Question

If in a right angle triangle the hypotenuse is four times as long as the perpendicular Drawn to Opposite vertex, then one of its acute angle is

Answer 1

Ref Above figure

$c^2=h^2+(a-x{)}^2$

$b^2=h^2+x^2$

on adding, we have $2h^2+x^2+(a-x{)}^2=a^2$ ..(1)

also $a=4h$ and $cosB=b/a=x/b$

so $x=b^2/a$

on solving (1), we have $h^2+x^2-4hx=0$

substituting above gives

$b^4+a^4/16-a^2.b^2=0..\left(2\right)$

as $cosB=b/a$

on solving eqn. (2) we have , $Co{s}^{2}B=0.933$ or $0.067$

$CosB=0.966$ or $0.259$

$B=15$ or $75$ degrees