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In square ABCD, P is on BC such that BP = 5 and PC = 2, and Q is on CD such that DQ = 4 and QC = 3. Find sin angle PAQ.

 Dec 23, 2023
 #1
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A                   B

                     5

                     P

                     2

D     4   Q  3 C

 

Let A = (0,7)

Let P = (7,2)

Let Q = (4,0)

 

AP  = sqrt ( 7^2 + 5^2) =  sqrt ( 74)

AQ = sqrt ( 4^2 + 7^2)  = sqrt (65)

PQ = sqrt (3^2 + 2^2) = sqrt (13)

 

Law of Cosines

 

PQ^2  = AP^2 + AQ^2  - 2 ( sqrt [AP * AQ ]) cosPAQ

13 = 74 + 65  - 2 sqrt (74 * 65) cos PAQ

cos PAQ =  [ 13 - 74 - 65 ] / [ -2 *sqrt( 74 *65) ]  =  63 / sqrt (4810)

 

sin PAQ =  sqrt [4810 - 63^2 ] /  sqrt [ 4810 ] = sqrt [841 / 4810 ] 

 

 

cool cool cool

 Dec 24, 2023

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