Question

please find this problem..

Answer 1

Hey friend, Harish here.

Here is your answer.

GIVEN :

1) ∠BAD = 3 ∠DBA

2) ∠ADB = 104°

3) ∠DCB =  65°

TO FIND :

∠CDB , ∠DBC and ∠ABC

SOLUTION :

Let ∠DBA = x.  Then ∠BAD = 3x.

Now in ΔABD , By angle sum property.

x + 3x + 104° = 180°  ⇔  4x + 104° = 180° ⇔ 4x = 76 ⇔ x = 19°

∴ ∠DBA = 19° and ∠BAD = 57°

Now in ΔABC, by angle sum property :

∠ABC + 57° + 65° = 180°

∴ ∠ABC = 58°

∠BDC = 180 - 104 = 76°  [ Because ADC is a straight line , so 180° ]

In ΔDBC , By exterior angle property

104° = 65° + ∠DBC

∴ ∠DBC = 39°

∴ $\boxed{\mathrm{ \angle ABC = 58 ^{\circ} , \angle CDB = 76^{\circ} , \angle DBC = 38^{\circ}}}$

________________________________________________

Hope my answer is helpful to you.