In triangle pqr, we have angle p = 30 degrees, angle rqp = 60 degrees, and angle r=90 degrees. point x is on pr such that qx bisects angle pqr. if pq = 12, then what is the area of triangle pqx?

PR = 12*sin60 = 12 * [sqrt(3)/2]  =  6 * sqrt(3)

 And we can find XR thusly : 

XR/ sin30  = 6/ sin60 

XR   = 6sin30/ sin60 

XR  = 6(1/2) / [sqrt(3) / 2]

XR  = 3*2/ sqrt(3) = 6/sqrt(3)

So PX  =  PR - XR =   6*sqrt(3) - 6/sqrt(3) =  [18 - 6] / sqrt(3)  =  

12/sqrt(3)And the area of PQX   =  (1/2) *PX* RQ  = (1/2) (12/sqrt(3)) * (6)   

=  36 / sqrt(3)  square units

I hope this helps you!

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step-by-step explanation:

. + λrh() e 1 1 2 4 4 4 4 4 3 4 4 4 4 4 + th−1(− s(x)2 dx ω + λrh−1() e 2 1 2 4 4

step-by-step explanation: