solve for the variable.
x = type your answer..
y = type your answer..

Question

Accepted Answer

# Explanation:
## Step1: Identify triangle type
The triangle $ \triangle WXY $ (or $ \triangle WYX $) has two equal sides (marked with ticks), so it's isosceles with $ \angle W = 52^\circ $, and the equal sides imply $ \angle X = \angle Y $? Wait, no—wait, the equal sides are $ WX $ and $ WY $? Wait, the marks: the two sides from $ W $ to $ X $ and $ W $ to $ Y $ are equal? Wait, looking at the triangle: vertices $ W $, $ X $, $ Y $. The sides $ WX $ and $ WY $ have ticks, so $ WX = WY $, making $ \triangle WXY $ isosceles with $ \angle X = \angle Y $? Wait, no—wait, in an isosceles triangle, the angles opposite equal sides are equal. So if $ WX = WY $, then the angles opposite them are $ \angle Y $ and $ \angle X $, respectively? Wait, no: side $ WX $ is opposite $ \angle Y $, and side $ WY $ is opposite $ \angle X $. So if $ WX = WY $, then $ \angle Y = \angle X $. Wait, but the angle at $ W $ is $ 52^\circ $, so the sum of angles in a triangle is $ 180^\circ $. So $ \angle W + \angle X + \angle Y = 180^\circ $. But if $ WX = WY $, then $ \angle X = \angle Y $? Wait, no—wait, maybe I got the equal sides wrong. Wait, the ticks are on $ WX $ and $ WY $? Wait, the diagram: $ W $ is the top, $ X $ and $ Y $ are the base? Wait, the angle at $ W $ is $ 52^\circ $, and the sides from $ W $ to $ X $ and $ W $ to $ Y $ have ticks, so $ WX = WY $, so the triangle is isosceles with $ WX = WY $, so the base angles are $ \angle X $ and $ \angle Y $? Wait, no—wait, in triangle $ WXY $, sides $ WX $ and $ WY $ are equal, so the angles opposite them are $ \angle Y $ (opposite $ WX $) and $ \angle X $ (opposite $ WY $). So $ \angle Y = \angle X $. Wait, but the angle at $ W $ is $ 52^\circ $, so:

$ 52^\circ + \angle X + \angle Y = 180^\circ $. But if $ \angle X = \angle Y $, then $ 52 + 2\angle X = 180 $, so $ 2\angle X = 180 - 52 = 128 $, so $ \angle X = 64^\circ $. Wait, but in the diagram, $ \angle X $ is $ (6y - 2)^\circ $, and $ \angle Y $ is $ (4x + 20)^\circ $. Wait, maybe I mixed up: maybe the equal sides are $ XW $ and $ YW $, so the base is $ XY $, and the equal sides are $ WX $ and $ WY $, so angles at $ X $ and $ Y $ are equal? Wait, no—wait, maybe the equal sides are $ WX $ and $ XY $? No, the ticks are on two sides from $ W $. Wait, maybe the triangle is isosceles with $ \angle W = 52^\circ $, and the other two angles (at $ X $ and $ Y $) are equal? Wait, but in the diagram, $ \angle X $ is $ (6y - 2) $ and $ \angle Y $ is $ (4x + 20) $. Wait, maybe I made a mistake: maybe the equal sides are $ WX $ and $ XY $? No, the ticks are on $ WX $ and $ WY $. Wait, let's re-express:

Wait, the triangle has vertices $ W $, $ X $, $ Y $. The sides $ WX $ and $ WY $ are equal (ticks), so $ \triangle WXY $ is isosceles with $ WX = WY $. Therefore, the angles opposite these sides are $ \angle Y $ (opposite $ WX $) and $ \angle X $ (opposite $ WY $). Therefore, $ \angle Y = \angle X $. Wait, but in the diagram, $ \angle Y $ is $ (4x + 20)^\circ $ and $ \angle X $ is $ (6y - 2)^\circ $. Wait, but also, the sum of angles in a triangle is $ 180^\circ $, so:

$ \angle W + \angle X + \angle Y = 180^\circ $

$ 52^\circ + (6y - 2)^\circ + (4x + 20)^\circ = 180^\circ $

But if $ WX = WY $, then $ \angle Y = \angle X $? Wait, no—wait, maybe the equal sides are $ WX $ and $ XY $, so $ \angle W = \angle Y $? Wait, that might be the case. Wait, maybe I misread the ticks. Let's look again: the two sides with ticks are $ WX $ and $ XY $? No, the ticks are on the two sides from $ W $ to $ X $ and from $ W $ to $ Y $. Wait, maybe the triangle is isosceles with $ \angle W = 52^\circ $, and the base is $ XY $, so the equal sides are $ WX $ and $ WY $, so angles at $ X $ and $ Y $ are equal. Wait, but then $ \angle X = \angle Y $, so $ 6y - 2 = 4x + 20 $. But also, sum of angles: $ 52 + (6y - 2) + (4x + 20) = 180 $. Let's simplify the sum:

$ 52 + 6y - 2 + 4x + 20 = 180 $

$ (52 - 2 + 20) + 4x + 6y = 180 $

$ 70 + 4x + 6y = 180 $

$ 4x + 6y = 110 $

But if $ \angle X = \angle Y $, then $ 6y - 2 = 4x + 20 $, so $ 6y = 4x + 22 $. Substitute into $ 4x + 6y = 110 $:

$ 4x + (4x + 22) = 110 $

$ 8x + 22 = 110 $

$ 8x = 88 $

$ x = 11 $

Then, for $ y $: $ 6y = 4(11) + 22 = 44 + 22 = 66 $, so $ y = 11 $? Wait, no—wait, that can't be. Wait, maybe the equal sides are $ WX $ and $ XY $, so $ \angle W = \angle Y $. Let's try that. If $ WX = XY $, then $ \angle W = \angle Y $. So $ \angle Y = 52^\circ $, so $ 4x + 20 = 52 $. Then solve for $ x $:

$ 4x + 20 = 52 $

$ 4x = 32 $

$ x = 8 $

Then, the sum of angles: $ 52 + (6y - 2) + 52 = 180 $

$ 102 + 6y - 2 = 180 $

$ 100 + 6y = 180 $

$ 6y = 80 $

$ y = 80/6 ≈ 13.33 $, but that seems odd. Wait, maybe the equal sides are $ WY $ and $ XY $, so $ \angle W = \angle X $. Then $ \angle X = 52^\circ $, so $ 6y - 2 = 52 $

$ 6y = 54 $

$ y = 9 $

Then sum of angles: $ 52 + 52 + (4x + 20) = 180 $

$ 104 + 4x + 20 = 180 $

$ 124 + 4x = 180 $

$ 4x = 56 $

$ x = 14 $

But this is confusing. Wait, let's look at the diagram again: the triangle has $ W $ at the top, $ X $ and $ Y $ at the base. The two sides from $ W $ (to $ X $ and to $ Y $) are equal (ticks), so $ WX = WY $, making $ \triangle WXY $ isosceles with $ \angle X = \angle Y $. Wait, but in the diagram, $ \angle Y $ is labeled $ (4x + 20)^\circ $ and $ \angle X $ is $ (6y - 2)^\circ $. Wait, maybe there's a typo, or maybe I misread. Wait, maybe the equal sides are $ WX $ and $ XY $, so $ \angle W = \angle Y $, so $ \angle Y = 52^\circ $, so $ 4x + 20 = 52 $, so $ x = 8 $. Then, since it's isosceles with $ WX = XY $, $ \angle X = 180 - 52 - 52 = 76^\circ $, so $ 6y - 2 = 76 $, so $ 6y = 78 $, $ y = 13 $. But that doesn't match. Wait, maybe the triangle is isosceles with $ \angle W = 52^\circ $, and the other two angles are equal, so $ \angle X = \angle Y $. Then:

$ 52 + 2\angle X = 180 $

$ 2\angle X = 128 $

$ \angle X = 64^\circ $

So $ 6y - 2 = 64 $

$ 6y = 66 $

$ y = 11 $

And $ \angle Y = 64^\circ $, so $ 4x + 20 = 64 $

$ 4x = 44 $

$ x = 11 $

Ah, that makes sense! So if $ \angle X = \angle Y = 64^\circ $, then:

For $ x $: $ 4x + 20 = 64 $

$ 4x = 44 $

$ x = 11 $

For $ y $: $ 6y - 2 = 64 $

$ 6y = 66 $

$ y = 11 $

Yes, that works. So the key is recognizing the isosceles triangle with $ \angle W = 52^\circ $, so the other two angles are equal, each $ (180 - 52)/2 = 64^\circ $. Then set the angle expressions equal to 64.

## Step2: Solve for $ x $
We have $ \angle Y = (4x + 20)^\circ = 64^\circ $

So:

$ 4x + 20 = 64 $

Subtract 20 from both sides:

$ 4x = 64 - 20 $

$ 4x = 44 $

Divide by 4:

$ x = \frac{44}{4} = 11 $

## Step3: Solve for $ y $
We have $ \angle X = (6y - 2)^\circ = 64^\circ $

So:

$ 6y - 2 = 64 $

Add 2 to both sides:

$ 6y = 64 + 2 $

$ 6y = 66 $

Divide by 6:

$ y = \frac{66}{6} = 11 $

# Answer:
$ x = 11 $, $ y = 11 $

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QUESTION IMAGE

Question

Explanation:

Step1: Identify triangle type

Answer: