In the given figure, AB || CD. If ∠BAC = (3x + 15)° and ∠ACD = (2x + 45)°, find the value of x.

Also, find the measures of ∠BAC and ∠ACD.

Given:

AB || CD

∠BAC = (3x + 15)°

∠ACD = (2x + 45)°

∠BAC and ∠ACD lie inside the parallel lines on the same side of transversal AC. So, they are co-interior angles.

Co-interior angles are supplementary:

∴ (3x + 15)° + (2x + 45)° = 180°

⇒ 3x° + 2x° + 15° + 45° = 180°

⇒ 5x° + 60° = 180°

⇒ 5x° = 180° - 60°

⇒ 5x° = 120°

⇒ x° = $\dfrac{120^{\circ}}{5}$

⇒ x° = 24°

Let's find each angle by substituting the value of x:

∠BAC = (3x + 15)° = (3(24) + 15)° = (72 + 15)° = 87°

∠ACD = (2x + 45)° = (2(24) + 45)° = (48 + 45)° = 93°

x° = 24°, ∠BAC = 87° and ∠ACD = 93°