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) \[\angle OBC={{50}^{{}^\circ }}\] \[\angle OCB={{40}^{{}^\circ }}\] \[\angle BOC={{180}^{{}^\circ }}-{{50}^{{}^\circ }}-{{40}^{{}^\circ }}={{90}^{{}^\circ }}\]You need to login to perform this action.
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