Question

Side $BC$ of $\Delta ABC$ has been produced to $D$ on left and to $E$ on right hand side of BC such that $\angle ABD={125}^{\circ }$ and $\angle ACE={130}^{\circ }$. Then $\angle A=$ ?
(a) ${50}^{\circ }$
(b) ${55}^{\circ }$
(c) ${65}^{\circ }$
(d) ${75}^{\circ }$

Answer 1

The correct option (d) ${75}^{\circ }$

DE is a straight line.​

$\therefore \angle ABD+\angle ABC={180}^{\circ }\left(Linearpair\right)$

$\Rightarrow {125}^{\circ }+\angle ABC={180}^{\circ }$

$\Rightarrow \angle ABC={55}^{\circ }$

Also,

$\therefore \angle ACE+\angle ACB={180}^{\circ }\left(LinearPair\right)$

$\Rightarrow {130}^{\circ }+\angle ACB={180}^{\circ }$

$\Rightarrow \angle ACB={50}^{\circ }$

Sum of the angles of a triangle $={180}^{\circ }$

$\angle BAC+\angle ABC+\angle ACB={180}^{\circ }$

$\Rightarrow \angle BAC+{55}^{\circ }+{50}^{\circ }={180}^{\circ }$

$\Rightarrow \angle BAC={75}^{\circ }$

$\therefore \angle A={75}^{\circ }$