Question

The lengths of the diagonals of a rhombus are 4.73 and 2.94. What are the measures of its angles? Round to the second decimal place.

Answer 1

116.27° and 63.73°

Explanation:

In a rhombus, the diagonals bisect each other at right angles, forming four congruent right-angled triangles.

If the diagonals of a rhombus are 4.73 and 2.94, then the legs of each congruent right triangle are equal to half the diagonals:

`\textsf{Leg\;1}=(4.73)/(2)=2.365`

`\textsf{Leg\;2}=(2.94)/(2)=1.47`

In a rhombus, opposite angles are equal due to the symmetry of the shape. Each angle in a rhombus is equal to twice the measure of an acute angle in the right-angled triangles formed by its diagonals. Therefore, to find the measures of the angles of the rhombus, we can substitute the legs of the right triangle into the tangent trigonometric ratio.

`\boxed{\begin{array}{l}\underline{\textsf{Tangent trigonometric ratio}}\\\\\sf \tan(\theta)=(O)/(A)\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$\theta$ is the angle.}\\\phantom{ww}\bullet\;\textsf{$O$ is the side opposite the angle.}\\\phantom{ww}\bullet\;\textsf{$A$ is the side adjacent the angle.}\end{array}}`

Angle 1

`\tan \theta_1=(2.365)/(1.47)`

`\theta_1=\arctan\left((2.365)/(1.47) \right)`

`\theta_1=58.136383753...^(\circ)`

Therefore, angle 1 is:

`2\cdot \theta_1=2\cdot 58.136383753...^(\circ)=\boxed{116.27^(\circ)\;\sf (2\;d.p.)}`

Angle 2

`\tan \theta_2=\frac{1.47} {2.365}`

`\theta_2=\arctan\left(\frac{1.47} {2.365} \right)`

`\theta_2=31.86361624...^(\circ)`

Therefore, angle 2 is:

`2\cdot \theta_2=2\cdot 31.86361624...^(\circ)=\boxed{63.73^(\circ)\;\sf (2\;d.p.)}`