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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?

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yapahyasharahla
Hi,

What u have to do with the question is to determine that you need to you equation so let's get to the answer.

So you have the famous 30-60-90 right-angled triangle, which has sides in the ratio 
1: √3: 2 
(memorize those values, easy to remember that the smallest side is opposite the smallest angle and the largest side is opposite the largest angle) 

so using simple ratios 
QR : PR : 12 = 1 : √ 3 : 2 

QR/1 = 12/2 ---> QR = 6 
PR/√3 = 12/2 ---> PR = 12√3/2 = 6√3 
(did you notice that we simply multiplied each of the ratio terms by 6, keeping our new triangle similar to 1:√3:2 ) 

area of PQR = (1/2)PR*QR 
= (1/2)(6√3)(6) = 18√3 

Since PXR has the same height, but only half the base, 
its area is 9√3

Hoped I Helped 

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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