Answer:
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Explanation:
From the given image of the triangle, let's go ahead and find HL by taking the sine of angle 45 degrees as shown below;
[tex]\begin{gathered} \sin 45=\frac{12}{HL} \\ HL=\frac{12}{\sin 45}=\frac{12}{\frac{1}{\sqrt[]{2}}}=12\times\sqrt[]{2} \\ HL=12\sqrt[]{2} \end{gathered}[/tex]
Let's find JH by taking the tangent of angle 45 degrees;
[tex]\begin{gathered} \tan 45=\frac{12}{JH} \\ JH=\frac{12}{\tan 45}=\frac{12}{1}=12 \end{gathered}[/tex]
Let's find JK by taking the tangent of angle 60 degrees;
[tex]\begin{gathered} \tan 60=\frac{JK}{12} \\ JK=12\tan 60 \\ JK=12\sqrt[]{3} \end{gathered}[/tex]
Let's find KH by taking the cosine of angle 60 degrees;
[tex]\begin{gathered} \cos 60=\frac{12}{KH} \\ KH=\frac{12}{\cos 60} \\ KH=\frac{12}{\frac{1}{2}}=12\times\frac{2}{1}=24 \\ KH=24 \end{gathered}[/tex]
