question_answer

In \[\Delta \mathbf{ABC},\angle \mathbf{B}=\mathbf{6}{{\mathbf{0}}^{{}^\circ }}\], and \[\angle \mathbf{C}=\mathbf{5}{{\mathbf{0}}^{{}^\circ }}\], AD and AE are respectively the bisector of \[\angle \mathbf{A}\] and perpendicular on BC. The measure of \[\angle \mathbf{EAD}\] is:

A)  \[{{11}^{{}^\circ }}\]                        

B)  \[{{5}^{{}^\circ }}\] 

C)  \[{{12}^{{}^\circ }}\]                        

D)  \[{{9}^{{}^\circ }}\]

Correct Answer: B

Solution :

(b): \[\angle B+C={{60}^{{}^\circ }}+{{50}^{{}^\circ }}={{110}^{{}^\circ }}\]  \[\therefore \]\[\angle A={{180}^{{}^\circ }}-{{110}^{{}^\circ }}={{70}^{{}^\circ }}\] \[\therefore \]\[\angle BAD={{35}^{{}^\circ }}\] In \[\Delta ABE.\] \[\angle AEB={{90}^{{}^\circ }}\] \[\therefore \] \[\angle BAE-{{180}^{{}^\circ }}-{{90}^{{}^\circ }}-{{60}^{{}^\circ }}-{{30}^{{}^\circ }}\] \[\therefore \]\[\angle EAD=\angle BAD-\angle BAE\] \[={{35}^{{}^\circ }}-{{30}^{{}^\circ }}={{5}^{{}^\circ }}\]