Altitudes AX and BY of acute triangle ABC intersect at H. If angle BAC = 61 and angle ABC = 82, then what is angle CHX?
H will be the orthocenter of the triangle
Let CZ be the other altitude
Then we have right triangle CZB with angle CZB = 90 angle ZBC = 82 and angle ZCB = 180 - (90 +82) =
And we have another right triangle CHX with angle HXC = 90 and angle HCX = 8
So angle CHX = 180 - (90 + 8) = 180 - 98 = 82°