Answer: Now, In △DBC, DB=DC (Given) hence, ∠DBC=∠DCB=36 ∘ (Isosceles triangle property) Sum of angles of triangle = 180 ∠DBC+∠DCB+∠CDB=180 36+36+∠CDB=180 ∠CDB=108 Now, ∠ADC=180−∠ADB ∠ADC=180−108 ∠ADC=72 ∘ in △ADC, ∠ADC=∠DAC=72 ∘ (Isosceles triangle property, AC= CD is given) Sum of angles = 180 ∠ADC+∠CAD+∠ACD=180 72+72+x=180 x=180−144 x=36 ∘ Reply
Answer:
Now, In △DBC,
DB=DC (Given)
hence, ∠DBC=∠DCB=36
∘
(Isosceles triangle property)
Sum of angles of triangle = 180
∠DBC+∠DCB+∠CDB=180
36+36+∠CDB=180
∠CDB=108
Now, ∠ADC=180−∠ADB
∠ADC=180−108
∠ADC=72
∘
in △ADC,
∠ADC=∠DAC=72
∘
(Isosceles triangle property, AC= CD is given)
Sum of angles = 180
∠ADC+∠CAD+∠ACD=180
72+72+x=180
x=180−144
x=36
∘