ABCD rhombus

ABCD is a rhombus with sides 10.5cm. If the length of the diagonal AC=15.8cm, using cosine formula.

a. calculate the length of the diagonal BD correct to the nearest cm
b. the angles of the rhombus to the nearest degree​.

Final Answer:

BD =  14 cm
α =  82 °
β =  98 °

Step-by-step explanation:

$$\begin{gathered} AB=10.5\text{cm} \\ BC=AB=10.5=\frac{21}{2}=10.5\text{cm} \\ AD=BC=10.5=\frac{21}{2}=10.5\text{cm} \\ CD=AB=10.5=\frac{21}{2}=10.5\text{cm} \\ AC=15.8\text{cm} \\ A{B}^2=(AC/2{)}^2+(BD/2{)}^2 \\ A{B}^2=A{C}^2/4+B{D}^2/4 \\ 4A{B}^2=A{C}^2+B{D}^2 \\ BD=\sqrt{4\cdot A{B}^2-A{C}^2}=\sqrt{4\cdot 10.5^2-15.8^2}=14\text{cm} \end{gathered}$$

$$\begin{gathered} B{D}^2=A{B}^2+A{D}^2-2\cdot AB\cdot AD\cdot \cos \alpha \\ B{D}^2=2A{B}^2-2\cdot A{B}^2\cdot \cos \alpha \\ A=\arccos (\frac{2\cdot A{B}^2-B{D}^2}{2\cdot A{B}^2})=\arccos (\frac{2\cdot 10.5^2-13.8333^2}{2\cdot 10.5^2})\doteq 1.4383\text{rad} \\ \alpha =A\to \text{°}=A\cdot \frac{180}{\pi }\text{°}=1.4383\cdot \frac{180}{\pi }\text{°}=82\text{°}=82^{\circ }=82°24^{\prime }21" \end{gathered}$$

$$\begin{gathered} B=\arccos (\frac{2\cdot A{B}^2-A{C}^2}{2\cdot A{B}^2})=\arccos (\frac{2\cdot 10.5^2-15.8^2}{2\cdot 10.5^2})\doteq 1.7033\text{rad} \\ \beta =B\to \text{°}=B\cdot \frac{180}{\pi }\text{°}=1.7033\cdot \frac{180}{\pi }\text{°}=98\text{°}=98^{\circ }=97°35^{\prime }39" \end{gathered}$$

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