You should be able to solve this

not certain but I think it's
[eqn]
\frac{3}{4} + 2(\sqrt{3} + \sqrt{10}) \sin(\arcsin\frac{3}{4} + \arccos{\frac{3}{7}}) \approx 9.74
[/eqn]

I fucked up towards the end actually
this isn't quite right

>8.416
did i win?

You know what FUCK TRIANGLES, why tf do we even need them anyways. Motherfucking "3 angles" scalene, half-assed looking ass motherfuckers.

okay fixed I think it's
[eqn]
3 - (\sqrt{7} + 2\sqrt{10}) \cos(\arcsin\frac{3}{4} + \arccos{\frac{3}{7}}) = \frac{30}{7}
[/eqn]
not sure how it simplifies to 30/7, but mathematica tells me it does

urruf fug I mean 183/28 not 30/7

okay
[eqn]
3 - (\sqrt{7} + 2\sqrt{10}) \cos(\arcsin\frac{3}{4} + \arccos{\frac{3}{7}}) = \frac{183}{28} \approx 6.53
[/eqn]
final answer I'm p sure

Looks correct to me. Only mistake you could have made is inputting stuff in the calculator. Your solution is neater than the one I thought of, I think you can call yourself an honorary little girl now.

Agreed

Approx. 6.91

T. Geometry Teacher

doing calculations on phone, come out roughly 6.5

is it correct?

it's not correct though
A is not longer than the length of the base of the triangle
it's not drawn exactly to scale, but it's pretty close
yeah this was quite similar to my solution

Don't have a cas on this computer
Solve
[eqn] \frac{ \sin \left( \arccos \left( \frac{3}{4} \right)+ \arccos \left( \frac{3}{7} \right)+ \arcsin \left( \frac{3}{4} \right) \right)}{x}= \frac{ \sin \left(a \right)}{ \sqrt{40}+ \sqrt{7}}= \frac{ \sin \left(180- \arccos \left( \frac{3}{4} \right)+ \arccos \left( \frac{3}{7} \right)+ \arcsin \left( \frac{3}{4} \right)-a \right)}{4}[/eqn]
Then the answer is
[eqn]A=x \sin \left( \arcsin \left( \frac{3}{4}-a \right) \right)[/eqn]

exact value

You guys are mistaken.

180- (arcsin root 7/4 +arcsin root 40/7)= approx 73.97

73.97 plus (arcsin 3/7) = apport 99.34

180-99.34

= 80.67

sin(80.67)*7= 6.91

Agreed

>it's not correct though
It's correct though. He likely just made a mistake in putting his number in the calculator, which is just a boring mistake.

oh fair enough sorry

K brainlets, here is my solution.

If you can't find a,b,c,x and L just kys.

The rest follows using a bit of trigonometry.

Oh, and y and u should be trivial to find. You need them to get z.

Am I a brainlet?

Am I a brainlet?

right you are.
bamboozled by my own shitty handwriting

[math] (2\sqrt10 + \sqrt7)sin(A+B-90)+3=L \\\
sin(A+B-90)=cos(A+B) \\\
cos(A+B) = sinAcosB-cosAsinB \\\
cos(A+B) = \frac{2\sqrt10}{4}\cdot \frac{3}{7}-\frac{3}{4} \cdot \frac{\sqrt7}{4} \\\
112L=(\sqrt7 +2\sqrt10)(24\sqrt10 - 21\sqrt7)+336 \\\
112L=24\sqrt70 -147 +480-42\sqrt70+336 \\\
L=\frac{669-18\sqrt70}{112}\approx4.62858... [/math]

shit i just realized where i fucked up

I actually made a lot of mistakes in that but it turns out it's 183/28 or around 6.5357...

pretty neat

might as well just do it

[math] (2\sqrt10 + \sqrt7)sin(A+B-90)+3=L \\\
sin(A+B-90)=-cos(A+B) \\\
cos(A+B) = cosAcosB-sinAsinB \\\
cos(A+B) = \frac{\sqrt7}{4}\cdot \frac{3}{7}-\frac{3}{4} \cdot \frac{2\sqrt10}{7} \\\
28L=(\sqrt7 +2\sqrt10)(6\sqrt10 - 3\sqrt7)+84 \\\
28L=6\sqrt70 +120 -21-6\sqrt70+84 \\\
L=\frac{183}{28}\approx 6.5357... [/math]

Had to fix a typo. Here it is. Drop the hypotenuse down, rest follows easily.

Looks like it's it's between 6.3 and 6.7, so I'm gonna say 6.5 because tests are made by humans and we like pretty numbers.

Took a different approach. Way too many fucking angle variables though. 183/28 is probably the best response.

This was effectively my method, except I used the angles in the bottom corners as A and B, so it meant looking for sin(B-A) instead of sin(A+B-90)=cos(A+B). The actual calculation works identically.

This is what I got as well
[eqn]3 + (\sqrt{7}+2\sqrt{10})\sin(\arccos\dfrac{\sqrt{7}}{4}+\arccos\dfrac{3}{7}-\dfrac{\pi}{2})[/eqn]

i got 6,99 lol i think it's 6,91 but with a bit wrong approximations
btw im in high school

then A = 3/7*(7-4)+sin(z)*7
z = pi - a - b
sin(z) = sin(a+b)
b=b1+b2
a+b = a + b1 + b2 = pi/2 + b2
sin(z) = sin(a+b) = sin(pi/2+b2) = cos(b2) = 3/4
so
A = 9/7+21/4

A=6.53

Used same method as this guy

>t. user who's not a brainlet
>Wrong result

kek

>"A geometry problem, huh..."
>brainlets going full trigonometry
fucking brainlets, man

How do you solve it without trig?

Embarrassing desu. No idea what's the mistake.

Not him, but

Extend the rightmost edge of the upper triangle, since that one is parallel to the length 4 edge.
The red line should be 12/7, by similar triangles, which gives 7+(12/7) as the hypotenuse of the right triangle with A as one of the sides.
This right triangle is similar to the small right triangle on the left, so
A / (7 + 12/7) = 3 / 4
Solving gives
A = 183/28

How will you ever enter a famous ladies attitude like this OP?

sorry for being a brainlet but how did you get 12/7?

Let r = length of the red line.
r / 3 = 4 / 7

I got 6.5357142857142857142857142857143 assuming these red triangles are similar.

Not sure I can say for certain they're similar but it feels right.

A = 3 + (sqrt(40) + sqrt(7))*sin(arcsin(3/4) - arcsin(3/7))

A= 6.5357

>Not sure I can say for certain they're similar but it feels right.

lets first look at the little triangle at the base of the big right triangle and see that it directly shares its right most angle with our main triangle, it also forms a vertical angle with the congruent triangle meaning that it has the same top angle.

since two of it's angles are the same the right most angles must be the same as well so they are similar

since the right most angles are the same and the little right triangle and the big right triangle are both right triangles, they indeed must be similar

meant left most in the second and third lines, but you get the idea

i have that exact same mechanical pencil.

good choice

well done.

Rest of you trash, hit the books you fucking brainlets.
This time: all the way back to elementary school. What an embarrassment

This manga was super cute

>(A+B-90)
explain

Man, similar triangles are useful. I just used trig, but this is way easier.

I find this the best anwser, but you don't really ned the complex plane