Question

7. In A.ABC. JB and KA are medians, JK = 10x - 12, AB = 9x + 18, JM = 21, KM = 23. AJ = 60. and

Answer 1

Answer:

A) JK and AB =

$[tex] \boxed{( \theta) = \f{12}{( \theta)} }$ So, =》 $\dfrac{10}{5} = \dfrac{1}{ \( \theta) }$ =》 $\( \theta) = \dfrac{5}{2}$[/tex]

B) AABC

$[tex] \boxed{( \theta) = {1}{( \theta)} }$ So, $23}{-18=\\Ans } \boxed$

C) AABM:

$[tex] e \fbox{: }$ The value of k for which $\sin(jm) = \cos(x)$ is $\dfrac{\p}{2}$ ans[/tex]

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