Question

guys help me Q.no 5......

Answer 1

GIVEN : angle ADB = 60° , AD = DC

In ∆ADC

angle C = angle DAC (angles opposite to equal sides are equal) -----(i)

now,
angle ADB + angle ADC = 180° ( linear pair axiom)
60° + angle ADC = 180°
angle ADC = 120°
let angle DAC = x° = angle C (eq.(i))

Now,
angle ADC + angle C + angle DAC° = 180° ( sum of all interior angles of a triangle is 180°)
120° + x° + x° = 180°
2x° = 60°
x° = 30°

angle C = angle DAC = 30°
•°• angle C or ACD = 30°

angle DAC = angle DAB (given that AD bisects angle A)

In ∆ADB
angle ADB + angle DAB + angle DBA = 180° ( sum of all interior angles of a triangle is 180°)
60° + 30° + angle DBA = 180°
angle DBA(or ABD) = 180° - 90° = 90°

Answer 2

hey buddy here isbur answer