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.

Question

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.

1. A. 235^
2. B. 245^ (Correct answer)
3. C. 225^
4. D. 230^

Correct Answer

Option B — 245^

Detailed Solution & Explanation

The correct answer is $245^\circ$.

Key Points

• Since $AC=BC$, the triangle is isosceles and $\angle ABC=\angle BAC=70^\circ$.
• Exterior angle $\angle ACD=\angle BAC+\angle ABC=70^\circ+70^\circ=140^\circ$.
• $2.5(140^\circ)-1.5(70^\circ)=350^\circ-105^\circ=245^\circ$.

Additional Information

• Two standard results combine here: the base angles of an isosceles triangle are equal, and an exterior angle equals the sum of the two opposite interior angles.
• The exterior-angle result follows from the angle sum: ∠ACD = 180° − ∠ACB = ∠BAC + ∠ABC.
• Note which sides are equal — AC = BC makes the angles opposite them, ∠ABC and ∠BAC, equal. Misreading which pair is equal is the usual error.
• Compute the angles fully before substituting into the final expression, so the arithmetic is done once.