I was reading a manga when I came across this problem. I want

>>9015777
Not soluble.

>>9015826
You can solve this problem just using law of sines and law of cosines.

>>9015826
I made some screw ups. I'd rather not make the picture again but, the the pi/2 - 2arcsin... angle should be arcsin(3/4) + arcsin(3/7) - arcsin(3/7) = arcsin(3/4).
The final answer comes to the much more sensible 7*3/4 + 9/7 = 183/28

>>9015777
[math] 3 + (\sqrt{5} + \sqrt{12}) \times \sin (-\frac{\pi}{2} + \cos^{-1}(\frac{3}{7}) + \sin^{-1}(\frac{3}{4})) [/math]

>>9015849
Your numerical answer is more than 8.29. However the triangle of the upper part of the segment A, the hypotenuse of the rotated triangle with length 7, is a right triangle then with hypotenuse 7 and another side with length 8.29 - 9/7 = 7.01

File: 1499278016426.png (46KB, 555x432px) Image search: [Google]

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>>9015859
could you carefully reword that? i have no idea what you're trying to say.

what i'm doing is finding the height of the triangle defined by the black line, purple angle and blue lengths in my pic.
the purple angle is the sum of the red and blue angles, and the blue lengths are defined by the lower triangle.

>>9015871
your sqrt 5 and sqrt 12 are wrong, though.

>>9015871
Ok. What I meant by upper leg of A was the part of A that is above the horizontal black written in line in this picture >>9015826, the part that has 7*cos.... blah blah next to it. If you claim the entire length of A is around 8.29, then that part of A should be 8.29 - 9/7, since the length of A below that horizontal line is 9/7 clearly. Thats a problem since the hypotenuse of the triangle that upper part of A forms is less than 8.29 - 9/7

>>9015882
oh, you're right. oops
should be [math] \sqrt{40} [/math], but i can't be arsed to rewrite it all

>>9015882
>>9015888
[math] \sqrt{7} [/math] and [math] \sqrt{40} [/math]
i mean. i am really blowing this arithmetic

3*(7/4) + (3 - 3*(4/7))
if you draw a line connecting the vertex of the upper triangle that is shared by 2 angles, such that that line is perpendicular to A and parallel to the bottom-most line segment, then you can see that the angle between the new line segment, and the "7" line segment of the lower triangle is complementary to the angle on the right of said vertex. The new line segment is perpendicular to A, so, the sum of the 3 angles is 90 degrees plus the angle between the "4" line segment and the "3" line segment. So, the orthogonal projection of the "7" line on the upper triangle onto the line segment A is 3*(7/4). Now, to get the rest of the length of A, we need to measure the lower segment of the "3" line, which is 3 - (upper segment) = 3 - 3*(4/7). So A = 3*(7/4) + 3 - 3*(4/7).

File: 2.png (48KB, 555x432px) Image search: [Google]

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>>9015902
I have no idea where you're getting these numbers from.

>>9015912
it's not a 3,4,5 triangle. look at where the right angle is. 4 is the hypotenuse, so the other leg is [math] \sqrt{4^2 - 3^2} = \sqrt{16 - 9} = \sqrt{7} [/math] and the other leg follows teh same rules to be [math] \sqrt{7^2 - 3^2} = \sqrt{49 - 9} = \sqrt{40} [/math]

>>9015912
I'm not him but already you have screwed up by putting 5 on the leg of the triangle with a hypotenuse of 4

File: 2.png (44KB, 555x432px) Image search: [Google]

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>>9015910
>So, the orthogonal projection of the "7" line on the upper triangle onto the line segment A is 3*(7/4).
I followed you until here, could you explain a bit?

>>9015915
>>9015916
Shit, you're right. I'm an idiot.

>>9015922
>upper triangle
>draws line from the lower triange
anon...

File: retardedperson1[1].jpg (306KB, 684x640px) Image search: [Google]

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>>9015929
>a line connecting the vertex of the upper triangle that is shared by 2 angles, such that that line is perpendicular to A and parallel to the bottom-most line segment
Maybe I'm double retarded, but where should the line be drawn?

>>9015942
just put your pen on the vertex that is physically closest to the number 7 as it appears in the image, and draw the line out so that it is perpendicular to A.

File: 2.png (44KB, 555x432px) Image search: [Google]

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>>9015943
That makes sense.
>the sum of the 3 angles is 90 degrees plus the angle between the "4" line segment and the "3" line segment.
what are the "three angles" talked about, here?

>>9015951
I mean the whole line out, don't just draw a segment of it. The third angle is the one on the left side.

>>9015951
>the one on the left side.
this is really vague. which angle on the left side?