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# Geometry question

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30
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In square ABCD, P is on BC such that BP = 4 and PC = 1, and Q is on CD such that DQ = 4 and QC = 1. Find sin angle PAQ.

Aug 9, 2023

#1
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Using the Law of Cosines

BC^2  =  DB^2  + DC^2  - 2(DB*DC) cos (60 degrees)

BC^2  = 2^2 + 5^2  - 2 (2*5)(1/2)

BC^2 = 29 - 10

BC = sqrt (19)

Using the Law of Sines

sin BCD / BD = sin BDC / BC

sin BCD / 2 = sqrt (3) / [ 2  sqrt (19) ]

sin BCD = sqrt (3/19)

BCD = arcsin (3/19) ≈ 23.4°

Angle ACE =  90 - 23.4  ≈ 66.6°

Angle CAE =  180 -60 - 66.6 ≈  53.4°

Using the Law of Sines again

CE / sin CAE =  AE /sin ACE

2 / sin  (53.4) = AE / sin (66.6)

AE =  2sin (66.6) / sin (53.4)   ≈  2.286

Aug 9, 2023
#2
+52
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You just copied CPhill's answer to my previous question...

choutowne  Aug 9, 2023
#3
+129270
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Since PC and QC = 1, then PQ = sqrt (2)

Let A = (0,,0)  and P = (4,5) so AP  =sqrt [ 4^2 + 5^2]  = sqrt (41)  = AQ

So....by Law of Cosines

PQ^2  = AP^2 + AQ^2 -2 (AP)(AQ) cos (PAQ)

2 = 41 + 41  - 2 (sqrt 41)( sqrt 41) cos (PAQ)

-80 / (- 2 * 41 ) =  cos PAQ

80 /82  =  40/41  cos PAQ

sin PAQ =   sqrt  [ 1 - (40/41)^2 ] =   sqrt  [ 41^2 - 40^2] / 41  =  sqrt (81) / 41 =  9 / 41

Aug 9, 2023
edited by CPhill  Aug 9, 2023
#4
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Thanks for the answer, it helped me a lot!

5sqrt(21)/41 wasn't correct, but your response helped me figure out what to do next.

I followed your solution and got mostly the same results except that I got 40/41 as cos(PAQ) instead of 34/41.

I plugged that into the relationship between the sine and cosine of PAQ and got 9/41 as the final answer, which turned out to be correct.

\( sin(PAQ) = \boxed{9/41}\)

choutowne  Aug 9, 2023
#5
+129270
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