Find the angles of a rhombus if the ratio of the angles that the diagonals form with a side is 6: 9.

The measures of the angles of the rhombus are 72° , 108° , 72° , 108°

Step-by-step explanation:

* Lets revise the properties of the rhombus

- It has 4 equal sides

- Each two opposite angles are equal

- Each two adjacent angles are supplementary (their sum = 180°)

- The diagonals are perpendicular to each other

- The diagonals bisect the vertices angles (divide each vertex into two

 equal parts)

* Now lets solve the problem

- Let the name of the rhombus is ABCD

- The two diagonals are AC and BD

- Let the diagonals AC and BD make the angles BAC and ABD with

 side AB

∵ The diagonals of the rhombus bisect the vertex angles

∴ m∠BAC is half m∠A

∴ m∠ABD is half m∠B

- There is a ratio between the measures of the angles BAC and ABD

∵ m∠BAC : m∠ABD = 6 : 9

∵ m∠A + m∠B = 180° ⇒ consecutive angles

∴ m∠BAC + m∠ABD = 1/2 × 180° = 90°

∵ m∠BAC : m∠ABD : their sum

         6     :       9      : 15⇒(6 + 9)

          ?    :        ?      : 90° ⇒(the sum of the angles)

∴ m∠BAC = 6/15 × 90° = 36°

∴ m∠ABD = 9/15 × 90° = 54°

∵ m∠BAC is half m∠A

∴ m∠A = 36° × 2 = 72°

∵ m∠ABD is half m∠B

∴ m∠B = 54° × 2 = 108°

* The measures of the angles of the rhombus are 72° , 108° , 72° , 108°

it could 14.28 but if its not that one then it has to be the 17 one. hope this ! : )

step-by-step explanation: