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In the diagram, ABCD is a square.  Find sin PAQ.

 

 Jun 4, 2024
 #1
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PQ  = sqrt [ 2^2 + 3^2 ] = sqrt [ 13]

AQ  = sqrt [1^2 + 4^2 ] sqrt [17]

PA  = sqrt  [2^2 + 4^2] = sqrt [ 20]  

 

Law of Cosines

 

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

13 = 17 + 20 - 2( sqrt 340) * cos (PAQ)

[ 13 - 17 -20 ] / [ -2sqrt 340 ]  = cos (PAQ) = 6/sqrt 85

 

sin PAQ =   sqrt [ 85 - 6^2 ] / sqrt 85  = sqrt [ 49] / sqrt 85  =  7 / sqrt 85

 

cool cool cool

 Jun 4, 2024

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