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.

Correct — B. AC = BC, so the angles opposite those sides are equal.

AC is opposite ∠ABC and BC is opposite ∠BAC, so ∠ABC = ∠BAC = 70°.

∠ACB = 180° − 70° − 70° = 40°

BC is produced to D, so ∠ACD and ∠ACB sit on a straight line:
∠ACD = 180° − 40° = 140°

2.5 × 140° = 350°
1.5 × 70° = 105°
350° − 105° = 245° → option (b)

Why the others are wrong

• (a)
  235° — 235° would need ∠ACD = 136° alongside ∠ABC = 70°. The exterior angle equals the two remote interior angles added, 70° + 70° = 140°, so 136° is not available.
• (c)
  225° — 225° would need ∠ABC = 83⅓° with ∠ACD = 140°. AC = BC forces the base angle to be exactly 70°.
• (d)
  230° — 230° is 350° − 1.5 × 80°, so it comes from an 80° base angle. With ∠BAC = 70° and AC = BC, ∠ABC is 70°.

Two standard results do all the work, and the trap is in matching sides to angles.

Equal sides face equal angles. In ∆ABC the side AC is opposite ∠ABC and BC is opposite ∠BAC, so AC = BC makes those two angles equal — not the angle at C.

A produced side makes an exterior angle. ∠ACD is on a straight line with ∠ACB, so the two add to 180°. The same figure gives the exterior angle theorem: ∠ACD = ∠BAC + ∠ABC.

After that the question is arithmetic, and the 2.5 and 1.5 multipliers exist only to punish a candidate who stops at 140°.

Key facts

• In ∆ABC the side opposite ∠ABC is AC, so AC = BC gives ∠ABC = ∠BAC.
• With ∠BAC = ∠ABC = 70°, the third angle ∠ACB = 40°.
• The exterior angle ∠ACD = ∠BAC + ∠ABC = 140°, which is also 180° − 40°.
• 2.5 × 140° − 1.5 × 70° = 350° − 105° = 245°.

Common traps

• Pairing AC = BC with ∠ACB, which is opposite AB.
• Reaching ∠ACD = 140° and then forgetting the 2.5 and 1.5 multipliers.
• Placing D between B and C, which would make the angle interior rather than exterior.

No directly related past PYQ was found.