Question

10 from unit 2, lesson 6 triangles acd and bcd are isosceles. angle dbc has a measure of 110° and angle bda has a measure of 22°. find the measure of angle bac

Explanation:

Step1: Find ∠BDC in △BCD

△BCD is isosceles with $\overline{BD} \cong \overline{BC}$, so base angles $\angle BDC = \angle BCD$.
Sum of angles in a triangle is $180^\circ$, so:
$\angle BDC = \frac{180^\circ - \angle DBC}{2} = \frac{180^\circ - 110^\circ}{2} = 35^\circ$

Step2: Calculate ∠ADC

$\angle ADC = \angle BDA + \angle BDC$
$\angle ADC = 22^\circ + 35^\circ = 57^\circ$

Step3: Find ∠ACD in △ACD

△ACD is isosceles with $\overline{AD} \cong \overline{AC}$, so $\angle ACD = \angle ADC = 57^\circ$

Step4: Calculate ∠DAC in △ACD

$\angle DAC = 180^\circ - 2\times\angle ADC$
$\angle DAC = 180^\circ - 2\times57^\circ = 66^\circ$

Step5: Find ∠BAD in △ABD

In △ABD, $\overline{AD} = \overline{AD}$ (common side), $\overline{BD} \cong \overline{BC}$, $\overline{AD} \cong \overline{AC}$, so $\triangle ABD \cong \triangle ABC$ (SSS). Thus $\angle BAD = \angle BAC$, and $\angle DAC = \angle BAD + \angle BAC = 2\angle BAC$

Step6: Solve for ∠BAC

$\angle BAC = \frac{\angle DAC}{2} = \frac{66^\circ}{2} = 33^\circ$

Answer:

$33^\circ$