Altitudes AD and BE of acute triangle ABC intersect at point H. If angle ABH = 30 and angle DAC = 45, then what is angle HCA in degrees?

Well we have right triangles ADC, ABD, ABE, BEC, BHD
triangle ABE is a 30-60-90 and ADC is a 45-45-90
and then im stuck, :(
some things i did notice were AHE and BHD are 45-45-90 triangles, and since ABE is a 30-60-90 triangle, BAH is 15 degrees,
...
i think i got it
make the altitude for triangle on ABC on segment AB, and since BAC is 60 degrees, HCA would be 30 degrees
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Well we have right triangles ADC, ABD, ABE, BEC, BHD
triangle ABE is a 30-60-90 and ADC is a 45-45-90
and then im stuck, :(
some things i did notice were AHE and BHD are 45-45-90 triangles, and since ABE is a 30-60-90 triangle, BAH is 15 degrees,
...
i think i got it
make the altitude for triangle on ABC on segment AB, and since BAC is 60 degrees, HCA would be 30 degrees
![]()
![]()
![]()