Consider the following triangle (not in scale):
We know that:
[tex]\begin{gathered} tan(A)=\frac{3}{4}tan(B)=\frac{x}{y} \\ tan(A)=\frac{sin(A)}{cos(A)} \\ tan(B)=\frac{sin(B)}{cos(B)} \\ \frac{s\imaginaryI n(A)}{cos(A)}=\frac{3}{4}\cdot\frac{s\imaginaryI n(B)}{cos(B)} \\ sin(A)=\frac{3}{4}\cdot\frac{sina(B)\cdot cos(A)}{cos(B)}=\frac{3}{4}\cdot\frac{\frac{y}{h}\cdot\frac{y}{h}}{\frac{x}{h}}=\frac{3}{4}\cdot\frac{y^2}{h\cdot x} \end{gathered}[/tex]
Applying the same steps for sin(B), cos (A) and cos(B), we have:
[tex]\begin{gathered} sin(B)=\frac{4}{3}\cdot\frac{x^2}{h\cdot y} \\ cos(A)=sin(B)=\frac{4}{3}\cdot\frac{x^2}{h\cdot y} \\ cos(B)=sin(A)=\frac{3}{4}\cdot\frac{y^2}{h\cdot x} \end{gathered}[/tex]
