Question

In rombul ABCD cu diagonalele AC=16cm si BD=12cm determinati sin A si tg C

Answer 1

Am atasat rezolvarea

Answer 2

[tex]\it \mathcal{A} = \dfrac{AC\cdot BD}{2} = \dfrac{16\cdot12}{2}= 8\cdot12=96\ cm^2\ \ \ \ (*) \\ \\ \\ Fie\ AC\cap BD=\{O\} \Rightarrow \begin{cases} \it AO = \dfrac{AC}{2} = \dfrac{16}{2} =8\ cm \\ \\ \\ \it BO = \dfrac{BD}{2} = \dfrac{12}{2} =6\ cm \end{cases}[/tex]

Cu teorema lui Pitagora în ΔOAB ⇒ AB = 10 cm.

Sau:

Tripletul (6, 8, 10) este pitagoreic, deci AB = 10 cm.

AB = BC = CD = AD = 10 cm 


[tex]\it \mathcal{A} = AB\cdot AD \cdot sinA} \stackrel{(*)}{\Longrightarrow} \ 96=10\cdot10\cdot sinA \Rightarrow 96 = 100\cdot sinA \Rightarrow \\ \\ \\ \Rightarrow sinA = \dfrac{96^{(4}}{100} \Rightarrow sinA = \dfrac{24}{25}[/tex]

Unghiurile opuse ale rombului sunt congruente, rezultă:

$\it \hat{C} =\hat{A} \Rightarrow tgC = tgA=$

Ducem înălțimea BF a rombului, cu F pe AD.

[tex]\it \mathcal{A} = AD\cdot BF \stackrel{(*)}{\Longrightarrow} 96 = 10\cdot BF \Rightarrow BF = \dfrac{96}{10} = 9,6\ cm \\ \\ \\ \Delta FAB -dreptunghic,\ m(\hat{F}) =90^o \stackrel{T.Pitagora}{\Longrightarrow }\ \ \ AF^2= AB^2-BF^2 \Rightarrow \\ \\ \\ \Rightarrow AF^2 = 10^2-9,6^2 =100- 92,16 = 7,84 =\dfrac{\ \ 784^{(4}}{100}=\dfrac{196}{25} \Rightarrow \\ \\ \\ \Rightarrow AF = \dfrac{^{2)}14}{\ 5} = \dfrac{28}{10} =2,8[/tex]

$\it tgA = \dfrac{BF}{AF} = \dfrac{^{10)}9,6}{\ \ 2,8} =\dfrac{96^{(4}}{28} = \dfrac{24}{7}$