In Euclidean geometry, the sum of the three angles of atriangle is 180°. How can we use this property when calculating theangles of a triangle?I. Proving the property In figure 1, ABC is any triangle. Line (x y) is the lineparallel to (AC), passing through B.∠ B1 and ∠A are alternate internals. The angle is the same, since lines(x y) and (AC) are parallel. Therefore, ∠B1 = ∠A. Likewise, ∠B3 and ∠C are the same. Therefore, ∠B3 = ∠C.We know that ∠B1 + ∠B2 + ∠B3 = 180°, since ∠ xBy is a straight angle.From this we see that ∠A + ∠B + ∠C = 180° in triangle ABC
II. Calculating the anglesA. In any triangle Example: We want to calculate ∠A of triangle ABC. We apply the rule ∠A + 114° + 25° = 180°.From this, we have the calculations: ∠A + 139° = 180° and ∠A = 180° - 139° = 41°.
B. In a right-angled triangle The sum of the two acute angles in a right-angled triangle is 90°.Triangle ABC has a right angle at A.So: ∠A = 90°. Therefore, 90° + ∠B + ∠C = 180°, which means ∠B + ∠C = 90°. Example: We want to calculate ∠B of triangle ABC, shown in figure 3, which has a right angle at A. We apply the rule stated previously: ∠B + ∠C = 90°. From this, ∠B + 57° = 90° and ∠B = 90° - 57° = 33°. ∠B is 33°.
C. In an isosceles triangle Example: We want to calculate ∠A and ∠B of isoscelestriangle ABC. We apply the rule ∠A + ∠B + 48° = 180°. As ABC is anisosceles triangle in C, we know that ∠A = ∠B;therefore ∠A + ∠A + 48° = 180°, 2∠A + 48° = 180°,2∠A = 180°– 48° = 132°, and ∠A (and ∠B ) is 66°.
D .In an equilateral triangle The three angles of an equilateral triangle are each 60°. The angles are the same because the triangle is equilateral and their sum is 180°. Therefore, they are each degrees, i.e., 60°.