If the angles of a quadrilateral are 4x, 3x + 10°, 2x + 10º and 4x + 15°, then find the
angles.
Answers
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°
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Angles of quadrilateral ABCD are:→
∠A = 4x
∠B = 3x + 10°
∠C = 2x + 10°
∠D = 4x + 15°
So therefore,,
solutions
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 = 360° - 35°
→ 13x = 325°
→ x = 325°/13
→ x = 25°
∠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°