The side BC of △ABC is produced to a point D. If AC = BC and ∠BAC = 70°, then find the value of 2.5∠ACD – 1.5∠ABC.
245°
Given: in △ABC, \(AC = BC\) and \(\angle BAC = 70^\circ\). Side \(BC\) is produced to \(D\).
Step 1 — Identify equal angles. In a triangle, angles opposite equal sides are equal. Side \(AC\) is opposite \(\angle B\) and side \(BC\) is opposite \(\angle A\). Since \(AC = BC\), we get \(\angle ABC = \angle BAC = 70^\circ\).
Step 2 — Find ∠ACB. \(\angle ACB = 180^\circ - 70^\circ - 70^\circ = 40^\circ\).
Step 3 — Exterior angle ∠ACD. Since \(BCD\) is a straight line, \(\angle ACD = 180^\circ - \angle ACB = 180^\circ - 40^\circ = 140^\circ\).
Step 4 — Evaluate. \(2.5\angle ACD - 1.5\angle ABC = 2.5(140^\circ) - 1.5(70^\circ) = 350^\circ - 105^\circ = 245^\circ\).
Hence the required value is 245°.
