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Sagot :
since they are similar triangle, we have
[tex]\begin{gathered} \frac{CA}{AB}=\frac{YW}{WX} \\ \frac{28}{24}\text{ = }\frac{16}{WX} \\ 28WX\text{ = 24}\times16 \\ WX\text{ = }\frac{384}{28}\text{ = 16} \\ \end{gathered}[/tex]To solve for AD
[tex]\begin{gathered} \frac{CA}{AD}\text{ = }\frac{AY}{WZ} \\ \frac{28}{AD}=\frac{16}{12} \\ 16AD\text{ = 28}\times12 \\ AD=\frac{336}{16}\text{ = 21} \\ \end{gathered}[/tex]To solve for YZ
[tex]\begin{gathered} \frac{CA}{CD}\text{ =}\frac{YW}{YZ} \\ \frac{28}{14}\text{ = }\frac{16}{YZ} \\ 2\text{ =}\frac{16}{YZ} \\ 2YZ\text{ = 16} \\ YZ\text{ =}\frac{16}{2}\text{ = 8} \end{gathered}[/tex]To solve for XZ, I will have to get the angle, given, check out this diagram.
From the diagram, i have to find, the value of X using cosine rule, which becomes.
[tex]\begin{gathered} \text{Cos W =}\frac{y^2+z^2-w^2}{2yz} \\ CosW\text{ =}\frac{y^2+z^2-w^2}{2yz}=\frac{12^2+16^2-8^2}{2\text{ }\times16\times12}\text{ = }\frac{144\text{ +256 - 64}}{384}\text{ = }\frac{336}{384}\text{ = 0.875} \\ Cos\text{ W = 0.875} \\ W=cos^{-1}(0.875) \\ W\text{ =28.95 }\approx29^{\circ} \\ \end{gathered}[/tex]Then to find XZ? we shall use cosine rule
[tex]\begin{gathered} XZ^2=WZ^2+WX^2\text{ - 2(}WZ\text{)(}WX\text{) COSXZ} \\ XZ^2=12^2+16^2\text{ - 2}\times12\times16\text{ COS 29} \\ XZ^2\text{ = 144 + 256 -384(0.874)} \\ XZ^2\text{ = 400 - 335.6} \\ XZ^2\text{ = 64.384} \\ XZ\text{ = }\sqrt[]{64.384\text{ }}\text{ = 8cm} \end{gathered}[/tex]To find DB?
[tex]\begin{gathered} \frac{21}{DB}\text{ =}\frac{12}{8} \\ 12DB\text{ = 21 }\times8 \\ \text{DB =}\frac{168}{12}\text{ = 14} \end{gathered}[/tex]
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