Answer:
A) JK and AB =
[tex][tex] \boxed{( \theta) = \f{12}{( \theta)} }[/tex] So, =》 [tex] \dfrac{10}{5} = \dfrac{1}{ \( \theta) } [/tex] =》 [tex] \( \theta) = \dfrac{5}{2} [/tex][/tex]
B) AABC
[tex][tex] \boxed{( \theta) = {1}{( \theta)} }[/tex] So, [tex] 23}{-18=\\Ans } \boxed[/tex]
C) AABM:
[tex][tex] e \fbox{: }[/tex] The value of k for which [tex] \sin(jm) = \cos(x) [/tex] is [tex] \dfrac{\p}{2} [/tex] ans[/tex]
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