# A straight line is drawn through the vertex B of the triangle ABC, parallel to the line AC.

**A straight line is drawn through the vertex B of the triangle ABC, parallel to the line AC. The resulting three angles with the vertex B are in the ratio 3: 10: 5. Find the angles of the triangle ABC.**

Let’s make the equation

3x + 10x + 5x = 180 (because the sum of those 3 angles is 180 degrees)

18x = 180

x = 180 ÷ 18 = 10

Hence

Angle ABC = 10 × x = 10 × 10 = 100 (degrees)

Find the sum of the bottom corners

Angle BAC + angle BAC = 180-100 = 80 (degrees)

Let’s finish the line AC (we know that it is parallel to the line from the top B)

Hence

СВН angle (admissible) = ACВ angle (because they are intersecting with ВН || AC and secant BC)

angle СВН = angle АСВ = 5x = 5 × 10 = 50 (degrees)

And we’ll find the last corner BAC

Angle BAC = 180- (100 + 50) = 30 (Degrees)

Answer: 100,50,30