Question

Answer (c)
Please answer this asap!

Answer 1

In a cyclic quadrilateral we have , sum of opposite angles is 180.
so, ∠ABC + ∠ADC =180°
⇒ 93° + ∠ADC = 180°  (∠ABC=93° GIVEN)
⇒∠ADC = 180°-93°
⇒∠ADC = 87°

∠ACE = 90°  (Tangents at any point on the circle is perpendicular to the radius of the circle  at the point of contact)
  So, ∠ACE=∠ACD+∠DCE
      ⇒ 90° = ∠ACD + 35°      (∠DCE = 35° GIVEN)
      ⇒90°- 35° = ∠ACD
      ⇒   55°   =  ∠ACD

Now, in triangle Δ ADC , we have ,
∠CAD + ∠ADC + ∠ACD = 180°        (Angle Sum Property of a triangle ).
∠CAD  + 87°  + 55°  = 180°
∠CAD    =   180° -(87°+55°)
∠CAD  = 180°- 142°
∠CAD = 38°

∴ ∠ADC = 87°
   ∠CAD = 38°
   ∠ACD = 55°