Question 46
In $$\triangle ABC, BE \perp AC, CDÂ \perp AB$$ and $$BE$$ and $$CD$$ intersect each other at O. The bisectors of $$\angle OBC$$ and $$\angle OCB$$ meet at P. If $$\angle BPC = 148^\circ$$, then what is the measure of $$\angle A$$?
Solution
$$\angle BPC = 148^\circ$$
In triangle BOC-
$$\angle OBC + \angle BCO +\angle BOC = 180$$
$$\angle BOC + 2(\angle PBC + \angle PCB) = 180$$
$$\angle BOC + 2(180 - 148) = 180$$
$$\angle BOC = 180 - 64 = 116$$
Create a FREE account and get:
- Free SSC Study Material - 18000 Questions
- 230+ SSC previous papers with solutions PDF
- 100+ SSC Online Tests for Free
SSC Quant Questions | SSC Quantitative Ability
SSC Data Interpretation QuestionsSSC Geometry QuestionsSSC Linear Equations QuestionsSSC Number Systems QuestionsSSC Quadratic Equations QuestionsSSC Trigonometry QuestionsSSC Ages QuestionsSSC Arithmetic Questions
SSC DILR Questions | LRDI Questions For SSC
SSC Analogy QuestionsSSC Coding Decoding QuestionsSSC Directions QuestionsSSC Meaningful Order QuestionsSSC Number Series QuestionsSSC Puzzles QuestionsSSC Syllogisms QuestionsSSC Words Formation QuestionsSSC Venn Diagram Questions
