Finding sides using trigonometry

• Jan 27th 2007, 05:30 AM
Tom G
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.
• Jan 27th 2007, 06:34 AM
earboth
Quote:

Originally Posted by Tom G
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
• Jan 27th 2007, 07:28 AM
CaptainBlack
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 AM
earboth
Quote:

Originally Posted by CaptainBlack
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
• Jan 29th 2007, 11:49 PM
Glaysher
Quote:

Originally Posted by earboth
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
• Jan 30th 2007, 03:39 AM
CaptainBlack
Quote:

Originally Posted by earboth
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
• Jan 30th 2007, 04:15 AM
earboth
Quote:

Originally Posted by Glaysher
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)
• Jan 30th 2007, 11:33 PM
Glaysher
Quote:

Originally Posted by earboth
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?
• Jan 31st 2007, 12:20 AM
earboth
Quote:

Originally Posted by Glaysher
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
• Jan 31st 2007, 04:00 AM
ThePerfectHacker
In German, Euclid is spelled "Euklid?"
• Jan 31st 2007, 04:31 AM
earboth
Quote:

Originally Posted by ThePerfectHacker
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