In triangle ABC, bisectors AN and BL are drawn which intersect at point O. Angle AOB is 170.

In triangle ABC, bisectors AN and BL are drawn which intersect at point O. Angle AOB is 170. Find the outer angle at vertex C.

The angles of any triangle add up to 180 °.

<BAO + <OBA + <AOB = 180 °;

<BAO + <OBA = 180 ° – <AOB = 180 ° – 170 ° = 10 °;

The bisector divides the angle in half:

<BAO = <BAC / 2; <BAC = 2 * <BAO;

<ABO = <ABC / 2; <ABC = 2 * <ABO;

<BAC + <ABC + <C = 180 °

<C = 180 ° – (<BAC + <ABC) = 180 – (2 * <BAO + 2 * <ABO) =

= 180 ° – 2 * (<BAO + <ABO) = 180 ° – 2 * 10 ° = 160 °.

Answer: <C = 160 °.