Question

If $ABCDE$ is a regular pentagon, then the angle $BDE$ is equal to

• A. ${90}^o$
• B. ${72}^o$
• C. ${60}^o$
• D. ${54}^o$

Answer 1

The correct option is B ${72}^o$

Number of sides $=5$
Each interior $\angle s=\left[\frac{(2n-4)\times 90}{n}\right]=\frac{(10-4)\times 90}{5}={108}^o$
In $△ABE,AB=AE,\angle 1=\angle 2={36}^o$
Similarly $\angle CDB={36}^o,\angle BDE=108-36={72}^o$