Deriving Corner Distance (C)

In "Making Tray Mats With Non-rectangular Bases", I gave the equation for determining the corner distance C. Now I'll prove it.

Definitions

Trigonometric functions are defined as the ratios between two sides of a right triangle (having a 90° angle). The sine of an angle θ is the length of the side opposite the angle divided by the hypotenuse. In Figure A1, sin(θ) = ^{Spread (S)} / _{Wall Width (W)}.

The tangent is the ratio between the opposite side and the adjacent side. In Figure A1,
tan(θ) = ^S / _{Depth (D)}

Derivation

In the upper right corner of Figure A2, we are trying to find C_A and C_B. C_A can be divided into C_A1 and C_A2, such that C_A= C_A1 + C_A2.

There are two right triangles of interest. The yellow triangle has sides of length C_B1, W_B - S_B, and C_A1. Based on the above definitions, we can see that

$$\sin \left(\phi \right)=\frac{W_B-S_B}{C_{\mathrm{A1}}}$$

and

$$\tan \left(\phi \right)=\frac{W_B-S_B}{C_{\mathrm{B1}}}\text{.}$$

From the green triangle, we can see that

$$\sin \left(\phi \right)=\frac{W_A-S_A}{C_{\mathrm{B2}}}$$

and

$$\tan \left(\phi \right)=\frac{W_A-S_A}{C_{\mathrm{A2}}}\text{.}$$

By rearranging to isolate the C's, we find that

$$C_{\mathrm{A1}}=\frac{W_B-S_B}{\sin \left(\phi \right)}$$

and

$$C_{\mathrm{A2}}=\frac{W_A-S_A}{\tan \left(\phi \right)}\text{,}$$

which means

$$C_A=\frac{W_B-S_B}{\sin \left(\phi \right)}+\frac{W_A-S_A}{\tan \left(\phi \right)}\text{.}$$

That's what I claimed in Step 6 of  Making Tray Mats With Non-rectangular Bases.