Geometry · Angle chasing · Isosceles triangles · Circumcentre · Inscribed angle theorem · 30 60 90 triangle

Problem 2, 2008

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NationalEnter the answer

In a triangle \(ABC\) we have \(\angle ABC = 45^\circ\) and \(\angle CAB = 15^\circ\). Let \(M\) be the point of the ray \(BC\) for which

\[ \overrightarrow{BM} = 3 \cdot \overrightarrow{BC} . \]

Determine the angles of the triangle \(ABM\).

A B C M
The marked angles are \(\angle ABC = 45^\circ\) and \(\angle CAB = 15^\circ\). The point \(M\) lies on the ray \(BC\) beyond \(C\), with \(BM = 3 \cdot BC\).

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Serbian National Competition (Drzavno takmicenje) 2008, high school grade I, category A, problem 2. Organized by the Mathematical Society of Serbia (DMS). Source