The following diagram shows a triangle with each of its sides produced

The following diagram shows a triangle with each of its sides produced both ways :
What is the sum of degree measures of the angles numbered ?

720
540
1080
900
This question was previously asked in
UPSC CAPF – 2009
The sum of the degree measures of the six angles numbered in the diagram is 720 degrees.
– The diagram shows a triangle with each side produced both ways. This forms pairs of vertically opposite angles at each vertex, outside the triangle.
– Let the interior angles of the triangle be A, B, and C. The sum of the interior angles of a triangle is 180 degrees (A + B + C = 180°).
– At each vertex, the two angles formed by producing the sides are vertically opposite, and thus equal. Let these pairs be (α1, α2) at vertex A, (β1, β2) at vertex B, and (γ1, γ2) at vertex C. The numbered angles are {α1, α2, β1, β2, γ1, γ2}.
– Also, each of these angles forms a linear pair with the corresponding interior angle. For example, α1 + A = 180°, β1 + B = 180°, γ1 + C = 180°. Since vertically opposite angles are equal, α1 = α2, β1 = β2, γ1 = γ2.
– The sum of the numbered angles is S = α1 + α2 + β1 + β2 + γ1 + γ2 = 2(α1 + β1 + γ1).
– Substituting the linear pair relationships: S = 2((180 – A) + (180 – B) + (180 – C)) = 2(540 – (A + B + C)).
– Since A + B + C = 180°, S = 2(540 – 180) = 2(360) = 720°.
– The sum of the exterior angles of any convex polygon, taken one at each vertex, is always 360 degrees. The angles numbered in this question are not the standard set of exterior angles (one at each vertex). They are pairs of vertically opposite angles to the exterior angles and also pairs of vertically opposite angles to each other.
– The six angles consist of three pairs of vertically opposite angles. The sum of angles around each vertex point is 360 degrees. The angles around vertex A are A + α1 + α2 = 360°. Since α1 = α2, A + 2α1 = 360°, so 2α1 = 360 – A. Similarly, 2β1 = 360 – B and 2γ1 = 360 – C. The sum of the numbered angles is S = 2α1 + 2β1 + 2γ1 = (360 – A) + (360 – B) + (360 – C) = 1080 – (A + B + C) = 1080 – 180 = 900°. Oh, re-calculating using the full angle around the point method. This is incorrect as α1 and α2 form a pair. Let’s stick to the linear pair method which is more direct. α1 is exterior angle, α2 is vertically opposite to exterior angle. Sum is α1+α2+β1+β2+γ1+γ2. Since α1=α2, β1=β2, γ1=γ2, sum = 2(α1+β1+γ1). α1=180-A, β1=180-B, γ1=180-C. Sum = 2(180-A + 180-B + 180-C) = 2(540-(A+B+C)) = 2(540-180) = 2*360 = 720. Yes, 720 is correct.
– Another perspective: At each vertex, say A, the angle inside is A. The two angles formed outside by extending sides are vertically opposite, let them be x and y. A + x + y + angle along the produced line = 360? No. The angles around vertex A are A, and the two numbered angles there, say α1 and α2. There are two straight lines intersecting at A. Angle A is inside the triangle. The angles vertically opposite to the interior angles are inside the area bounded by the produced lines, let’s call them A’, B’, C’. A’=A, B’=B, C’=C. The angles numbered are those outside the triangle. At vertex A, the angles are α1 and α2. α1 and A form a linear pair with the angle on the other side of the line, which is vertically opposite to A. No, this is getting complicated. The first method is the clearest. α1 and α2 are vertically opposite, so α1 = α2. α1 and A form a linear pair only if α1 is the exterior angle. Yes, if the side is produced. Angle A + exterior angle E_A = 180. The two numbered angles at A are E_A and its vertical opposite. So the sum of the two numbered angles at A is 2 * E_A = 2 * (180 – A). Similarly, at B, 2 * (180 – B), and at C, 2 * (180 – C). The total sum is 2(180-A) + 2(180-B) + 2(180-C) = 540*2 – 2(A+B+C) = 1080 – 2*180 = 1080 – 360 = 720°. This re-confirms 720.
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