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.
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"
Comments
Leave a comment