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Math Help - Finding sides using trigonometry

  1. #1
    Newbie Tom G's Avatar
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    Finding sides using trigonometry

    I've tried to do this question but I keep getting strange answers that don't 'look' right. Can someone explain how it is done please.

    In a quadrilateral PQRS, PQ=10cm, QR=7cm, RS=6cm and <PQR=65 and <PSR=98. Find the length PS.
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  2. #2
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    Quote Originally Posted by Tom G View Post
    I've tried to do this question but I keep getting strange answers that don't 'look' right. Can someone explain how it is done please.

    In a quadrilateral PQRS, PQ=10cm, QR=7cm, RS=6cm and <PQR=65 and <PSR=98. Find the length PS.
    Hello, Tom,

    first draw a sketch (see attachment).

    First calculate the line PR. Use Cosine rule:

    (|\overline{PR}|)^2=10^2+7^2-2 \cdot 10 \cdot 7 \cdot \cos(65^\circ) \approx 89.8334433...

    Calculate the side PS inthe triangle PRS. Use Cosine rule again (only for convenience I use PS = x):

    (|\overline{PR}|)^2=x^2+6^2-2 \cdot 6 \cdot x \cdot \cos(98^\circ). This is a quadratic equation:

    x^2+0.2783462 x-53.8334433=0. Now use the formula to solve quadratic equations. I've got:

    x=7.19927...\ \vee \ \underbrace{x=-7.4...}_{\text{not very realistic here}}

    So the side PS is nearly 7.2 cm long.

    EB
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  3. #3
    Grand Panjandrum
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    Thanks to earboths diagram it is obvious that there is insufficient information
    to solve this.

    For if we drew a circle with PR as a chord, and S also on the circle, we could
    replace S by a point S' on the arc PSR and the angle PS'R would still be 98
    degrees and so quadrilateral PQRS' would satisfy all the conditions of the
    problem but length of line segment PS != length of line segment PS'.

    The problem does not have a unique solution, in fact a quadrilateral with
    length of PS taking any value from 0 to length of PR could be constructed.

    RonL
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  4. #4
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    Quote Originally Posted by CaptainBlack View Post
    Thanks to earboths diagram it is obvious that there is insufficient information
    to solve this....

    RonL
    Hello, CaptainBlack,

    I'm awfully sorry that my sketch is the base of your argumentation(?) and the reason why it is not true. In the original text the line RS = 6 cm was given. So the left upper triangle is determined by two lines and one angle, which will give an unique solution.

    My apologies!

    EB
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  5. #5
    Member Glaysher's Avatar
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    Quote Originally Posted by earboth View Post
    Hello, Tom,

    first draw a sketch (see attachment).

    First calculate the line PR. Use Cosine rule:

    (|\overline{PR}|)^2=10^2+7^2-2 \cdot 10 \cdot 7 \cdot \cos(65^\circ) \approx 89.8334433...

    Calculate the side PS inthe triangle PRS. Use Cosine rule again (only for convenience I use PS = x):

    (|\overline{PR}|)^2=x^2+6^2-2 \cdot 6 \cdot x \cdot \cos(98^\circ). This is a quadratic equation:

    x^2+0.2783462 x-53.8334433=0. Now use the formula to solve quadratic equations. I've got:

    x=7.19927...\ \vee \ \underbrace{x=-7.4...}_{\text{not very realistic here}}

    So the side PS is nearly 7.2 cm long.

    EB
    Haven't worked all the way through this answer but 7.2 is not the one in the mark scheme to this question which takes a different approach of using the sine rule a couple of times instead and the fact that angles in a triangle add up to 180 degrees
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by earboth View Post
    Hello, CaptainBlack,

    I'm awfully sorry that my sketch is the base of your argumentation(?) and the reason why it is not true. In the original text the line RS = 6 cm was given. So the left upper triangle is determined by two lines and one angle, which will give an unique solution.

    My apologies!

    EB
    Grr.., and it was such a nice demonstration of unsolvability too.

    RonL
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  7. #7
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    Quote Originally Posted by Glaysher View Post
    Haven't worked all the way through this answer but 7.2 is not the one in the mark scheme to this question ...
    Hello,

    you are right - and I've found my mistake. The first steps are all OK. But in the quadratic equation I used a wrong factor. It follows now the correct equations:

    x^2+1.670077 x-53.8334433=0. Now use the formula to solve quadratic equations. I've got:

    x=6.5494...\ \vee \ \underbrace{x=-8.219...}_{\text{not very realistic here}}

    So the side PS is nearly 6.55 cm long.

    EB

    PS: To prove my calculations I've done an exact drawing of the quadrilateral. (See attachment)
    Attached Thumbnails Attached Thumbnails Finding sides using trigonometry-viereck_kontrolle.gif  
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  8. #8
    Member Glaysher's Avatar
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    Quote Originally Posted by earboth View Post
    Hello,

    PS: To prove my calculations I've done an exact drawing of the quadrilateral. (See attachment)
    What program did you use to do that drawing?
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  9. #9
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    Quote Originally Posted by Glaysher View Post
    What program did you use to do that drawing?
    Hello,

    have a look here: EUKLID DynaGeo Header Page

    I've attached the complete construction including all necessary steps

    EB
    Attached Thumbnails Attached Thumbnails Finding sides using trigonometry-viereck_glaysher.gif  
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  10. #10
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    In German, Euclid is spelled "Euklid?"
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  11. #11
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    Quote Originally Posted by ThePerfectHacker View Post
    In German, Euclid is spelled "Euklid?"
    Hell, TPH,

    in Greek you spell Euclid's name so: Ευκλείδης

    As you may see it is used the letter kappa, so the German version "Euklid" is a little bit nearer to the original.

    EB
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