Question

in △xyz provided, w is a point on \overline{yz} such that \overrightarrow{xw} is an angle bisector of ∠ zxy. what is m∠z ?
?
6a°
a°+20
a 80°
b 90°
c 100°
d 110°

Explanation:

Step1: Use exterior angle theorem

The exterior angle at W is $6a^\circ$, equal to $a^\circ + (a^\circ + 20^\circ)$. So $6a = a + a + 20$.

Step2: Solve for a

Simplify: $6a = 2a + 20 \implies 4a = 20 \implies a = 5$.

Step3: Find angles in △XYZ

∠YXZ = $2a = 10^\circ$, ∠XYZ = $a + 20 = 25^\circ$. Sum of angles: $180^\circ - 10^\circ - 25^\circ = 145^\circ$? Wait, no—wait exterior angle is at W: ∠XWY = $6a$, which is exterior to △XWZ, so ∠XWY = ∠WXZ + ∠XZW? No, correct exterior angle: ∠XWY (on extension) = ∠WXZ + ∠XZW? Wait original: ∠XWZ's exterior is $6a$, so ∠XWZ = $180 - 6a$. In △XWZ: $a + (180 - 6a) + ∠Z = 180 \implies -5a + ∠Z = 0 \implies ∠Z = 5a$. In △XYZ: $2a + (a + 20) + 5a = 180 \implies 8a + 20 = 180 \implies 8a = 160 \implies a = 20$. Then ∠Z = $5*20 = 100^\circ$.

Answer:

C. 100°