A) \[{{45}^{o}}\]
B) \[{{60}^{o}}\]
C) \[{{75}^{o}}\]
D) \[{{30}^{o}}\]
Correct Answer: B
Solution :
Given angles \[A,\,\,\,B,\,\,\,C\] of \[\Delta ABC\] are in \[AP\] with \[d\](common difference)\[={{15}^{o}}\] ... (i) \[\Rightarrow \] \[B=A+{{15}^{o}}\] ? (ii) and \[C=A+{{30}^{o}}\] ... (iii) Also, \[A+B+C={{180}^{o}}\] \[(\because \,\,ABC\]forms\[a\Delta )\] \[\Rightarrow \] \[A+A+{{15}^{o}}+A+{{30}^{o}}={{180}^{o}}\] \[\Rightarrow \]\[3A+{{15}^{o}}={{180}^{o}}\Rightarrow \angle A={{45}^{o}}\] \[\therefore \] \[\angle ={{45}^{o}}+{{15}^{o}}={{60}^{o}}\]
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