question_answer

In figure, ABCD is \[\parallel \mathbf{gm}\] with \[\angle \mathbf{A}=\mathbf{8}{{\mathbf{0}}^{{}^\circ }},\]the internal bisectors of \[\angle \mathbf{B}\] and \[\angle \mathbf{C}\] meet at 0. Find the measure of three angles of ABCO

A)  \[{{30}^{{}^\circ }},{{60}^{{}^\circ }},{{90}^{{}^\circ }}\]                    

B)  \[{{40}^{{}^\circ }},{{50}^{{}^\circ }},{{90}^{{}^\circ }}\]

C)  \[{{60}^{{}^\circ }},{{60}^{{}^\circ }},{{60}^{{}^\circ }}\]                    

D)  \[{{70}^{{}^\circ }},{{50}^{{}^\circ }},{{60}^{{}^\circ }}\]

Correct Answer: B

Solution :

(b): \[\angle B={{100}^{{}^\circ }}\] (by property of parallelogram) \[\angle C={{80}^{{}^\circ }}\] (by internal bisector property) \[\angle OBC={{50}^{{}^\circ }}\] \[\angle OCB={{40}^{{}^\circ }}\] \[\angle BOC={{180}^{{}^\circ }}-{{50}^{{}^\circ }}-{{40}^{{}^\circ }}={{90}^{{}^\circ }}\]