The side BC of triangle ABC is produced to a point D. If AC = BC and ∠BAC = 70°, then find the value of 2.5∠ACD − 1.5∠ABC.

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Accepted Answer

Since AC = BC, triangle ABC is isosceles with angles opposite these sides equal. Thus, ∠ABC = ∠BAC = 70°.  
When BC is extended to D, ∠ACD is the exterior angle at C, so by the exterior angle theorem:  
$$
\angle ACD = \angle BAC + \angle ABC = 70^\circ + 70^\circ = 140^\circ
$$
Now, calculate the required expression:  
$$
2.5 	imes \angle ACD - 1.5 	imes \angle ABC = 2.5 	imes 140^\circ - 1.5 	imes 70^\circ = 350^\circ - 105^\circ = 245^\circ
$$
So, the answer is 245°.

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