question_answer

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

\[\angle EAB+\angle ABE=\angle BES\]

\[\Rightarrow \]   \[c{}^\circ +b{}^\circ +a{}^\circ =x{}^\circ \]

\[\therefore \]      \[x{}^\circ =a{}^\circ +b{}^\circ +c{}^\circ \]