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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.

 Jan 29, 2025
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AP = sqrt [5^2 + 4^2]  = sqrt 41

PQ = sqrt [2^2 + 1^1] =  sqrt 5

AQ = sqrt [ 5^2 + 2^2] = sqrt 29

 

Law of Cosines

 

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

 

5  = 41 + 29 - 2 [sqrt [41] *sqrt[29] cos (PAQ)

 

cos (PAQ)  =  [ 5 - 41 - 29 ] / [ - 2sqrt [41]sqrt[29] ]  = 65  / sqrt [ 4756 ]

 

sin PAQ =   sqrt [ 1 - 65^2 / 4756 ]  =  3sqrt [ 70151] / 2378

 

cool cool cool

 Jan 30, 2025

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