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# Geometry

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+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
+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
+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

Jun 14, 2024