Math Help

Discussion in 'Physics & Math' started by ash64449, May 25, 2014.

  1. ash64449 Registered Senior Member

    Messages:
    795
    In triangle PQR, if 3 sinP + 4 cosQ = 6
    and 4 sinQ + 3 cosP = 1, then the angle R is ___...

    also please explain all the steps you have carried out to arrive at the answer...
     
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  3. rpenner Fully Wired Valued Senior Member

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    4,833
    For a triangle, all angles are between 0° and 180°, thus in all cases \(\sin \theta = +\sqrt{1 - \cos^2 \theta}\).

    So if set \(p = \cos P, q = \cos Q\) then the equations are: \(3 \sqrt{1 - p^2} + 4 q = 6, 4 \sqrt{1 - q^2} + 3 p = 1\). The first equation is simple to rearrange for an expression for q, the second is easy to make an expression for p: \(q = \frac{3}{2} - \frac{3}{4} \sqrt{1 - p^2}, p = \frac{1}{3} - \frac{4}{3} \sqrt{1 - q^2}\), then it's trivial to substitute these expressions into the re-written second and first equations respectively so that we have two expressions in one variable: \(p = \frac{1}{3} - \frac{4}{3} \sqrt{1 - \left( \frac{3}{2} - \frac{3}{4} \sqrt{1 - p^2} \right)^2}, q = \frac{3}{2} - \frac{3}{4} \sqrt{1 - \left(\frac{1}{3} - \frac{4}{3} \sqrt{1 - q^2}\right)^2}\) *

    For both equations, add constants to both sides to make sure that nothing remains on the right side but the radical and then square both sides, an operation that may lead to some unwanted extra solutions: \(\left( p - \frac{1}{3} \right)^2 = \frac{16}{9} \left(1 - \left( \frac{3}{2} - \frac{3}{4} \sqrt{1 - p^2} \right)^2\right), \left(q - \frac{3}{2} \right)^2 = \frac{9}{16} \left( 1 - \left(\frac{1}{3} - \frac{4}{3} \sqrt{1 - q^2}\right)^2\right)\)

    Expand the left and right sides: \(p^2 - \frac{2}{3} p + \frac{1}{9} = \frac{16}{9} -4 + 4 \sqrt{1-p^2} + p^2 - 1, q^2 - 3 q + \frac{9}{4} = \frac{9}{16} - \frac{1}{16} + \frac{1}{2} \sqrt{1 - q^2} + q^2 - 1 \)

    Add terms to left and right sides to ensure that only radicals are on the right side, then square again: \(\left(\frac{10}{3} - \frac{2}{3} p \right)^2 = 16 - 16p^2 , \left( \frac{11}{4} - 3 q \right)^2 = \frac{1}{4} - \frac{1}{4} q^2 \)

    Expand the left sides and add terms to both sides until the right sides are zero. Divide out common terms. Now you have two quadratic equations in one variable:
    \(-\frac{44}{9} - \frac{40}{9} p + \frac{148}{9} p^2 = 0 \Rightarrow -11 - 10 p + 37 p^2 = 0 \Rightarrow p = \frac{5 \pm 12 \sqrt{3}}{37} \frac{117}{16} - \frac{66}{4} q + \frac{37}{4} q^2 \Rightarrow 117 -264 q+148 q^2= 0 \Rightarrow q = \frac{66 \pm 3 \sqrt{3}}{74}\)

    By reexamination of the two solutions of these quadratic equations with the equations marked with * we see the actual solutions are:

    \(\cos P = \frac{5 - 12 \sqrt{3}}{37} \\ \cos Q = \frac{66 - 3 \sqrt{3}}{74} \\ \sin P = \frac{30 + 2 \sqrt{3}}{37} \\ \sin Q = \frac{11 + 18 \sqrt{3}}{74}\)

    By angle addition \(- \cos R = \cos ( P + Q ) = \cos P \cos Q - \sin P \sin Q = \frac{438-807 \sqrt{3}}{2738} - \frac{438+562 sqrt{3}}{2738} = - \frac{\sqrt{3}}{2}\) thus \(R = 30^{\circ} = \frac{\pi}{6}\).
     
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  5. rpenner Fully Wired Valued Senior Member

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    Alternate method:

    Given \(3 \sin P + 4 \cos Q = 6, 4 \sin Q + 3 \cos P = 1\) it follows that \(( 3 \sin P + 4 \cos Q ) ^2 + (4 \sin Q + 3 \cos P)^2 = 9 \sin^2 P + 24 \sin P \cos Q + 16 \cos^2 Q + 16 \sin^2 Q + 24 \sin Q \cos P + 9 \cos^2 P = 25 + 24 \sin (P + Q) = 25 + 24 \sin R\) Thus \(6^2 + 1^2 = 25 + 24 \sin R\) or \(\sin R = \frac{1}{2}\)

    And, by the law of sines, the sides of the triangle are proportional to the sines of the opposite angles, and so, by the law of cosines:
    \(\cos R = \frac{\sin^2 P + \sin^2 Q - \sin^2 R}{2 \sin P \, \sin Q } = \frac{ \left( 2 - \frac{4}{3} \cos Q \right)^2 + \sin^2 Q - \frac{1}{4}}{4 \sin Q - \frac{8}{3} \cos Q \sin Q} \gt 0\)

    Thus \(\cos R = + \frac{\sqrt{3}}{2}\)
     
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  7. ash64449 Registered Senior Member

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    795
    Thank you rpenner for devoting some time to answer this question.

    I understood your 1st way of solving and 1st alternative method- understood the logic behind the 1st alternative method..
     
  8. ash64449 Registered Senior Member

    Messages:
    795
    I didn't understand how did you arrive at this result from earlier steps
    ..
     
  9. rpenner Fully Wired Valued Senior Member

    Messages:
    4,833
    \(3 \sin P + 4 \cos Q = 6 \\ \Rightarrow \sin P + \frac{4}{3} \cos Q = 2 \\ \Rightarrow \sin P = 2 - \frac{4}{3} \cos Q\)

    Thus \(\frac{\sin^2 P + \sin^2 Q - \sin^2 R}{2 \sin P \, \sin Q } = \frac{ \left( 2 - \frac{4}{3} \cos Q \right)^2 + \sin^2 Q - \frac{1}{4}}{4 \sin Q - \frac{8}{3} \cos Q \sin Q} = \frac{ \frac{15}{4} - \frac{16}{3} \cos Q + \frac{16}{9} \cos^2 Q + \sin^2 Q}{2 (2 - \frac{4}{3} \cos Q)\, \sin Q} = \frac{28 (1 - \cos Q)^2+ 136 (1-\cos Q) +7}{72 (2 - \frac{4}{3} \cos Q) \, \sin Q} > 0\)


    One possible triangle is:


    P ( 54 + 11 √3, 11 + 18 √3 )
    Q ( 120 + 8 √3, 0 )
    R ( 0, 0 )

    | P - Q | = 74
    | P - R | = 22 + 36 √3
    | Q - R | = 120 + 8 √3
    cos P = (Q - P)⋅ (R - P) / ( | Q - P | × | R - P | ) = (5-12 √3)/37
    cos Q = (P - Q)⋅ (R - Q) / ( | P - Q | × | R - Q | ) = (66 - 3√3)/74
    cos R = (P - R)⋅ (Q - R) / ( | P - R | × | Q - R | ) = √3/2
    sin P = | Q - R | / 148 = ( 30 + 2 √3)/37
    sin Q = | P - R | / 148 = ( 11 + 18 √3)/74
    sin R = | P - Q | / 148 = 1/2

    There are many rational relationships between the sines and cosines of these angles:
    cos P + 6 sin P = 5
    12 cos Q + 2 sin Q = 11
    3 sin P + 4 cos Q = 6
    4 sin Q + 3 cos P = 1
    45 cos P + 11 sin P + 28 cos R = 15
    6 cos Q + sin Q + 3 sin R = 7
    cos P + 6 sin P - 4 sin R = 3
    5 cos Q + 44 sin Q - 21 cos R = 11
    9 cos P + 2 cos Q + 6 cos R = 3
    9 sin P - 2 sin Q - 6 sin R = 4
     
    Last edited: May 26, 2014

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