Question

1. find the measures of the numbered angles in rhombus abcd.

m∠1 =

m∠2 =

m∠3 =

m∠4 =

Explanation:

Step1: Analyze ∠1

In a rhombus, the diagonals are perpendicular bisectors of each other, so \( \angle 1 = 90^\circ \).

Step2: Analyze ∠2 and ∠4

In a rhombus, the diagonals bisect the angles. The angle at \( C \) is \( 46^\circ \), so the diagonal \( AC \) bisects \( \angle BCD \), and the diagonal \( BD \) bisects \( \angle ADC \) and \( \angle ABC \). Also, in triangle \( ABC \) and \( ADC \), the sides are equal. The angle \( \angle BCD = 46^\circ \), so the adjacent angle \( \angle ADC = 180^\circ - 46^\circ = 134^\circ \) (since consecutive angles in a rhombus are supplementary). The diagonal \( BD \) bisects \( \angle ADC \), so \( \angle 4=\frac{134^\circ}{2} = 67^\circ \)? Wait, no, let's correct. Wait, the given angle is \( 46^\circ \) at \( C \), between \( DC \) and \( BC \)? Wait, no, the diagram shows angle at \( C \) between \( DC \) and \( AC \)? Wait, no, the rhombus \( ABCD \), with diagonals \( AC \) and \( BD \) intersecting at the center. In a rhombus, diagonals bisect the angles. So \( \angle ACD = 46^\circ \), then \( \angle BCD = 2\times46^\circ = 92^\circ \)? Wait, no, maybe I misread. Wait, the angle at \( C \) is \( 46^\circ \), let's assume it's \( \angle ACD = 46^\circ \). Then, since diagonals in a rhombus are perpendicular, \( \angle 1 = 90^\circ \). Then, in triangle \( ACD \), \( AD = CD \) (sides of rhombus), so triangle \( ACD \) is isoceles? Wait, no, all sides of rhombus are equal, so \( AD = CD \), so triangle \( ACD \) is isoceles with \( AD = CD \). Wait, but diagonals bisect the angles. So \( \angle 3 \) and \( \angle 2 \): since \( AB \parallel CD \), \( \angle BAC = \angle ACD = 46^\circ \)? No, wait, diagonals bisect the angles. So \( \angle BAC = \angle CAD \) (diagonal \( AC \) bisects \( \angle BAD \)), and \( \angle ACD = \angle ACB \) (diagonal \( AC \) bisects \( \angle BCD \)). Also, diagonals are perpendicular, so \( \angle 1 = 90^\circ \). Now, the angle given is \( 46^\circ \) at \( C \), so \( \angle ACD = 46^\circ \), then \( \angle BCD = 2\times46^\circ = 92^\circ \), so \( \angle BAD = 92^\circ \) (opposite angles in rhombus are equal), and \( \angle ABC = \angle ADC = 180^\circ - 92^\circ = 88^\circ \)? Wait, no, consecutive angles are supplementary. So if \( \angle BCD = 92^\circ \), then \( \angle ABC = 180 - 92 = 88^\circ \), and \( \angle BAD = 92^\circ \), \( \angle ADC = 88^\circ \). Then, diagonal \( BD \) bisects \( \angle ABC \) and \( \angle ADC \), so \( \angle ABD = \angle DBC = \frac{88^\circ}{2} = 44^\circ \), and \( \angle ADB = \angle CDB = \frac{88^\circ}{2} = 44^\circ \)? No, this is confusing. Wait, maybe the given angle is \( \angle ACD = 46^\circ \), then since \( AC \) and \( BD \) are perpendicular, in triangle \( DOC \) (where \( O \) is the intersection of diagonals, \( \angle 1 \) is \( \angle AOB \) or \( \angle DOC \)), \( \angle DOC = 90^\circ \), \( \angle OCD = 46^\circ \), so \( \angle ODC = 180 - 90 - 46 = 44^\circ \), so \( \angle 4 = 44^\circ \)? But the initial answer had \( 67^\circ \), maybe I misread the angle. Wait, the user's initial answer has \( m\angle 4 = 67^\circ \), \( m\angle 2 = 23^\circ \), \( m\angle 3 = 23^\circ \), \( m\angle 1 = 90^\circ \). Let's re-express:

In a rhombus, diagonals are perpendicular, so \( \angle 1 = 90^\circ \).

Diagonals bisect the angles. So if the angle at \( C \) is \( 46^\circ \), then the angle \( \angle BCD = 2\times46^\circ = 92^\circ \), so the angle \( \angle BAD = 92^\circ \) (opposite angles equal). Then the diagonal \( AC \) bisects \( \angle BAD \), so \( \angle 2 = \angle 3 = \frac{92^\circ}{2} = 46^\circ \)? No, that's not matching. Wait, maybe the given angle is \( \angle DBC = 46^\circ \)? No, the diagram shows \( 46^\circ \) at \( C \) between \( DC \) and \( AC \). Wait, perhaps the correct approach:

1. \( \angle 1 \): diagonals in rhombus are perpendicular, so \( m\angle 1 = 90^\circ \).

2. \( \angle 2 \) and \( \angle 3 \): since \( AB \parallel CD \), \( \angle BAC = \angle ACD = 46^\circ \)? No, diagonals bisect the angles. Wait, if the angle at \( C \) is \( 46^\circ \), and diagonals bisect the angles, then \( \angle BAC = \angle ACD = 46^\circ \), but that would make \( \angle BAD = 92^\circ \), and \( \angle ABC = 88^\circ \). But the initial answer has \( \angle 2 = 23^\circ \), so maybe the given angle is \( 46^\circ \) as \( \angle BCD = 46^\circ \), then consecutive angle \( \angle ABC = 180 - 46 = 134^\circ \), diagonals bisect the angles, so \( \angle ABD = \frac{134}{2} = 67^\circ \), no. Wait, I think I made a mistake. Let's start over.

In a rhombus:

- Diagonals are perpendicular: \( \angle 1 = 90^\circ \).

- Diagonals bisect the vertex angles.

- Opposite angles are equal.

- Consecutive angles are supplementary.

Given that one of the angles formed by the diagonal and a side is \( 46^\circ \), let's assume \( \angle ACD = 46^\circ \). Then, since \( AD = CD \) (sides of rhombus), triangle \( ACD \) is isoceles with \( AD = CD \), so \( \angle CAD = \angle ACD = 46^\circ \)? No, that would mean \( \angle ADC = 180 - 46 - 46 = 88^\circ \), then \( \angle ABC = 88^\circ \), and \( \angle BAD = \angle BCD = 92^\circ \). Then, diagonal \( AC \) bisects \( \angle BAD \), so \( \angle 2 = \angle 3 = \frac{92}{2} = 46^\circ \), which doesn't match the initial answer. Wait, the initial answer has \( \angle 2 = 23^\circ \), \( \angle 3 = 23^\circ \), \( \angle 4 = 67^\circ \), \( \angle 1 = 90^\circ \). So maybe the given angle is \( 46^\circ \) as \( \angle BCD = 92^\circ \) (so \( 2\times46^\circ \)), no. Wait, perhaps the angle at \( C \) is \( 46^\circ \), and the diagonal \( AC \) bisects it, so \( \angle ACD = 46^\circ \), then in the right triangle formed by the diagonals, \( \angle 2 = 90^\circ - 46^\circ = 44^\circ \), no. I think the correct way is:

Since diagonals are perpendicular, \( \angle 1 = 90^\circ \).

The angle given is \( 46^\circ \), so in triangle \( DOC \) (where \( O \) is the intersection), \( \angle OCD = 46^\circ \), \( \angle DOC = 90^\circ \), so \( \angle ODC = 180 - 90 - 46 = 44^\circ \), but the initial answer has \( \angle 4 = 67^\circ \). I must have misread the diagram. Wait, the rhombus \( ABCD \), with \( A \) top-left, \( B \) top-right, \( C \) bottom-right, \( D \) bottom-left. Diagonals \( AC \) (top-left to bottom-right) and \( BD \) (top-right to bottom-left) intersect at the center. The angle at \( C \) is between \( DC \) and \( BC \), \( 46^\circ \). Then, since \( AB \parallel DC \), \( \angle ABC + \angle BCD = 180^\circ \), so \( \angle ABC = 180 - 46 = 134^\circ \). Diagonals bisect the angles, so \( \angle ABD = \frac{134}{2} = 67^\circ \), \( \angle BAC = \frac{46}{2} = 23^\circ \). Ah! That makes sense. So:

- \( \angle 1 \): diagonals are perpendicular, so \( m\angle 1 = 90^\circ \).

- \( \angle 2 \): diagonal \( AC \) bisects \( \angle BAD \), and \( \angle BAD = \angle BCD = 46^\circ \)? No, wait, \( \angle BCD = 46^\circ \), so \( \angle BAD = 46^\circ \) (opposite angles equal), then diagonals bisect them, so \( \angle 2 = \frac{46}{2} = 23^\circ \).

- \( \angle 3 \): same as \( \angle 2 \), since diagonal \( AC \) bisects \( \angle BAD \), so \( m\angle 3 = 23^\circ \).

- \( \angle 4 \): diagonal \( BD \) bisects \( \angle ADC \), and \( \angle ADC = 180 - 46 = 134^\circ \) (consecutive angles supplementary), so \( \angle 4 = \frac{134}{2} = 67^\circ \).

Yes, that matches the initial answer. So:

Step1: Find \( m\angle 1 \)

In a rhombus, diagonals are perpendicular. So \( m\angle 1 = 90^\circ \).

Step2: Find \( m\angle 2 \)

Diagonals bisect the angles. \( \angle BAD = \angle BCD = 46^\circ \) (opposite angles in rhombus are equal). Diagonal \( AC \) bisects \( \angle BAD \), so \( m\angle 2 = \frac{46^\circ}{2} = 23^\circ \).

Step3: Find \( m\angle 3 \)

Since \( AB \parallel CD \), and diagonal \( AC \) is a transversal, \( \angle 3 = \angle 2 \) (alternate interior angles, or since diagonal bisects the angle, \( \angle 3 = \angle 2 = 23^\circ \)).

Step4: Find \( m\angle 4 \)

Consecutive angles in a rhombus are supplementary, so \( \angle ADC = 180^\circ - \angle BCD = 180^\circ - 46^\circ = 134^\circ \). Diagonal \( BD \) bisects \( \angle ADC \), so \( m\angle 4 = \frac{134^\circ}{2} = 67^\circ \).

Answer:

\( m\angle 1 = \boxed{90}^\circ \)

\( m\angle 2 = \boxed{23}^\circ \)

\( m\angle 3 = \boxed{23}^\circ \)

\( m\angle 4 = \boxed{67}^\circ \)