One angle of a rhombus measures 110, and the shorter diagonal is 4 inches long. How long is the side of the rhombus?

Respuesta :

side of rhombus: 2.44 inch

Use the sine rule:

[tex]\sf \dfrac{sin(A)}{a} = \dfrac{sin(B)}{b}[/tex]

=============                       Let the side be "b"

[tex]\rightarrow \sf \dfrac{sin(110)}{4} = \dfrac{sin(35)}{b}[/tex]

[tex]\rightarrow \sf b = \dfrac{sin(35)*4}{sin(110)}[/tex]

[tex]\rightarrow \sf b = 2.441549178[/tex]

[tex]\rightarrow \sf b =2.44 \ in[/tex]

Ver imagen fieryanswererft

Answer:

3.5 in (nearest tenth)

Step-by-step explanation:

Properties of a rhombus:

  • Quadrilateral (four sides & four interior angles)
  • Parallelogram (opposite sides are parallel)
  • All sides are equal in length
  • Opposite angles are equal in measure
  • Diagonals bisect each other at right angles
  • Interior angles sum to 360°
  • Adjacent angles are supplementary (sum to 180°)
  • Diagonals bisect interior angles

Therefore, a rhombus is made up of 4 congruent right triangles.

** see attached diagram **

To find the side length of the rhombus, we need to calculate the hypotenuse of the right triangle.

As the shorter diagonal is 4 in, the base of the right triangle is 2 in

The angles that measure 110° are the angles by the shorter diagonal.  Therefore, the base angle of the right triangle is 55°

Using cos trig ratio:

[tex]\sf \cos(\theta)=\dfrac{A}{H}[/tex]

where:

  • [tex]\theta[/tex] is the angle
  • A is the side adjacent the angle
  • H is the hypotenuse

Given:

  • [tex]\theta[/tex] = 55°
  • A = 2
  • H = x

[tex]\implies \sf \cos(55^{\circ})=\dfrac{2}{x}[/tex]

[tex]\implies \sf x=\dfrac{2}{\cos(55^{\circ})}[/tex]

[tex]\implies \sf x=3.486893591...[/tex]

Therefore, the side of the rhombus is 3.5 in (nearest tenth)

Ver imagen semsee45