## Trigonometry (11th Edition) Clone

$h=k(\tan B-\tan A)$
Name $x$ = the leg opposite to the angle $A$. Using TOA in SOHCAHTOA to find x, in the triangle where x is opposite to A: $\displaystyle \tan A=\frac{x}{k}$ $x=k\tan A$ in the triangle where (x+h) is opposite to B: $\displaystyle \tan B=\frac{h+x}{k}$ $k\tan B=h+x$ $x=k\tan B-h$ Equate the two expressions for x and solve for h: $k\tan A=k\tan B-h$ $h=k\tan B-k\tan A$ $h=k(\tan B-\tan A)$