Question

In the figure ∠BAD = 3 ∠DBA, find ∠CDB. ∠DBC and ∠ABC.

Answer 1

In ΔBCD, ∠DBC + ∠BCD = ∠BDA (exterior angle property)

 ∠DBC + 65° -104°

 ∴ ∠DBC = 104° – 65° = 39°

 Also ∠BDA + ∠BDC = 180° (linear pair of angles) 

104° + ∠BDC = 180° 

∠CDB or ∠BDC = 180° – 104° = 76° 

Now in ΔABD, 

∠BAD + ∠ADB + ∠DHA = 180°

3∠DBA + ∠DBA + 104° = 180° (given) 

4∠DBA + 104° – 180° 

∴ 4∠DBA = 180° – 104°= 76° 

∴∠DBA =  76°/4 = 19° 

Now ∠ABC = ∠DBA + ∠DBC = 19° + 39° = 58°