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
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