The angles \[x{}^\circ ,\]\[a{}^\circ ,\]\[c{}^\circ \]and \[(\pi -b{}^\circ )\]are indicated in the figure given below. Which one of the following is correct? |
A) \[x{}^\circ =a{}^\circ +c{}^\circ -b{}^\circ \]
B) \[x{}^\circ =b{}^\circ -a{}^\circ -c{}^\circ \]
C) \[x{}^\circ =a{}^\circ +b{}^\circ +c{}^\circ \]
D) \[x{}^\circ =a{}^\circ -b{}^\circ +c{}^\circ \]
Correct Answer: C
Solution :
\[\because \]\[\angle PCT+\angle PCB=\pi \] [linear pair] |
\[\angle PCB=\pi -(\pi -b{}^\circ )=b{}^\circ \] ... (i) |
In \[\Delta BPC,\] |
\[\angle PCB+\angle BPC+\angle PBC=\pi \] |
\[\Rightarrow \]\[\angle PBC=\pi -\angle PCB+\angle BPC\] |
\[=\pi -b{}^\circ -a{}^\circ \] (ii) |
\[\because \]\[\angle ABE+\angle EBC=\pi \] \[[\because \angle PBC=\angle EBC]\] |
[linear pair] |
\[\Rightarrow \]\[\angle ABE=\pi -\angle PBC=\pi -(\pi -b{}^\circ -a{}^\circ )\] |
\[=a{}^\circ +b{}^\circ \] (iii) |
Now, in \[\Delta ABE,\] |
Sum of two interior angles = Exterior angle |
\[\angle EAB+\angle ABE=\angle BES\] |
\[\Rightarrow \] \[c{}^\circ +b{}^\circ +a{}^\circ =x{}^\circ \] |
\[\therefore \] \[x{}^\circ =a{}^\circ +b{}^\circ +c{}^\circ \] |
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