You should be able to solve this

>>159684258
I am sorry sensei, I can't.

>>159684258
what bugs me more is that nearly all the time capital letters denote corners and small letters denote lines

>>159684258
Is this it?

(7-4)sin(arcsin(3/7)) + 7sin((pi)-(arccos(3/4)+arccos(3/7)+arcsin(3/7)))

6.535714285714285714285714285714285714285714285714285714285...

>>159685554
40/7 maybe

>>159685277
And that is the correct way of doing it
The lines are also signed after the corner opposite to them

I google "geometry triangle theorems" and go from there.
>implying people need to actually know anything these days

Also Chio chan is cute!

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I'm able to solve this. You?

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What about this one, though?

>>159688775
180

Yes, but I'd solve it with trigonometric functions and linear algebra, instead of whichever way you're supposed to.

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>>159684258

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>>159688775

>>159689288
It's a ten-sided figure.
Just do 180 * (10-2)

>>159684258
>tfw Geometry was always my worst math subject

I did well in literally everything else, but whenever shapes and angles came up my mind tended to just hit a roadblock.

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>>159684258
I'll leave the calculations to my students for practice.

>>159685554
That's exactly what I got, though I used a different approach.

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>>159689727
>>159689781

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Two for the price of one.

>>159689727
Well fuck, I thought congruent meant same shape, but can be different size.
How would you solve A if that was the case?

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I got 183/28

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>>159689781
>>159685554
Yeah, seems about right.

>>159690944
Wait, I coulda just said 4/3 = (61/7)/A

>>159690338
Let x^x^x... = f(x)
f(x)*Ln(x) = Ln(2)
2Ln(x) = Ln(2)
x=sqrt(2)
x^2 = 2

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>>159690944

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>>159690683
>I thought congruent meant same shape, but can be different size.
Nah, you were thinking of Similar

Funny

>>159688982
17!
probably

>>159695942
oh fuck I see the funny trick now.

>>159688982
3x = 30; x= 10

2b = 20-10; b = 5

2c = 9-5; c = 2;

5 + (2*10) = 25;

>>159688982
16

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>>159684258
First I solved the 3|7 (red) angle then the 4|3 (blue) to find the 4|7 (green) angle
that subtracted from a straight 180° line to find the purple angle
then add the purple angle back to the red angle and subtract 90° to create the teal angle
then reverse trig to find the a segment length of A given we know the long side is 7
Then I made a sub triangle which has the same angles as 3|7 and found the length of one side (green) given known lengths
reverse trig again to find the side length which corresponds to the remaining segment of A

>>159699896
I got lazy with my maths and there are compounding rounding errors in this so its only accurate to the tenths place I think?

>>159688982
About tree fiddy