[SOLVED] cosine rule help!

• Feb 10th 2009, 12:41 AM
xch3m_is_l0v3x
[SOLVED] cosine rule help!
I have racked my brain trying to think of what I've done wrong, but I just can't. (Headbang)

Jim and Tony leave the same point at the same time with Jim walking away at the speed of 1.4m/s and tony at a speed of 1.7m/s, the two directions of travel making a angle of 50 degrees with eachother. If they continue on the same straight line paths how far apart are they after 8 seconds?

x^2=6.3^2+7.2^2-2(6.3)(7.2)cos17
x^2=91.53-90.72Xcos17
x^2=91.53-86.76
x^2=4.77
x=2.18(Angry)

Can someone please help me? I want to know what I've done wrong! and thank you for taking the time to read this post =]

xxx
marika
• Feb 10th 2009, 12:49 AM
earboth
Quote:

Originally Posted by xch3m_is_l0v3x
I have racked my brain trying to think of what I've done wrong, but I just can't. (Headbang)

Jim and Tony leave the same point at the same time with Jim walking away at the speed of 1.4m/s and tony at a speed of 1.7m/s, the two directions of travel making a angle of 50 degrees with eachother. If they continue on the same straight line paths how far apart are they after 8 seconds?

x^2=6.3^2+7.2^2-2(6.3)(7.2)cos17
x^2=91.53-90.72Xcos17
x^2=91.53-86.76
x^2=4.77
x=2.18(Angry)

Can someone please help me? I want to know what I've done wrong! and thank you for taking the time to read this post =]

xxx
marika

I really don't understand where from you've got these values (Surprised)

1. Distance walked by Jim: $8\ s \cdot 1.4\ \frac ms = 11.2\ m$

2. Distance walked by Tony: $8\ s \cdot 1.7\ \frac ms = 13.6\ m$

3. Angle included by the two directions: 50°

4. Therefore the distance between them will be:

$x^2=11.2^2+13.6^2-2\cdot 11.2 \cdot 13.6 \cdot \cos(50^\circ)$

I've got $x\approx 10.7\ m$
• Feb 10th 2009, 12:53 AM
Cosine Rule
Hello xch3m_is_l0v3x
Quote:

Originally Posted by xch3m_is_l0v3x
I have racked my brain trying to think of what I've done wrong, but I just can't. (Headbang)

Jim and Tony leave the same point at the same time with Jim walking away at the speed of 1.4m/s and tony at a speed of 1.7m/s, the two directions of travel making a angle of 50 degrees with eachother. If they continue on the same straight line paths how far apart are they after 8 seconds?

x^2=6.3^2+7.2^2-2(6.3)(7.2)cos17
x^2=91.53-90.72Xcos17
x^2=91.53-86.76
x^2=4.77
x=2.18(Angry)

Can someone please help me? I want to know what I've done wrong! and thank you for taking the time to read this post =]

xxx
marika

Thanks for showing us your working, but I don't understand where you get your numbers from. Is this the same question? After 8 sec Jim has walked 1.4 x 8 = 11.2 m and Tony 1.7 x 8 = 13.6 m. The angle between their paths is 50 degrees. So

$x^2 = 11.2^2 + 13.6^2 - 2\times 11.2 \times 13.6 \cos 50^o$

• Feb 10th 2009, 07:44 AM
nardu
hi i like to use the sine law
thats sine equals opposite over hypothenuse
thats sine 50 times 13.6 (8x1.7)
sine 50 = 0.766
0.766x13.6=10.42

hope you agree
thanks
• Feb 10th 2009, 09:54 AM
Cosine Rule
Hello nardu
Quote:

Originally Posted by nardu
hi i like to use the sine law
thats sine equals opposite over hypothenuse
thats sine 50 times 13.6 (8x1.7)
sine 50 = 0.766
0.766x13.6=10.42

hope you agree
thanks

You can only use sine $= \frac{\text{opposite}}{\text{hypotenuse}}$ when you have a right-angled triangle, and I'm afraid that the triangle in this question isn't right-angled. In a case like this, then, the formulae you use are rather different. You'll find a summary of them here: Sine and Cosine Formulae

Good luck.