2}-\frac{1}{\sqrt 2}\frac{1}{2}\right)}{\frac1{\sqrt 2}}=\color{red}{\sqrt 3-1}$$
similarly, considering second & third, 
$$\frac{y}{\sin 30^\circ}=\frac{2}{\sin 135^\circ}$$
 $$y=\frac{2\sin 30^\circ}{\sin 135^\circ}=\frac{2\cdot \frac{1}{2}}{\frac{1}{\sqrt 2}}=\color{red}{\sqrt 2}$$
A: Draw a 30-60-90 triangle with hypotenuse =2, height = 1, and base = $\sqrt 3$.  
Inside it construct a 90-45-45 triangle so that the 90-45-45- and 90-60-30 share the altitude equal =1.  The base is 1.  And the hypotenuse is $\sqrt 2$.  The line on the opposite side of the this base, extending to the base of the original 30-60-90 triangle will be of length $\sqrt 3 - 1$.
Now remove the 90-45-45 from the 90=60-30 and you are left with a 15-30-135 triangle.  This is your triangle.  
The side opposite the 135 is the original hypontenuse = 2.  The side opposite the 30 is the hypotenuse of the 90-45-45 and is equal to $\sqrt 2$.  The remaining side is $\sqrt 3 - 1$, the base of the 30-60-90 minus the base of the 90-45-45..
A: Hint:  Draw the triangle in a coordinate plane.  Call the triangle $ABC$ with angle $A$ equal to 135 degrees and at the origin.  Let $B$ be the 30 degree angle and position it along the positive $x$-axis.  Let $C$ be the 15 degree angle.  $C$ will be in the Second Quadrant because of the 135 degree angle.  The side $BC$ is 2 feet.  Drop the perpendicular from vertex $C$ down to the $x$-axis.  This forms a 30-60-90 right triangle $BCD$ with angle $D$ the 90 degree angle.  From there, you can use this triangle to help you figure out the sides.
