Answer to Question #224503 in Geometry for S.S

Question

The drawing below shows a right-angled triangle. A straight line crosses the triangle parallel to the line z and encloses an angle of α. The lengths x and y of the bottom and top line segments as well as the angle α are given. Find an equation for the length z.

Answer 1

lets make the points ABCDE as shown below

$DE||BC (Given)$

$\angle ACB=90\degree\space given\\\therefore\angle AED=\angle ACB=90\degree(as\space DE||BC)\\\angle ABC=\angle ADE=\alpha(corresponding \space angles\space as\space DE||BC)\\\therefore in\triangle ADE;\angle AED=90\degree\\\therefore DE=ADcos\alpha=xcosx\\DE=xcos\alpha\\AE=ADsin\alpha=xsin\alpha\\AE=nsin\alpha\\now \space for \triangle ADE\space and \space \triangle ABC\\\angle A \space is\space common\\\angle AED=\angle ACB=90\degree\\\therefore\triangle ADE\backsim\triangle ABC\space (by AA \space rule)\\\therefore \frac{AE}{AC}=\frac{DE}{BC}\implies\frac{xsin\alpha}{xsin\alpha+y}=\frac{xcos\alpha}{z}\\\implies z=\frac{coa\alpha}{sin\alpha}(yxsin\alpha)\\\therefore z=cot\alpha(y+xsin\alpha)\\z=ycot\alpha+xcos\alpha$