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avatar+1981 

In square ABCD, P is on BC such that BP = 4 and PC = 1, and Q is on line CD such that DQ = 2 and QC = 3. Find sin angle PAQ.

 Jun 13, 2024
 #1
avatar+1252 
+1

First, we have to find the values of PA, PQ, and AQ.

 

Using the pythagorean theorem for all 3 of them, we get

 \(AP = \sqrt{4^2+5^2}=\sqrt{41}\)

\(PQ = \sqrt{1^2+3^2}=\sqrt{10}\)

\(AQ=\sqrt{1^2+5^2} = \sqrt{26}\)

 

Now, we apply the Law of Cosines. According to the law, 

\(PQ^2 = AQ^2 + PA^2 - 2(AQ * PA) * cos (PAQ)\)

 

Plugging in all the values of AP, PQ, and AQ we figured out, we get

\(10=26+41-2(\sqrt{26}*\sqrt{41})*cos(PAQ)\\ 10=67-2\sqrt{1066}*cos(PAQ)\\ cos(PAQ) = \frac{-57}{-2\sqrt{1066}}\)

 

Now, sin(PAQ) is just 1 - cos(PAQ), so we get

\(sin(PAQ)=1- \frac{-57}{-2\sqrt{1066}}\\ sin(PAQ)=1+ \frac{57}{-2\sqrt{1066}}\\ sin(PAQ) = \frac{-2\sqrt{1066}+57}{-2\sqrt{1066}}\)

 

This is approxaimtely \(0.12709645414\)

 

So our answer is \(\frac{-2\sqrt{1066}+57}{-2\sqrt{1066}}\)

 

Thanks! :)

 Jun 13, 2024
 #2
avatar+129733 
+1

A                         B

                           4

                           P

                           1                          

D    2     Q    3    C

 

PA =  sqrt [ 5^2 + 4^2] = sqrt 41

PQ = sqrt [ 3^2 + 1^2] = sqrt 10

AQ = sqrt [ 5^2 + 2^2]  = sqrt 29

 

Law of Cosines

 

PQ^2 = PA^2 + AQ^2 - 2(PA * AQ) cos (PAQ)

 

10 = 41 + 29  - 2 ( sqrt [41 * 29] ) cos (PAQ)

 

cos (PAQ)  =  (10 - 41 - 29 ) /  (-2 *sqrt [ 41 * 29 ] )  =  30 / sqrt 1189

 

sin (PAQ)  =  sqrt [1189  - 30^2 ] / sqrt 1189   = sqrt [289] / sqrt 1189  =  17 / sqrt 1189

 

 

cool cool cool

 Jun 14, 2024

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