Angles in a Acute Triangle

Since angle AXH = angle AEH = 90 degree, we can conclude that AEHX are vertices of a cyclic quadrilateral.

Hence, since the opposite angles of a cyclic quadrilateral are supplementary, angle EHX = 146 degree

Consider the quadrilateral AEHX, we know that sum of all the interior angles of a quadrilateral is 360degrees. It is given that E, and X are the perpendiculars, also given that ∠CAB=34degrees. So, ∠EHX=360-[90+90+34] = 146degrees.

XHEA is a quadrilateral and it's given that HX is perpendicular to AB and forms a 90 deg angle, and same goes with HE to AC. Angle A is 34 deg.

The sum of the interior angles of a quadrilateral is 360 deg angle AXH + angle XHE + angle HEA + angle A = 360

90 + angle XHE + 90 + 34 = 360

giving us angle XHE = 146

∠CAB=34°, ∠AEB=90° beacuse line EB is perpendicular to line AC, ∠AXH=90° because line XH is perpendicular to line AB

Supposed, ∠CAB=a ∠AEB=b ∠AXH=c & ∠EHX=d

a+b+c+d=360°

34°+90°+90°+d=360°

d=360°-214°

d=146°

360 - 90 - 90 - 34 = 146

sum of angle of an rectangle is 360 degree, so, angle EHX = 360 - (90 + 90 + 34) = 146

(Angles ABC and ACB are not needed to answer this question) The sum of angles in a Quadrilateral AXHE= 360 degrees. AXHE is a Quadrilateral. Line BE is a perpendicular line so angle AEH=90. Line HX is also a perpendicular so angle AXH is also 90 degrees. Therefore, angle EHX =360-(34+90+90)= 146 degrees.

$EBA = 90^o - 34^o = 56^o$ ;

$XHB = 90^o - 56^o = 34^o$

$EHX = 180^o - 34^o = 146^o$