Question

pls ans second question

Answer 1

Hope this helps.......

Answer 2

Given,

∠ADC = 85°

∠CAD = 30°

∠ABD = 60°

We know  that  sum interior angles of  a Δ is 180°

IN ΔADC

∠ADC + ∠CAD + ∠DCA = 180°

85° + 30° + y = 180°

115° + y = 180°

y = 180° - 115°

y = 65°

At point D,

∠ADC + ∠ADB = 180°

85° + ∠ADB = 180°

∠ADB = 180° - 85°

x = 95

IN ΔABD,

∠ABD  + ∠ADB + ∠BAD = 180°

60° +  95° + z = 180°

z + 155° = 180°

z = 180° - 155°

z = 25°