Question

what is the measure of ∠wzy?
w
x
(23x - 4)°
(8x - 1)°
z
27°
y
© gina wilson (all things algebra), 2014

Explanation:

Step1: Identify isosceles triangle property

In $\triangle WYZ$, $WZ=WY$, so $\angle WZY = \angle WYZ$. Let $\angle WZY = y$, then $\angle WYZ = y$.

Step2: Calculate $\angle ZWY$

$\angle ZWY = 180^\circ - 2y$ (sum of angles in triangle)

Step3: Use parallel line angle rule

$WX \parallel ZY$, so $\angle XWY = \angle ZYW = y$ (alternate interior angles). Also, $\angle XWY = (23x - 4)^\circ$, so $y = 23x - 4$.

Step4: Relate angles at point W

$\angle XWZ = \angle XWY + \angle ZWY$
$\angle XWZ$ is a straight line? No: $\angle XWZ = (23x-4)^\circ + (180-2y)^\circ$, and $\angle XWZ$ is also equal to $(23x-4)^\circ + (8x-1)^\circ$. So:
$$(23x-4) + (8x-1) = (23x-4) + (180 - 2y)$$
Simplify: $8x - 1 = 180 - 2y$

Step5: Substitute $y=23x-4$

$$8x - 1 = 180 - 2(23x - 4)$$
$$8x - 1 = 180 - 46x + 8$$
$$8x + 46x = 188 + 1$$
$$54x = 189$$
$$x = \frac{189}{54} = 3.5$$

Step6: Calculate $\angle WZY$

$y = 23x - 4 = 23(3.5) - 4 = 80.5 - 4 = 76.5$

Answer:

$76.5^\circ$