To find the missing angles in the rhombus, let's break down the problem step-by-step:

Properties of a Rhombus:

1. All sides are equal.
2. Diagonals bisect each other at right angles (90°).
3. Diagonals bisect the angles at each vertex.

Given that one angle $m\angle 1$ is labeled as $55^\circ$, we can use the following steps:

Step 1: Find $m\angle 1$.

Since $m\angle 1 = 55^\circ$ is already given, we can proceed to find the other angles.

Step 2: Find $m\angle 2$.

The diagonals of the rhombus intersect at $90^\circ$, so $m\angle 2$ is part of the same right triangle as $m\angle 1$ and the right angle.

In this right triangle: $m\angle 1 + m\angle 2 = 90^\circ$ $55^\circ + m\angle 2 = 90^\circ$ $m\angle 2 = 90^\circ - 55^\circ = 35^\circ$

Step 3: Find $m\angle 3$.

Since diagonals bisect opposite angles of a rhombus, $m\angle 3$ is the supplement of $2 \times m\angle 1$, because the opposite angles in a rhombus are equal, and the sum of adjacent angles is $180^\circ$.

So: $m\angle 3 = 180^\circ - 2 \times m\angle 1$ $m\angle 3 = 180^\circ - 2 \times 55^\circ$ $m\angle 3 = 180^\circ - 110^\circ = 70^\circ$

Conclusion:

• $m\angle 1 = 55^\circ$
• $m\angle 2 = 35^\circ$
• $m\angle 3 = 70^\circ$

Would you like further details or have any questions?

Here are some related questions to expand your understanding:

1. How do the properties of a rhombus help in solving geometric problems?
2. What is the significance of diagonal bisection in a rhombus?
3. How do you find the area of a rhombus using its diagonals?
4. How do supplementary angles relate to the properties of quadrilaterals?
5. What are other types of quadrilaterals where diagonals bisect each other?

Tip: Remember that in any rhombus, diagonals always bisect the angles and meet at right angles, which can simplify many geometric calculations.