Question

In a trapezium ABCD, AB || DC, AB = AD, ∠ADC = 64° and ∠BCD = 54°. Find ∠DBC.

(a) 64° 

(b) 72° 

(c) 94° 

(d) 116°

Answer 1

(c) 94°

∠DAB = 180° –∠ADC= 180°– 64° = 116° 

(AB || DC, co-int. ∠s are supp.) 

In ΔDAB, DA = AB ⇒∠ABD=∠ADB    (isos. Δ prop.)

∴ ∠ADB = \(\frac{180°-116°}{2}=\frac{64°}{2}=32°\)

∴ ∠BDC = 64° – 32° = 32° 

Hence, in ΔDBC,

∠DBC= 180° – (∠BDC+∠BCD) 

= 180° – (32° + 54°) =180° – 86° = 94°.