Since, ΔAOC is an isoceles triangle. ∴ ∠OAC=∠OCA=x (let) lnΔAOC x+x+118°=180° (by angle sum property of triangle) 2x=180°−118° ⇒ 2x=62° ⇒ x=31° Now, we know that tangent is always perpendicular to the radius, at point of contact, ∴ ∠OCD=90° Now, ∠ACD=∠OCD−∠OCA ∠ACD=90°−31°(∵∠OCA=x=31°) ⇒ ∠ACD=59°