finding x

[pliant jolt]

#1015578016606343218

[fleet mason]

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[weary patrol]

How'd you try

[jagged falcon]

Angle E is 90°

[weary patrol]

Why? E can't be 90°

[jagged falcon]

Why?

Look at the picture

[past vault]

2y+z=150

[pliant jolt]

That's not how it works

[past vault]

2z+y-x=330

[past vault]

Let him yap lmao, this post is gonna attract a lot of attention
Geometry problems like this do

[pliant jolt]

Lol

[past vault]

x+y+180-z = 180

x+y=z

Bruh

[past vault]

3y+x=150

[pliant jolt]

Lemme figure it out on paper

[past vault]

Nope fail

The equations are the same lmao

[past vault]

Nah don’t

[pliant jolt]

Wait did you extended smthing?

[past vault]

No no constructions

I tried elementary angle chasing

That didn’t work sooo expect a lot of attention

I’ll try primitive cosine and sine bashing

[weary patrol]

Eh? Just use the fact that ADE is isosceles too. You could relate it's angles with the bigger triangle.

[pliant jolt]

Uga buga my brain hurts

[past vault]

That’s what i did

[pliant jolt]

Sane

[past vault]

Angle chasing won’t work here

[pliant jolt]

Same

[past vault]

Aight gotta unpack a dishwasher

[weary patrol]

I mean, it did for me

[pliant jolt]

I am just gonna assume its 30

[weary patrol]

It's not

The top angle of ABC is decreased by 30° to get ADC triangle. So how much should the bottom angles increase?

This should get you to the answer

Both triangles are isosceles

[pliant jolt]

Hmm

[jagged falcon]

To find the value of ( x ) in the given triangle, we can use the properties of isosceles triangles and angle relationships.

Given:

• ( AB = AC ) (isosceles triangle ( ABC ))
• ( \angle BAD = 30^\circ )
• ( AE = AD )

First, since ( AB = AC ), the base angles ( \angle ABC ) and ( \angle ACB ) of triangle ( ABC ) are equal. Let's call each of these angles ( \theta ).

In triangle ( ABD ):

• ( \angle BAD = 30^\circ )
• ( AE = AD ), which means triangle ( ADE ) is isosceles with ( \angle EAD = \angle ADE ).

Now, let's consider the exterior angle theorem for triangle ( ABD ):
[ \angle ADB = \angle ABC + \angle BAD ]
Since ( \angle ABC = \theta ) and ( \angle BAD = 30^\circ ):
[ \angle ADB = \theta + 30^\circ ]

For triangle ( ADE ):
Since ( AE = AD ):
[ \angle ADE = \angle EAD ]
[ \angle ADE = \angle EAD = x

[pliant jolt]

AB = AC, means that the red angles are equal.
AE = AD, means that the blue angles are equal.

For triangle ABD:
Red + 30° + (180° - Blue - x) = 180°.
This gives Red - Blue - x + 30° = 0
For triangle DEC:
x + (180° - Blue) + Red = 180°.
This gives Red - Blue + x = 0

Equate: Red - Blue - x + 30° = Red - Blue + x
2x = 30.
x = 15°.
My friend pulled this

[weary patrol]

That's correct

[weary patrol]

I believe this one's easier to understand tho

[weary patrol]

Is this ai?

[past vault]

Yeah it’s chatGPT

and I feel like an idiot now

Bruh I was co-ord bashing

[weary patrol]

Lol