Question

In a ΔАВС, ∠C = 3 ∠B = 2(∠A + ∠B). Find all angles of ΔАВС.

Answer 1

Given: In triangle ABC, ∠C = 3 ∠B = 2(∠A + ∠B).

Step-wise calculation:

1. Let ∠B = x.

2. From the given, ∠C = 3x.

3. Also given, ∠C = 2(∠A + ∠B) = 2(∠A + x).

4. Equate the two expressions for ∠C:

3x = 2(∠A + x)

5. Expand the right side:

3x = 2∠A + 2x

6. Rearranged:

3x – 2x = 2∠A

x = 2∠A

7. So, ∠B = x and `∠A = x/2`.

8. Sum of angles in triangle:

∠A + ∠B + ∠C = 180°

9. Substitute values:

`(x/2) + x + 3x = 180^circ`

10. Simplify:

`(1/2)x + x + 3x = 180^circ`

`(1/2 + 1 + 3)x = 180^circ`

(4.5)x = 180°

11. Solve for x:

`x = 180/4.5`

x = 40°

12. Find other angles:

`∠A = x/2`

`∠A = 40^circ/2`

∠A = 20° 

∠B = x

∠B = 40°

∠C = 3x

∠C = 3 × 40° 

∠C = 120°

The angles of triangle ABC are ∠A = 20°, ∠B = 40°, ∠C = 120°.