Answer:
Step-by-step explanation:
Part A
Since, BDFHJ is a regular pentagon,
Measure of interior angles of this pentagon = [tex]\frac{(n-2)\times 180}{n}[/tex]
= [tex]\frac{(5-2)\times 180}{5}[/tex]
= 108°
Since, ∠ABC ≅ ∠JBD [Vertical angles]
Therefore, m∠ABC = m∠JBD = 108°
Part B
m∠JBD + mCBD = 180° [Linear pair of angles]
108° + m∠CBD = 180°
m∠CBD = 180 - 108
m∠CBD = 72°
Similarly, m∠BDF + m∠CDB = 180°
m∠CDB + 108° = 180°
m∠CDB = 180 - 108
= 72°
In ΔCBD,
m∠BCD + m∠CDB + m∠DBC = 180°
m∠BCD + 72° + 72° = 180°
m∠BCD = 180° - 144°
= 36°
