In triangle \(ABC\), angle \(BAC = (2x + 10)^\circ\), angle \(ABC = (x + 5)^\circ\) and angle \(ACB = (3x - 15)^\circ\). Side \(BC\) is extended to \(D\). A...

This question builds a full angle-sum proof for a triangle in algebraic form, then verifies the result using an independent parallel-line argument.

(a) Using the angle sum of triangle \(ABC\):

[3]

(b) Substituting \(x=30\): angle \(BAC = 2(30)+10 = 70^\circ\), angle \(ABC = 30+5 = 35^\circ\), angle \(ACB = 3(30)-15 = 75^\circ\). [2]

Check: \(70+35+75=180^\circ\).

(c) Angle \(ACD\) is the exterior angle at \(C\), on the straight line \(BD\) extended beyond \(C\), so it is supplementary to angle \(ACB\): \(180^\circ - 75^\circ = 105^\circ\). This agrees with the exterior angle theorem, which gives it directly as the sum of the two opposite interior angles: \(70^\circ + 35^\circ = 105^\circ\). [2]

(d) A line through \(A\) parallel to \(BC\) creates alternate angles with the transversal \(AB\); the angle alternate to angle \(ABC\) is equal to it, since alternate angles between parallel lines are equal, so it is \(35^\circ\). [2]

(e) At vertex \(A\), three angles now lie along the straight line formed by the parallel through \(A\): the angle alternate to \(B\) (\(35^\circ\)), angle \(BAC\) itself (\(70^\circ\)), and the angle alternate to \(C\) (\(75^\circ\)). Since these three angles lie on a straight line, they must sum to \(180^\circ\). But the alternate angles equal angle \(B\) and angle \(C\) exactly, so this sum is the same as angle \(A\) + angle \(B\) + angle \(C\), which proves the triangle's three angles must add to \(180^\circ\). [2]

Part (e) is the classic parallel-line proof of the triangle angle sum rule; it is worth understanding this construction (a line through one vertex, parallel to the opposite side) as the reason the \(180^\circ\) rule is always true, not just something to accept.