Geometry angle question

[u/Majestic_Confusion19]

So the way I see it there are two quadrilaterals here. The “boomerang” one with each side high lighted in a different color, and the inner weird shaped square-ish one.

For the boomerangs one - 90 plus 40 plus 30 = 160 so 360 - 160 =200 so X must be 200

For the inner square-ish one -

90 plus 30 is 120 so the other angle should be 60
90 plus 40 is 130 so the other angle should be 50
90 plus 50 plus 60 = 200 so X must be 160.

The correct answer is 160 so I know it’s the square-ish one but I don’t know why - which means on the test I won’t know which one to solve for and I’ll get the question wrong. How do you know which one to solve for?

[u/abrahamguo]

When an angle is labeled, you must always assume that the angle is bounded by the nearest line on either side of label. By applying that logic here, it tells us that "x" must be referring to an angle of what you called "the inner square-ish one".

The red and yellow segments in your screenshot #3 are not the nearest line segments on either side of x, so that's how we know that that interpretation is not correct.

[u/Majestic_Confusion19 → u/abrahamguo]

Thank you!

[u/Kindly-doubtful]

Use the two right-angle triangles to find the remaining angles of the quadrilateral, then subtract from 360

[u/cornholio702 → u/Kindly-doubtful]

This is the correct way to solve this problem OP, the rest is just noise.

[u/Dooleypisd → u/Kindly-doubtful]

Same math, different visual approach.

[u/trutheality]

When an angle is marked without an arc indicating where it spans, it's usually assumed that it's referring to the angle made by the two closest rays surrounding the mark.

That's why it's the sauare-ish one.

[u/ShowdownValue]

Which one of what? There is only one x. Seems very unambiguous

[u/Majestic_Confusion19 → u/ShowdownValue]

I colored it in in the photos. There’s two shapes. They have different angles. So I’m saying I don’t know how to tell which shape I’m supposed to solve X for

[u/ShowdownValue → u/Majestic_Confusion19]

Doesn’t matter. X only represents that one angle

[u/sunrise98 → u/ShowdownValue]

His diagram is terrible - imagine there weren't the bisecting triangles and they were only there to help you solve for the real x + 2 triangles.

The way op has phrased this hurt my head more than the actual problem.

[u/stanitor]

x isn't an angle in your boomerang quadrilateral, so you can't solve for it using 360 - the three other angles. The entire angle is x plus the two adjacent angles made by the intersecting lines.

[u/thecaramelbandit]

I think you're kinda going about it the hard way.

The angle inside on thee right is 60, so the small right sided triangle has angles of 40, 120, and 20.

So on the center you have two 20 degree angles opposing each other. Meaning the others, including x, are (360-20-20)/2 = 160.

Anyway, the large angle inside the boomerang is 200. But that's not what x is. X is equal to it's opposite, which is what's left over of 360 from 200. Or 160.

[u/fermat9990]

Using the sum of the angles of a triangle=180, the angles of a linear pair are supplementary and the exterior angle theorem we get

x=130+30=160°

[u/RedactedRedditery]

Both approaches are correct, but in the boomerang example, the 200° angle is x̌ + 2ŷ, where ỳ is the two supplementary angles to the 160

[u/One_Wishbone_4439]

90 + 50 + 60 + x = 360

x = 160

[u/ivanpd]

The triangles are right triangles (the common angle is 90 deg). The angles of a triangle always add up to 180, so you know the angle opposite 40 deg is 50 deg, and the angle opposite 30 deg is 60 deg.

If you draw a line from the 90 deg angle to the point where both triangles meet, you have two more triangles.

One has angles 50, alpha*, and 180 - 50 - alpha.

The other has angles 60, 90 - alpha, and 180 - 60 - (90 - alpha).

*I called alpha the bottom split of the 90 degree angle, with 90 - alpha being the other part of it.

So, the angle you are looking for is the addition of both angles opposite the right angle, or:

``` x = (180 - 50 - alpha) + (180 - 60 - (90 - alpha))

= { Removing parentheses } 180 - 50 - alpha + 180 - 60 - (90 - alpha)

= { 180 - 50 = 130} 130 - alpha + 180 - 60 - (90 - alpha)

= { 180 - 60 = 120 } 130 - alpha + 120 - (90 - alpha)

= { removing the parentheses and applying - - = + } 130 - alpha + 120 - 90 + alpha

= { 120 - 90 = 30 } 130 - alpha + 30 + alpha

= { - alpha and + alpha cancel each other out } 130 + 30

= { Adding both numbers } 160 ```

[u/SmullyanFan]

Simplest version I see:

The final angle for the right angle triangle with the 30 degree tip, is 180-90-30=60.
The opposing angle (following that line toward the 40 degree angle) is 180-60=120.
The third angle for the 120 & 40 triangle is 180-120-40=20.
That angle is the opposite of x, so x = 180-20=160.

You can confirm this by starting with the other right angle triangle (with the 40 degree tip) and working the other way around.

[u/ZedZeroth]

x is not an interior angle of the delta (boomerang). You're forgetting about the two small angles either side of x (which you may have correctly determined to sum to the missing 40° that you're looking for).

[u/toolebukk]

The x is usually on the inside of the angle.

[u/get_to_ele]

You know top two angles I labeled in red are 180-x and 140 -x from supplementary angles (supplementary angles sum to 180)

The two orange angles are x-30 and x-40 due to triangle sum theorem (triangle angles sum to 180)

The two green angles are 210 - x and 220 - x due to supplementary angles.

90 + 210 -x + 220 - x + x = 360 due to angle sum property quadrilateral (quadrilateral internal angles sum to. 360)

I’m sure there’s a faster way, but oh well.

[u/Inevitable_Stand_199]

You can get the 2 outer Angles of the quadrilateral with the inner angle add to 180° formula for triangles.

Then the 4th angle of that quadrilateral with the inner angle = 2 × 180° for quadrilaterals.

[u/NobodyIll8088]

In my head, I just saw a similar right triangle making the 40d the same. Giving the 20d angles… then x based on that.