In the diagram, \(ABC\) is a triangle and \(BCD\) is a straight line. Angle \(BAC=(x+15)^{\circ}\), angle \(ABC=(2x-10)^{\circ}\) and the exterior angle \(A...

The key geometry fact is the exterior angle theorem: the exterior angle of a triangle equals the sum of the two interior angles not adjacent to it. Here \(ACD\) is the exterior angle at \(C\), and the two opposite interior angles are \(BAC\) and \(ABC\).

(a) Show that \(x=30\)

1. Apply the theorem: \(4x-25=x+15+2x-10\) [M1] for equating the exterior angle to the sum of the two opposite interior angles.
2. Collect like terms on the right: \(x+2x=3x\) and \(15-10=+5\), giving \(4x-25=3x+5\) [M1].
3. Subtract \(3x\) from both sides: \(x-25=5\), then add 25 to both sides: \(x=30\) [A1].

An equally valid route uses angles on a straight line at \(C\): angle \(ACB=180-(4x-25)=205-4x\), and the three interior angles sum to \(180^{\circ}\). That gives the same value of \(x\).

(b) Substitute \(x=30\) into each expression.

The third angle is found either from the angle sum of the triangle, \(180-45-50=85\), or from the straight line \(BCD\), since the exterior angle is \(4(30)-25=95^{\circ}\) and \(180-95=85^{\circ}\). Both give \(85^{\circ}\), and the check \(45+50+85=180\) confirms the work.

The frequent error is treating \(ACD\) as if it were an interior angle and including it in the \(180^{\circ}\) sum. Only \(BAC\), \(ABC\) and \(ACB\) are interior to the triangle.