Question

In the pentagon ABCDE,. $\angle A={120}^{\circ }$, $\angle B={138}^{\circ }$ and $\angle D=\angle E$. The sides AB and DC, when produced, meet at right angle. Calculate $\angle BCD$ and $\angle E$.

• A. ${133}^{\circ }$ and ${75}^{\circ }$
• B. ${162}^{\circ }$ and ${75}^{\circ }$
• C. ${132}^{\circ }$ and ${75}^{\circ }$
• D. ${132}^{\circ }$ and ${25}^{\circ }$

Answer 1

The correct option is C ${132}^{\circ }$ and ${75}^{\circ }$

Let AB and Dc meet at P

So in angle $BCPis90AndABC$ is the extienar angle

So angle $BPC+BCP=angleABC$ (since sum of the opposite interior angle =the exterior

Then $90+AngleBCP=138orAngleBCP=48$

Now Angle $(BCD+BCP)=180$ (linear pair

Then Angle $BCD+48=180$

Or angle $BCD=180-48=132$

Now Angle $D=angleE\left(given\right)\left(letisx\right)$

Angle $(A+B+C+D+E)=540$

Then $120+138+132+x+x=540$

Or $390+2x=540$

Or $2x=540-390$

Or $x=\frac{150}{2}=75$

Hence Angle $BCD={132}^0$ and $\angle E={75}^0$