A) \[63{}^\circ \]
B) \[35{}^\circ \]
C) \[27{}^\circ \]
D) \[54{}^\circ \]
Correct Answer: D
Solution :
1 Grade \[=\frac{1}{100}\] of right angle \[=\left( \frac{90}{100} \right)o\] \[={{0.9}^{o}}\] degree Let the two angles be \[\alpha \] and \[\beta \] \[\therefore \] \[\alpha +\beta =\frac{890}{9}\] grade \[=\frac{890}{9}\times \frac{9}{10}={{89}^{o}}\] and \[\alpha -\beta ={{19}^{o}}\] \[\therefore \] \[2\alpha =89+19={{108}^{o}}\] \[\therefore \] Greater angle \[\alpha =\frac{1}{2}({{108}^{o}})={{54}^{o}}\]
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