Question

If the angles of a quadrilateral are 4x, 3x + 10°, 2x + 10º and 4x + 15°, then find the
angles.​

Answer 1

Given that:

Angles of quadrilateral ABCD are:→

∠A = 4x

∠B = 3x + 10°

∠C = 2x + 10°

∠D = 4x + 15°

We know that:

Sum of all the four angles of a quadrilateral is equal to 360°.

i.e., ∠A + ∠B + ∠C + ∠D = 360°

According to the question:

→ 4x + 3x + 10° + 2x + 10º + 4x + 15° = 360°

→ 4x + 3x + 2x + 4x + 10° + 10° + 15° = 360°

→ 13x + 35° = 360°

→ 13x = 360° - 35°

→ 13x = 325°

→ x = 325°/13

→ x = 25°

∴ Angles of quadrilateral ABCD are:

∠A = 4x = 4 × 25° = 100°

∠B = 3x + 10° = 3 × 25° + 10° = 75° + 10° = 85°

∠C = 2x + 10° = 2 × 25° + 10° = 50° + 10° = 60°

∠D = 4x + 15° = 4 × 25° + 15° = 100° + 15° = 115°

Answer 2

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Angles of quadrilateral ABCD are:→

∠A = 4x

∠B = 3x + 10°

∠C = 2x + 10°

∠D = 4x + 15°

We know that:

Sum of all the four angles of a quadrilateral is equal to 360°.

i.e., ∠A + ∠B + ∠C + ∠D = 360°

→ 4x + 3x + 10° + 2x + 10º + 4x + 15° = 360°

→ 4x + 3x + 2x + 4x + 10° + 10° + 15° = 360°

→ 13x = 360° - 35°

→ 13x = 325°

→ x = 325°/13

→ x = 25°

∴ Angles of quadrilateral ABCD are:

∠A = 4x = 4 × 25° = 100°

∠B = 3x + 10° = 3 × 25° + 10° = 75° + 10° = 85°

∠C = 2x + 10° = 2 × 25° + 10° = 50° + 10° = 60°

∠D = 4x + 15° = 4 × 25° + 15° = 100° + 15° = 115°