Question

1. $bc$ bisects $\angle acd$

find $m\angle1 = \underline{quad}$, $m\angle2 = \underline{quad}$, $m\angle3 = \underline{quad}$, $m\angle4 = \underline{quad}$

Explanation:

Step1: Identify right angle at C

$\angle 2 + m\angle 3 = 90^\circ$

Step2: Find $\angle 1$ via triangle sum

In $\triangle ABC$, $m\angle 1 = 180^\circ - 48^\circ - 90^\circ = 42^\circ$

Step3: Find $\angle 3$ via $\triangle BCD$ sum

In $\triangle BCD$, $m\angle 3 = 180^\circ - 83^\circ - m\angle 4$. Since $BC$ bisects $\angle ACD$, $m\angle 3 = 45^\circ$

Step4: Calculate $\angle 2$

$m\angle 2 = 90^\circ - 45^\circ = 45^\circ$

Step5: Calculate $\angle 4$

$m\angle 4 = 180^\circ - 83^\circ - 45^\circ = 52^\circ$

Answer:

$m\angle 1 = 42^\circ$
$m\angle 2 = 45^\circ$
$m\angle 3 = 45^\circ$
$m\angle 4 = 52^\circ$