Hello, beefsupreme!

The angles of a triangle are 50°, 60° and 70°,

and a circle touches the sides at $\displaystyle A, B, C.$

Calculate the angles of triangle $\displaystyle ABC$. Code:

P
o
/ \
/50°\
/ \
/ \
/ \
/ * * * \
/* *\
* *
A o- - - - - - - - -o B
/ \ / \
/* \ / *\
/ * \ o / * \
/ * \ O / * \
/ \ / \
/ * \ / * \
/ * \ / * \
/ 60° * \ / * 70° \
Q o - - - - - - - * o * - - - - - - - o R
C

We have $\displaystyle \Delta PQR$ with $\displaystyle \angle P = 50^o,\,\angle Q = 60^o,\:\angle R = 70^o$

Circle $\displaystyle O$ is inscribed in the triangle

. . and is tangent to $\displaystyle PQ, PR, QR$ at $\displaystyle A,B,C,$ respectively.

Draw segments $\displaystyle OP, OQ, OR$ and radii $\displaystyle OA, OB, OC.$

$\displaystyle OA,OB,OC$ are penpendicular to the sides of the triangle.

Since O is the center of the inscribed circle,

. . $\displaystyle OP, OQ, OR$ bisect $\displaystyle \angle P, \angle Q, \angle R$, resp.

We have: .$\displaystyle \angle APO = 25^o \quad\Rightarrow\quad \angle POA = 65^o \quad\Rightarrow\quad \angle AOB = 130^o$

Since $\displaystyle \Delta AOB$ is isosceles: .$\displaystyle \angle OAB = \angle OBA = 25^o$

Similarly, we find that: .$\displaystyle \angle OAC = \angle OCA = 30^o,\:\angle OBC = \angle OCB = 35^o $

Therefore: .$\displaystyle \angle A = 55^o,\:\angle B = 60^o,\:\angle C = 65^o$