1. Name – Ashwani Kumar

2. In Euclidean geometry, the sum of the three angles of a
triangle is 180°. How can we use this property when calculating the
angles of a triangle?
I. Proving the property
In figure 1, ABC is any triangle. Line (x y) is the line
parallel 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

3. II. Calculating the angles
A. 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°.

4. 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°.

5. C. In an isosceles triangle
Example: We want to calculate ∠A and ∠B of isosceles
triangle ABC.
We apply the rule ∠A + ∠B + 48° = 180°. As ABC is an
isosceles 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°.

6. 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°.