Question 1156150
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Problem:
Given triangle PQR is acute with 
<ul><li>QR = 9.7</li><li>PR = 12.2</li><li>angle P = 43 degrees</li></ul>find the measure of angle Q


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The diagram would look something like this
<img width="50%" src = "https://i.imgur.com/ckOZpkc.png">


Use the law of sines to solve for angle Q
sin(A)/a = sin(B)/b
sin(P)/p = sin(Q)/q
sin(P)/QR = sin(Q)/PR
sin(43)/9.7 = sin(Q)/12.2
sin(Q)/12.2 = sin(43)/9.7
sin(Q) = 12.2*(sin(43)/9.7)
sin(Q) = 0.85777113327449
Q = arcsin(0.85777113327449)
Q = 59.067239465619
Q = 59
Make sure your calculator is in degree mode. Arcsine is the same as inverse sine. Remember to round to the nearest degree. 


Note: There is another triangle possible, but that would mean angle Q is over 90 degrees. This is because Q = 180-arcsin(0.85777113327449) = 180-59.067239465619 = 120.932760534381. So we ignore this possible Q value. Recall that triangle PQR is acute meaning that all 3 angles must be less than 90. 


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Answer: <font color=red>approximately 59 degrees</font>
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