A rhombus 4

A rhombus has a side length of 10 cm. Find the angles at each corner of the rhombus if the shorter of the two diagonals measures 7 cm. Give your answers to the nearest degree and give clear geometric reasoning at each stage of your solution.

Final Answer:

α =  139 °
β =  41 °
γ =  139 °
δ =  41 °

Step-by-step explanation:

$$\begin{gathered} a=10\text{cm} \\ d=7\text{cm} \\ {a}^2=(d/2{)}^2+(f/2{)}^2 \\ f=2\cdot \sqrt{a^2-(d/2{)}^2}=2\cdot \sqrt{10^2-(7/2{)}^2}=3\sqrt{39}\text{cm}\doteq 18.735\text{cm} \\ \sin \phi =d/2:a \\ \phi =\frac{180°}{\pi }\cdot \arcsin ((f/2)/a)=\frac{180°}{\pi }\cdot \arcsin ((18.735/2)/10)\doteq 69.5127{}^{\circ } \\ \alpha =2\cdot \phi =2\cdot 69.5127=139^{\circ }=139°1^{\prime }31" \end{gathered}$$

$\beta =180-\alpha =180-139.0254=41^{\circ }=40°58^{\prime }29"$

$\gamma =\alpha =139.0254=139^{\circ }$

$\delta =\beta =40.9746=41^{\circ }$

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