Acute CODE

Since the angles of a triangle add up to $180 ^ \circ$ , we get $\angle ACB = 78 ^ \circ$ . The incenter of a triangle is the intersection of the angle bisectors from each of the vertices. So $\angle OBA = \angle OBC = 33 ^ \circ$ and $\angle OCB = \angle OCA = 39 ^ \circ$ . Using triangle interior angle sums again, we get $\angle BOC = 108 ^ \circ$ , $\angle BOD = 57 ^ \circ$ , and $\angle COE = 51 ^ \circ$ . This leaves us with $\angle DOE = 144 ^ \circ$ . Since $OD$ and $OE$ are radii, $\bigtriangleup ODE$ is isosceles, giving us $\angle DEO = \angle EDO = 18 ^ \circ$ . We use triangle interior angle sums one last time on $\bigtriangleup CPE$ to find our acute angle, which is $33 ^ \circ$ .

∠ACB + ∠BAC + ∠ABC = 180∘

∠ACB + 36∘ + 66∘ = 180∘

∠ACB = 78∘

∠ACO = 1/2 x ∠ACB = 1/2 x 78∘ = 39∘

ADE is an isosceles triangle , AD = AE , we get ∠AED = ∠ADE = (180∘ - ∠DAE)/2

∠AED = (180∘ - 36∘)/2 = 72∘ then , ∠CED = 180 - ∠AED

∠CED = 180∘ - 72∘ = 108∘

Let P be the intersection of extended ED and CO

Triangle CEP : ∠CEP + ∠ CPE + ∠ ECP = 180∘

108∘ + ∠ CPE + 39∘ = 180∘ ∠ CPE = 180∘ - (108∘ + 39∘) = 33∘

the acute angle between lines CO and DE = 33∘

AD = AE which implies <ADE=72 degrees,let N be the point of intersection of DE and CO extended. now <NCB=78/2=39,let CO intersect AB at L which implies <BLC = <NLA =75,therefore the required angle is 180 - 72 - 75 = 33!

OD line perpendicular to the line AB, AC line perpendicular to the line AC and angle BAC = 36 degrees. Because the total angle of ADOE is 360 degrees then we will get the angle DOE = 114 degrees. If we draw a line from point D to point E then we will get an isosceles triangle DOE and angle ODE = 33 degrees. Because we can draw a straight line from point C to point O and point D so the acute angle between lines CO and DE equal to angle ODE = 33 degrees

In triangle ACN, angle A = 36, angle ACN = (180 - (36 + 66)/2 = 39 so angle ANC = 180 - 39 - 36 = 105

In triangle ADE, => ADE = AED = 72

In triangle DXN, XND = 180 - ANC = 75 and XDN = 72

=> NXD = 180 - 72 - 75 = 33

firstly,

we take the third angle 180-(66+36)=78. we know that incenter is the point where angle bisectors meet

we make a triangle and then the CO and DE are joined and by angle sum property we can calculate the angle

by that way we get answer 33