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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- Jan 27th 2007, 05:30 AMTom GFinding 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. - Jan 27th 2007, 06:34 AMearboth
Hello, Tom,

first draw a sketch (see attachment).

First calculate the line PR. Use Cosine rule:

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

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

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

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

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

So the side PS is nearly 7.2 cm long.

EB - Jan 27th 2007, 07:28 AMCaptainBlack
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 - Jan 29th 2007, 12:52 AMearboth
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 - Jan 29th 2007, 11:49 PMGlaysher
- Jan 30th 2007, 03:39 AMCaptainBlack
- Jan 30th 2007, 04:15 AMearboth
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:

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

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

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) - Jan 30th 2007, 11:33 PMGlaysher
- Jan 31st 2007, 12:20 AMearboth
Hello,

have a look here: EUKLID DynaGeo Header Page

I've attached the complete construction including all necessary steps

EB - Jan 31st 2007, 04:00 AMThePerfectHacker
In German, Euclid is spelled "Euklid?"

- Jan 31st 2007, 04:31 AMearboth