\mathsf{
\angle BOC(\theta) - \angle AOB(\alpha) = 30^o\implies \theta =30^o +\alpha(I)\\
Bissetriz: ~\angle AOF= \angle FOC\implies \frac{\alpha+\theta}{2}\\
\angle BOF = \theta - \angle FOC =...

\mathsf{
\angle AOB = \alpha \implies Bissetriz \angle AOB \implies\angle AOE = \angle EOB =\frac{\alpha}{2}\\
\angle BOC = \theta = 40^o\\
Bissetriz~\angle AOC\implies \angle AOJ = \angle JOB =...

\mathsf{
x+y =62^o\\
z+z+y+y=62^\implies 2(x+y)=62\\
\boxed{\color{red}\therefore (z+y) = 31^o}

}

\mathsf{
x+z+y=?\\
2y+2z = 70^o\\
2x+2z=70^o\\
\therefore2x+2y+2z =140 \implies 2(x+y+z)=140\\
\therefore \boxed{\color{red} x+y+z= 70^o}

}

Sendo os ângulos x e y
\mathsf{90-x + 180 - y = 140^o \implies x+y = 130^o\\
180-(x+y) = 180 -130 = \boxed{\color{red}50^o}}

\mathsf{x; 180^o-x\\
(180-x)+35 = 8(x-35)\implies x = 55^o\\
\therefore180-55 = 125^o\\
\therefore \boxed{\color{red}125-55 = 70^o } }

\mathsf{
90-(180-(180-(90-x)=15\implies x =15^o\\
S = 180-(90-(90-(180-x) \implies \boxed{\color{red}S=x=15^o}
}

\mathsf{\angle MOC = z+z-y = 2z - y\\
\angle NOD = y\\
\therefore 2z-y-y = 83 \implies z-y = 41,5^o\\
mas ~\angle COL = z- y \therefore \boxed{\color{red}\angle COL = 41,5^o = 41^o30'}

}

Juntando as duas figuras e distribuindo os ângulos:
\mathsf{a+b = 3x\\
a+b+x+10+x = 180 \implies 5x = 170\\
\therefore \boxed{\color{red}x = 34^o }}

Prolongue a reta m. Veja que será formada um triângulo com ângulos \theta, 180°-x e 67°

\theta + 67° -x = 0°

O valor de x será inteiro se o valor de \theta for inteiro. Logo \theta = 51°...

\angle BOC - \angle AOB=60^o\\
a+x -(a-x) = 60\implies 2x=60\\
\therefore \boxed{\color{red}x = 30^o}

Traçando a bissetriz do ângulo \angle BOC , temos que \angle BOM = \angle MOC = 30° .
O ângulo \angle DOM = \angle DOB+\angle BOM = 80°+30° = \boxed{110°}

\mathsf{x - \frac{2}{3}(90-x) = 52 \implies \boxed{\color{red}x = 67,2^o}}

\mathsf{x - \frac{3(90-x)}{15}=6 \implies \boxed{\color{red}x=20^o}}

C_n = 90°-n . Logo:

2C_n = 180°-2n
C_{2{C_n}}90° - (180°-2n) \implies 2C_{2C_n} = 180° -360+4n
C_{2C_{2C_n}} = 90º - 180° +360°-4n

Daí:

180°-2n+90°-180°+2n+90°-180°+360°-4n = 160° \implies...

\mathsf{
3 =2 \implies 270 = 180+4x\\
\therefore \boxed{\color{red}x = \frac{90}{4} = 22,5^o=22^o30'}
}

Segue a imagem da solução....

\mathsf{
Prolongar ~EF~até~(B=EF \cap m)\\
\triangle DEB : \angle B = 2\theta\\
T.Sarrus :X = 2\theta+2\theta = 4\theta\\
EFGIJ: 180-4\theta+4\theta+90+180-4\theta+180-\theta = 540 \implies \theta =...

\mathsf{
x+ y=140\\
i)2(90-x) = 180-2x\\
ii)2y\\
180-2y\\
90-(180-2y) = -90+2y\\
90-(-90+2y) =180-2y\\
3(180-2y) = 540-6y \\
i=ii: 180-2x=540-6y \implies 6y-2x = 360\\
3y-x = 180\\...

\mathsf {
\frac{5}{6} (180-a)-(90-\frac{a}{2})=2(90-a)+\frac{a}{15}\\ \implies
a = 75^o \therefore 180-(90-75) =\boxed{\color{red}165^o}
}

\mathsf{
\angle BOH = a-c\\
a-c+a+b-2c=40 \implies 2a+b-3c=40(I)\\
\therefore 2(a-c)+b-c = 40\\
\therefore 2\angle BOH+b-c = 40^o(II)\\
mas~2a = 2b+2c-b-c \implies 2a=b+c\\
De(I):b+c+b-3c = 40...

\mathsf{
x - \frac{x}{4}+ = \frac{1}{3} \\
5x = 60 \therefore \boxed{\color{red}x = 12^o}

}

\mathsf{
x =(90-y)+60\implies x = 150-y(I)\\
y = \frac{180-x}{2} \implies 2y = 180 -x(II)\\
De(I)e(II): y = 30^o, x = 120^o\\
\therefore 90 - \frac{(120-30)}{2}=\boxed{\color{red}45^o}

}

\mathsf{

\angle XOC = 2a-b\\
\angle XOD = 2a+b\\
\angle BOZ = a-c\\
\therefore 2a-b+2a+b-4(a-c) = 80 \implies c = 20^o\\
\therefore \boxed{\color{red}\angle AOB = 2c = 40^o}
}

Seja A\hat{O}B = 2\alpha e B\hat{O}C = 2\theta .
Então, A\hat{O}P = P\hat{O}B = \alpha e B\hat{O}Q= C\hat{O}Q =\theta .
Como 2\alpha + 2\theta = 300° \implies \alpha+ \theta = 150° .

A\hat{O}R =...

OH é bissetria AOB

\mathsf{\angle HOE=?\\
HOD =180^o\\
\therefore
\angle HOB+\angle BOC +\angle COD=180^o\\
\frac{k}{2}+3k+4k = 180^o \implies k = 24\\
\angle HOE + \angle OED = 180^0\\
\angle OED...

\mathsf{\angle MVF = 63^o\\\angle KFJ =90-2\alpha-27 = 63^o-2\alpha\\
\angle KJF = 27+2\alpha \implies 4\alpha+3\alpha +27+2\alpha = 180^o\\
\therefore \alpha = 17^0\implies \angle KJF = 61^o,3\alpha...

Utilizando o teorema

COmo MN é paralelo a OA percemos que há uma volta completa portanto
\mathsf{15\alpha=360 \implies \boxed{\color{red}\alpha = 24^o}

}

\mathsf{
\frac{x}{3}=\frac{y}{5}=\frac{z}{7} =k\implies x+y+z = 15k\\
90-(15k) = 15 \implies k = 5 \\
\therefore 7.5 = \boxed{\color{red}35^o}
}