Question

Triangle ABC ~ triangle WXY

Answer 1

since they are similar triangle, we have

`\begin{gathered} (CA)/(AB)=(YW)/(WX) \\ (28)/(24)\text{ = }(16)/(WX) \\ 28WX\text{ = 24}*16 \\ WX\text{ = }(384)/(28)\text{ = 16} \\ \end{gathered}`

To solve for AD

`\begin{gathered} (CA)/(AD)\text{ = }(AY)/(WZ) \\ (28)/(AD)=(16)/(12) \\ 16AD\text{ = 28}*12 \\ AD=(336)/(16)\text{ = 21} \\ \end{gathered}`

To solve for YZ

`\begin{gathered} (CA)/(CD)\text{ =}(YW)/(YZ) \\ (28)/(14)\text{ = }(16)/(YZ) \\ 2\text{ =}(16)/(YZ) \\ 2YZ\text{ = 16} \\ YZ\text{ =}(16)/(2)\text{ = 8} \end{gathered}`

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.

`\begin{gathered} \text{Cos W =}(y^2+z^2-w^2)/(2yz) \\ CosW\text{ =}(y^2+z^2-w^2)/(2yz)=\frac{12^2+16^2-8^2}{2\text{ }*16*12}\text{ = }\frac{144\text{ +256 - 64}}{384}\text{ = }(336)/(384)\text{ = 0.875} \\ Cos\text{ W = 0.875} \\ W=cos^(-1)(0.875) \\ W\text{ =28.95 }\approx29^(\circ) \\ \end{gathered}`

Then to find XZ? we shall use cosine rule

`\begin{gathered} XZ^2=WZ^2+WX^2\text{ - 2(}WZ\text{)(}WX\text{) COSXZ} \\ XZ^2=12^2+16^2\text{ - 2}*12*16\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}`

To find DB?

`\begin{gathered} (21)/(DB)\text{ =}(12)/(8) \\ 12DB\text{ = 21 }*8 \\ \text{DB =}(168)/(12)\text{ = 14} \end{gathered}`