Question

The angles of a quadrilateral are (4x – 15)°, 3x°, (2x + 5)° & (x + 40)°. Find the measure

of each angle of the quadrilateral.​

Answer 1

Answer:

$we \: know$

Sum of angles of a quadrilateral = 360 degree

Here, 1st angle = ( 4x - 15 )°

2nd angle = 3x°

3rd angle = ( 2x + 5 )°

4th angle = ( x + 40 )°

$(4x - 15) + 3x + (2x + 5) + (x + 40) = 360 \\ \\ = > 4x - 15 + 3x + 2x + 5 + x + 40 = 360 \\ \\ = > 10x - 15 + 5 + 40 = 360 \\ \\ = > 10x - 15 + 45 = 360 \\ \\ = > 10x + 30 = 360 \\ \\ = > 10x = 360 - 30 \\ \\ = > 10x = 330 \\ \\ = > x = \frac{330}{10} \\ \\ = > x = 33 \: degree$

So putting the value of x in all the angles we get :

$1 ^{st} \: angle \: 4x - 15 \\ \\ = 4(33) - 15 \\ \\ = 132 - 15 \\ \\ = 117 \\ \\ {2}^{nd} \: angle \: 3x \\ \\ = 3 \times 33 \\ \\ = 99 \\ \\ {3}^{rd} \: angle \: 2x + 5 \\ \\ = 2 \times 33 + 5 \\ \\ = 66 + 5 \\ \\ = 71 \\ \\ {4}^{th} \: angle \: x + 40 \\ \\ = 33 + 40 \\ \\ = 73$

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Answer 2

Answer:

(4x – 15)° + (3x)° + (2x + 5)° + (x + 40)° = 360° (Sum of angles of a quadrilateral is 360°)

4x – 15 + 2x + 5 + 3x + x + 40 = 360

10x + 30 = 360

10x = 360 - 30

10x = 330

x = $\frac{330}{10}$

x = 33°

∠1st = 4 × 33 - 15 = 117°

∠2nd = 3 × 33 = 99°

∠3rd = 2 × 33 + 5 = 71°

∠4th = 33 + 40 = 73°