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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.

 May 3, 2024
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Let's first notice that both triangle BPQ and triangle CQP are both right triangles. 

 

In order for the sin of angle PAQ, we need to find the length of side PQ and PA and then use the ratio for the sin.

 

Using the pythagorean thereom, we can find that the length of PQ is \(PQ^2 = BP^2 + PC^2 PQ^2 = 4^2 + 1^2 PQ^2 = 17 PQ = \sqrt{17} \)

Use the Pythagorean Theorem, we can find the length of PA is \(PA^2 = CQ^2 + QP^2 PA^2 = 4^2 + \sqrt{17}^2 PA^2 = 33 PA = \sqrt{33}\)

 

Now, we simply take the ratio of the two numbers getting us \(\frac{\sqrt{17}}{\sqrt{33}} = 17/33\)

 

So the sin of PAQ is simply \(\sqrt{\frac{17}{33}}\)!

 

Thanks!

 May 3, 2024
edited by NotThatSmart  May 3, 2024

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