Can Veeky Forums solve a simple geometry puzzle?

Well sci?

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wolframalpha.com/input/?i=sin(arcsin(3/4) + arccos(3/7) - pi/2) * (2*sqrt(10) + sqrt(7)) + 3
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Anyone with AutoCAD can.

Pretty trivial application of interuniversal geometry

cmon, this is high school level

5

did I do something wrong, I get [math]\frac{183}{28}[/math]

wolframalpha.com/input/?i=sin(arcsin(3/4) + arccos(3/7) - pi/2) * (2*sqrt(10) + sqrt(7)) + 3

>7

Greater than seven....

i get 141/28 - 6*sqrt(70)/7

...but i might have made a slight mistake since its a bit negative...

ok did it again, found 183/28 like

183/28 for me this is brainlet tier fuck you for posting this

I was estimating, I'm an engineer.

...

Typo, should be 183/28

same

4:7 = (7+x) : (7-4)
As soon as you had found x
4:3 = (7+x) : A

i don't want to use the calcutlator but
-use pythagoras to find the mising side
-use cos/sin law to find the angles
-prolong the necessary side to for a right triangle with side A
-this in turn will reveal all the angles within this triangle and in turn allow us to set up some sort of system of equation or something, I dunno,
these sort of problems usually have a catch, if we could find out that the original triangles are right triangles it mighth help us
Any ideas?

nevermind i didn't notice something really dumb excuse me your all right

[code]
v_1 = [sqrt(4^2 - 3^2), 3]';
v_2 = [sqrt(7^2 - 3^2) + sqrt(4^2 - 3^2), 0]';
phi = asind(3/4) + acosd(3/7) - 90;
R = [cosd(phi), -sind(phi)
sind(phi), cosd(phi)];
v = v_1 + R*v_2;
A = v(2);
[/code]

Square root of 58