Question

ABC is a right triangle right angled at A. D lies on BA produced and DE perpendicular to BC intersecting AC at F. If $\angle AFE={130}^{\circ }$, find
(i) $\angle BDE$ (ii) $\angle BCA$ (iii) $\angle ABC$

Answer 1

(i) Here,
$\angle BAF+\angle FAD={180}^{\circ }\left(Linearpair\right)$
$\angle FAD={180}^{\circ }-\angle BAF={180}^{\circ }-{90}^{\circ }={90}^{\circ }$
Also from the figure,
$\angle AFE=\angle ADF+\angle FAD\left(Exteriorangleproperty\right)$
$\angle ADF+{90}^{\circ }={130}^{\circ }$
$\angle ADF={130}^{\circ }-{90}^{\circ }={40}^{\circ }$
$\angle BDE={40}^{\circ }$
(ii) We know that the sum of all the angles of a triangle is 180o.
Therefore, for $\Delta BDE,$ we have
$\angle BDE+\angle BED+\angle DBE={180}^{\circ }$
$\angle DBE={180}^{\circ }-\angle BDE-\angle BED$
$\angle DBE={180}^{\circ }-{40}^{\circ }-{90}^{\circ }={50}^{\circ }....Equation\left(i\right)$
Again from the figure we have,
$\angle FAD=\angle ABC+\angle ACB\left(Exteriorangleproperty\right)$
${90}^{\circ }={50}^{\circ }+\angle ACB$
$\angle ACB={90}^{\circ }-{50}^{\circ }={40}^{\circ }$
(iii) From equation we have
$\angle ACB=\angle DBE={50}^{\circ }$