SOLUTION: In the diagram, ABCD is a square. Find sin angle PAQ. P is on BC such that BP = 2 and PC = 3, and Q is on CD such that CQ = 2 and QD = 3.

Algebra ->  Pythagorean-theorem -> SOLUTION: In the diagram, ABCD is a square. Find sin angle PAQ. P is on BC such that BP = 2 and PC = 3, and Q is on CD such that CQ = 2 and QD = 3.      Log On


   

Question 1207589: In the diagram, ABCD is a square. Find sin angle PAQ.
P is on BC such that BP = 2 and PC = 3, and Q is on CD such that CQ = 2 and QD = 3.
Found 3 solutions by math_tutor2020, Edwin McCravy, ikleyn:
Answer by math_tutor2020(3816) About Me  (Show Source):
You can put this solution on YOUR website!

This is possibly what the diagram looks like


Focus on right triangle PAB.
We'll use the tangent trig ratio to determine the following.
tan(angle) = opposite/adjacent
tan(angle PAB) = PB/AB
tan(angle PAB) = 2/5
angle PAB = arctan(2/5)
angle PAB = 21.801409 degrees approximately.

Now move your attention to right triangle DAQ.
tan(angle) = opposite/adjacent
tan(angle DAQ) = DQ/AD
tan(angle DAQ) = 3/5
angle DAQ = arctan(3/5)
angle DAQ = 30.963757 degrees approximately.

Then,
(angleDAQ) + (anglePAQ) + (anglePAB) = angleDAB
(angleDAQ) + (anglePAQ) + (anglePAB) = 90
anglePAQ = 90 - (angleDAQ + anglePAB)
anglePAQ = 90 - (30.963757 + 21.801409)
anglePAQ = 37.234834 degrees approximately
sin(angle PAQ) = sin(37.234834) = 0.605083 approximately


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Another approach.

Use the Pythagorean Theorem to determine these lengths
QA = sqrt(5^2+3^2) = 5.830952 approximately
PA = sqrt(5^2+2^2) = 5.385165 approximately
QP = sqrt(2^2+3^2) = 3.605551 approximately

Focus on triangle PAQ of the diagram shown above.
Use the Law of Cosines to find angle PAQ.
a^2 = b^2+c^2-2*b*c*cos(A)
(QP)^2 = (QA)^2+(PA)^2-2*(QA)*(PA)*cos(angle PAQ)
(3.605551)^2 = (5.830952)^2+(5.385165)^2-2*(5.830952)*(5.385165)*cos(angle PAQ)
12.999998 = 63.000003 -62.801277*cos(angle PAQ)
12.999998 - 63.000003 = -62.801277cos(angle PAQ)
-50.000005 = -62.801277cos(angle PAQ)
cos(angle PAQ) = -50.000005/(-62.801277)
cos(angle PAQ) = 0.796162
angle PAQ = 37.234852
There appears to be some slight rounding error going on.
Compare this value to the previous angle PAQ measure found in the section above.

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!

Maybe you want exact values, not decimal approximations.

 

Use the properties of a square and the Pythagorean theorem
on the 3 right triangles to get all the measures in red.

Then use the law of cosines to find the cos(PAQ):

PQ%5E2%22%22=%22%22AP%5E2%2BAQ%5E2-2%2AAP%2AAQ%2Acos%28PAQ%29

sqrt%2813%29%5E2%22%22=%22%22sqrt%2829%29%5E2%2Bsqrt%2834%29%5E2-2%2Asqrt%2829%29%2Asqrt%2834%29%2Acos%28PAQ%29

13%22%22=%22%2229%2B34-2%2Asqrt%2834%29%2Asqrt%2829%29%2Acos%28PAQ%29

13%22%22=%22%2263-2%2Asqrt%28986%29cos%28PAQ%29

2%2Asqrt%28986%29%2Acos%28PAQ%29%22%22=%22%2263-13

2%2Asqrt%28986%29%2Acos%28PAQ%29%22%22=%22%2250

cos%28PAQ%29%22%22=%22%2250%2F%282%2Asqrt%28986%29%29

cos%28PAQ%29%22%22=%22%2225%2Fsqrt%28986%29


Then use the identity

sin%5E2%28PAQ%29%22%22=%22%221-cos%5E2%28PAQ%29

sin%5E2%28PAQ%29%22%22=%22%221-%2825%2Fsqrt%28986%29%29%5E2

sin%5E2%28PAQ%29%22%22=%22%221-625%2F986

sin%5E2%28PAQ%29%22%22=%22%22361%2F986

sin%28PAQ%29%22%22=%22%22sqrt%28361%2F986%29

sin%28PAQ%29%22%22=%22%22sqrt%28expr%28361%2F986%29%2Aexpr%28986%2F986%29%29

sin%28PAQ%29%22%22=%22%22sqrt%28361%2A986%29%2F986%29%29

sin%28PAQ%29%22%22=%22%22%2819%2Asqrt%28986%29%29%2F986%29%29

That is approximately 0.6050832675, which agrees with the other tutor.

Edwin

Answer by ikleyn(52780) About Me  (Show Source):
You can put this solution on YOUR website!
.
In the diagram, ABCD is a square. Find sin angle PAQ.
P is on BC such that BP = 2 and PC = 3, and Q is on CD such that CQ = 2 and QD = 3.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


        This problem is a plain routine and can be solved easily.
        There is no need to make long boring calculations.
        True solution is very short and straightforward. 

 


Using Pythagoras, from triangle ABP find  AP = sqrt%285%5E2%2B2%5E2%29 = sqrt%2829%29.
Next,  sin%28alpha%29 = 2%2Fsqrt%2829%29,  cos%28alpha%29 = 5%2Fsqrt%2829%29.



Using Pythagoras, from triangle ADQ find  AQ = sqrt%285%5E2%2B3%5E2%29 = sqrt%2834%29.
Next,  sin%28beta%29 = 3%2Fsqrt%2834%29,  cos%28beta%29 = 5%2Fsqrt%2834%29.



OK.  Now,  ∠PAQ = 90° - alpha - beta;  therefore

    sin(∠PAQ) = sin%2890%5Eo-alpha-beta%29 = cos%28alpha%2Bbeta%29 = 

              = cos%28alpha%29%2Acos%28beta%29+-+sin%28alpha%29%2Asin%28beta%29 = %285%2Fsqrt%2829%29%29%2A%285%2Fsqrt%2834%29%29 - %282%2Fsqrt%2829%29%29%2A%283%2Fsqrt%2834%29%29 = 

              = 25%2Fsqrt%28986%29 - 6%2Fsqrt%28986%29 = 19%2Fsqrt%28986%29.

Solved.

It is as short as a child's checkmate in a chess game.