4. wxyz is a kite. angle wxy has a measure of 133° and ∠zwx has a measure of 60°. find the measure of ∠zyw.

review problems

5. triangles acd and bcd are isosceles. angle dbc has a measure of 84° and ∠bda has a measure of 24°. find the measure of ∠bac.

6. reflect right triangle abc across $\overline{ab}$. classify $\triangle cac$ according to its side lengths. explain how you know.

Question

4. wxyz is a kite. angle wxy has a measure of 133° and ∠zwx has a measure of 60°. find the measure of ∠zyw.

review problems

5. triangles acd and bcd are isosceles. angle dbc has a measure of 84° and ∠bda has a measure of 24°. find the measure of ∠bac.

6. reflect right triangle abc across $\overline{ab}$. classify $\triangle cac$ according to its side lengths. explain how you know.

Accepted Answer

### Problem 4:
# Explanation:
## Step1: Recall kite properties
A kite has two distinct pairs of adjacent sides equal, and one diagonal bisects the angles at its endpoints. Also, the diagonal connecting the vertices between the unequal sides bisects the other diagonal and bisects the angles. In kite $ WXYZ $, $ WX = WZ $ and $ XY = ZY $ (assuming the kite is labeled with $ W $ and $ Y $ as the vertices between the unequal sides). The diagonal $ WY $ bisects $ \angle ZWX $ and $ \angle ZYX $. Wait, first, let's find the angle at $ W $. Wait, $ \angle ZWX = 60^\circ $, and since $ WY $ bisects $ \angle ZWX $? Wait, no, maybe the diagonal $ WY $ bisects $ \angle ZWX $ and $ \angle ZYX $. Wait, in a kite, one diagonal bisects the two angles at its endpoints. So if $ WXYZ $ is a kite with $ WX = WZ $ and $ XY = ZY $, then diagonal $ WY $ bisects $ \angle ZWX $ and $ \angle ZYX $. Wait, but we know $ \angle WXY = 133^\circ $, and $ \angle ZWX = 60^\circ $. Wait, maybe we need to consider triangle $ WXY $ or $ WZY $. Wait, no, let's correct: in a kite, the sum of adjacent angles? Wait, no, the key is that the diagonal $ WY $ splits the kite into two isosceles triangles? Wait, no, $ WX = WZ $ and $ XY = ZY $, so triangles $ WXY $ and $ WZY $ are congruent? Wait, $ WX = WZ $, $ XY = ZY $, $ WY = WY $, so by SSS, $ \triangle WXY \cong \triangle WZY $. Therefore, $ \angle ZYW = \angle XYW $. Now, in triangle $ WXY $, we know $ \angle WXY = 133^\circ $, $ \angle ZWX = 60^\circ $? Wait, no, $ \angle ZWX $ is at $ W $, between $ ZW $ and $ WX $. Wait, maybe the angle at $ W $ is $ \angle ZWX = 60^\circ $, and angle at $ X $ is $ \angle WXY = 133^\circ $. Then, in quadrilateral $ WXYZ $, the sum of interior angles is $ 360^\circ $. So $ \angle ZWX + \angle WXY + \angle XYZ + \angle YZW = 360^\circ $. But since it's a kite, $ \angle YZW = \angle ZWX = 60^\circ $? No, wait, in a kite, one pair of opposite angles (the ones between the unequal sides) are equal? Wait, no, the correct property is that one diagonal bisects the other diagonal and bisects the angles at its endpoints. So if $ WY $ is the diagonal that bisects $ \angle ZWX $ and $ \angle ZYX $, then $ \angle ZYW = \angle XYW $, and $ \angle ZWY = \angle XWY $. Wait, maybe I made a mistake. Let's start over.

In kite $ WXYZ $, $ WX = WZ $ and $ XY = ZY $. So sides $ WX $ and $ WZ $ are equal, $ XY $ and $ ZY $ are equal. The diagonal $ WY $ is common to triangles $ WXY $ and $ WZY $, so $ \triangle WXY \cong \triangle WZY $ (SSS). Therefore, corresponding angles are equal: $ \angle WXY = \angle WZY = 133^\circ $, $ \angle ZWX = \angle ZYX = 60^\circ $? Wait, no, that can't be, because the sum of angles in a quadrilateral is $ 360^\circ $, so $ 133 + 60 + 133 + 60 = 386 $, which is more than $ 360 $. So my labeling is wrong. Maybe $ WX = XY $ and $ WZ = ZY $, so the kite has two pairs of adjacent equal sides: $ WX = XY $ and $ WZ = ZY $. Then the diagonal $ WY $ bisects $ \angle W $ and $ \angle Y $. Wait, the problem says $ \angle WXY = 133^\circ $ (at $ X $) and $ \angle ZWX = 60^\circ $ (at $ W $). So at vertex $ W $, angle $ \angle ZWX = 60^\circ $, at vertex $ X $, angle $ \angle WXY = 133^\circ $. Since it's a kite, the diagonal $ WY $ bisects $ \angle ZWX $ into two equal angles? Wait, no, the diagonal that connects the vertices between the unequal sides bisects the angles. Wait, maybe the diagonal $ WY $ splits the kite into two triangles, $ \triangle WXY $ and $ \triangle WZY $, where $ WX = WZ $, $ XY = ZY $, $ WY = WY $, so they are congruent. Therefore, $ \angle ZYW = \angle XYW $. Now, in triangle $ WXY $, we can find $ \angle XYW $. The sum of angles in a triangle is $ 180^\circ $. So $ \angle WXY + \angle ZWX + \angle XYW + \angle ZYW = 360^\circ $? No, quadrilateral. Wait, quadrilateral $ WXYZ $: angles at $ W $: $ \angle ZWX = 60^\circ $, at $ X $: $ \angle WXY = 133^\circ $, at $ Y $: $ \angle XYZ $, at $ Z $: $ \angle YZW $. In a kite, $ \angle ZWX = \angle YZW $ and $ \angle WXY = \angle XYZ $? No, that's not the property. The correct property is that one pair of opposite angles (the ones between the unequal sides) are equal. Wait, maybe I should look at the diagonal $ WY $ bisecting $ \angle ZWX $ and $ \angle XYZ $. Wait, the problem is to find $ \angle ZYW $. Let's assume that diagonal $ WY $ bisects $ \angle ZWX $ into two angles of $ 30^\circ $ each (since $ \angle ZWX = 60^\circ $). Then, in triangle $ WZY $, we have $ WZ = WX $, $ ZY = XY $, so it's isosceles? Wait, no, maybe the key is that in the kite, the diagonal $ WY $ is the angle bisector of $ \angle ZWX $ and $ \angle XYZ $. Wait, I think I messed up. Let's use the fact that in a kite, one diagonal bisects the other and bisects the angles at its endpoints. So if $ WY $ is the diagonal that bisects $ \angle ZWX $ and $ \angle ZYX $, then $ \angle ZWY = \angle XWY = \frac{60^\circ}{2} = 30^\circ $, and $ \angle ZYW = \angle XYW $. Now, in triangle $ WXY $, angles sum to $ 180^\circ $: $ \angle WXY + \angle XWY + \angle XYW = 180^\circ $. So $ 133^\circ + 30^\circ + \angle XYW = 180^\circ $. Then $ \angle XYW = 180 - 133 - 30 = 17^\circ $. Since $ \angle ZYW = \angle XYW $ (because of congruent triangles), $ \angle ZYW = 17^\circ $. Wait, is that right? Let's check: $ 133 + 30 + 17 = 180 $, yes. Then, since $ \triangle WXY \cong \triangle WZY $, $ \angle ZYW = \angle XYW = 17^\circ $.

## Step1: Find $ \angle XWY $
Since $ WY $ bisects $ \angle ZWX $, $ \angle XWY = \frac{\angle ZWX}{2} = \frac{60^\circ}{2} = 30^\circ $.

## Step2: Find $ \angle XYW $ in $ \triangle WXY $
Sum of angles in a triangle: $ \angle WXY + \angle XWY + \angle XYW = 180^\circ $.
Substitute known values: $ 133^\circ + 30^\circ + \angle XYW = 180^\circ $.
Solve for $ \angle XYW $: $ \angle XYW = 180^\circ - 133^\circ - 30^\circ = 17^\circ $.

## Step3: Conclude $ \angle ZYW $
In kite $ WXYZ $, $ \triangle WXY \cong \triangle WZY $ (SSS), so $ \angle ZYW = \angle XYW = 17^\circ $.

# Answer:
$ 17^\circ $

### Problem 5:
# Explanation:
## Step1: Analyze $ \triangle BCD $
$ \triangle BCD $ is isosceles with $ BD \cong BC $ (given $ \overline{BD} \cong \overline{BC} $). So $ \angle BDC = \angle BCD $.
Sum of angles in $ \triangle BCD $: $ \angle DBC + \angle BDC + \angle BCD = 180^\circ $.
Given $ \angle DBC = 84^\circ $, let $ \angle BDC = \angle BCD = x $.
So $ 84^\circ + x + x = 180^\circ $ → $ 2x = 96^\circ $ → $ x = 48^\circ $. Thus, $ \angle BDC = 48^\circ $.

## Step2: Find $ \angle ADB $
Given $ \angle BDA = 24^\circ $ (wait, the problem says $ \angle BDA $ has a measure? Wait, the original problem: "Angle $ DBC $ has a measure of $ 84^\circ $ and $ \angle BDA $ has a measure of $ 24^\circ $"? Wait, maybe a typo, but assuming $ \angle BDA = 24^\circ $, then $ \angle ADC = \angle ADB + \angle BDC = 24^\circ + 48^\circ = 72^\circ $.

## Step3: Analyze $ \triangle ACD $
$ \triangle ACD $ is isosceles with $ AD \cong AC $ (given $ \overline{AD} \cong \overline{AC} $). So $ \angle ADC = \angle ACD = 72^\circ $.
Sum of angles in $ \triangle ACD $: $ \angle DAC + \angle ADC + \angle ACD = 180^\circ $.
Thus, $ \angle DAC = 180^\circ - 72^\circ - 72^\circ = 36^\circ $.

## Step4: Find $ \angle BAC $
We need to find $ \angle BAC $. Wait, maybe $ AB $ is a segment. Wait, maybe $ \triangle ABD $ or $ \triangle ABC $. Wait, maybe $ \angle BAC = \angle DAC - \angle DAB $, but we need more info. Wait, maybe $ \angle BDA = 24^\circ $, and in $ \triangle ABD $, $ \angle ABD $ is related. Wait, maybe I misread. Let's re-express:

Wait, the problem says "Triangles $ ACD $ and $ BCD $ are isosceles. Angle $ DBC $ has a measure of $ 84^\circ $ and $ \angle BDA $ has a measure of $ 24^\circ $. Find the measure of $ \angle BAC $."

Given $ BD = BC $, so $ \triangle BCD $ is isosceles with $ \angle BDC = \angle BCD $. As before, $ \angle BDC = 48^\circ $. Then $ \angle ADB = 24^\circ $, so $ \angle ADC = 24 + 48 = 72^\circ $. $ AD = AC $, so $ \triangle ACD $ is isosceles with $ \angle ADC = \angle ACD = 72^\circ $, so $ \angle DAC = 36^\circ $. Now, maybe $ AB $ is the angle bisector or $ \triangle ABD $ is isosceles? Wait, maybe $ \angle ABD = \angle DBC - \angle ABC $, no. Wait, maybe $ \angle BAC = \angle DAC - \angle DAB $, and $ \angle DAB = \angle ADB = 24^\circ $ (if $ \triangle ABD $ is isosceles with $ AD = AB $), but that's not given. Wait, maybe the diagram shows $ B $ on $ DC $. If $ B $ is on $ DC $, then $ \angle DBC = 84^\circ $ is a straight angle? No, that can't be. Wait, maybe the diagram has $ D $, $ B $, $ C $ with $ B $ between $ D $ and $ C $. Then $ \angle DBC = 180^\circ - 84^\circ = 96^\circ $, but no, the problem says $ \angle DBC = 84^\circ $. I think I made a mistake. Let's try again.

Wait, $ \triangle BCD $ is isosceles with $ BD = BC $, so $ \angle BDC = \angle BCD $. $ \angle DBC = 84^\circ $, so $ \angle BDC = (180 - 84)/2 = 48^\circ $. Then $ \angle ADB = 24^\circ $, so $ \angle ADC = \angle ADB + \angle BDC = 24 + 48 = 72^\circ $. $ \triangle ACD $ is isosceles with $ AD = AC $, so $ \angle ACD = \angle ADC = 72^\circ $, so $ \angle DAC = 180 - 72 - 72 = 36^\circ $. Now, if $ AB $ is such that $ \angle BAC = \angle DAC - \angle DAB $, and $ \angle DAB = \angle ADB = 24^\circ $ (since $ \triangle ABD $ has $ \angle ADB = 24^\circ $ and maybe $ AD = AB $), then $ \angle BAC = 36 - 24 = 12^\circ $? No, that doesn't make sense. Wait, maybe $ \angle BAC = \angle DAC / 2 $? No. Wait, maybe the diagram shows $ AB $ bisecting $ \angle DAC $. If $ \angle DAC = 36^\circ $, and $ AB $ bisects it, then $ \angle BAC = 18^\circ $. But I need to check the diagram. Since the diagram has $ D $, $ B $, $ C $ with $ B $ inside, maybe $ \angle BAC = 18^\circ $. Wait, maybe my initial steps are wrong. Let's assume that $ \angle BDA = 24^\circ $, $ \angle BDC = 48^\circ $, so $ \angle ADC = 72^\circ $, $ AD = AC $, so $ \angle DAC = 36^\circ $. Then, if $ AB $ is the angle bisector, $ \angle BAC = 18^\circ $. But I think the correct answer is $ 18^\circ $. Wait, let's recalculate:

$ \triangle BCD $: $ BD = BC $, $ \angle DBC = 84^\circ $, so $ \angle BDC = (180 - 84)/2 = 48^\circ $.
$ \angle ADB = 24^\circ $, so $ \angle ADC = 24 + 48 = 72^\circ $.
$ \triangle ACD $: $ AD = AC $, so $ \angle ACD = 72^\circ $, $ \angle DAC = 180 - 72 - 72 = 36^\circ $.
Now, if $ AB $ is such that $ \angle BAC = \angle DAC - \angle DAB $, and $ \angle DAB = \angle ADB = 24^\circ $ (since $ \triangle ABD $ has $ \angle ADB = 24^\circ $ and $ AD = AB $? No, $ AD = AC $, not necessarily $ AD = AB $. Wait, maybe $ \angle BAC = \angle DAC / 2 $ because $ AB $ is the median or something. Wait, the diagram shows $ B $ on

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QUESTION IMAGE

Question

Explanation:

Problem 4:

Step1: Recall kite properties

Step1: Analyze \( \triangle BCD \)

Step2: Find \( \angle ADB \)

Step3: Analyze \( \triangle ACD \)

Step4: Find \( \angle BAC \)

Answer:

Problem 5: