A) \[\angle Q+\angle R\]
B) \[{{90}^{o}}+\frac{1}{2}\angle Q\]
C) \[{{90}^{o}}-\frac{1}{2}\angle R\]
D) \[\frac{1}{2}(\angle Q-\angle R)\]
Correct Answer: D
Solution :
\[\angle 1+\angle 2=\angle 3\] \[\angle Q={{90}^{o}}-\angle 1\] \[\angle R={{90}^{o}}-\angle 2-\angle 3\] So, \[\angle Q-\angle R=({{90}^{o}}-\angle 1)\] \[-({{90}^{o}}-\angle 2-\angle 3)\] \[\Rightarrow \] \[\angle Q-\angle R=({{90}^{o}}-\angle 1)\] From (i) \[=\angle 2+(\angle 1+\angle 2)-\angle 1\] \[\Rightarrow \] \[\angle Q-\angle R=2\angle 2\] \[\Rightarrow \] \[\frac{1}{2}(\angle Q-\angle R)=\angle TPS\]You need to login to perform this action.
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