Solve Cos B Homework - Trig & Geometry

[um0123]

Homework Statement

Find Cos B

Hint: use trigonometry + Geometry

http://img210.imageshack.us/img210/4596/pow.png

Homework Equations

trigonometry + Geometry

The Attempt at a Solution

I don't see how i could use trig to find any of the angles, and since none of the lins go through the center it doesn't help me find the raudius. I don't even know how to begin solving this.

 

[tiny-tim]

Hi um0123!

Hint: It's a circle, so B = 180º - D.

 

[um0123]

i may be daft, but may i have a little more help? i see how that information would be really useful but i can't seem to grap what i need to do.

 

[tiny-tim]

Hint: AC² = … ?

 

[um0123]

$$AC^2 = 1^2 + 9^2 - 4(1)(9)cos(180-D)$$

$$AC^2 = 82 - 34cos(180-D)$$

$$\frac{AC^2 - 82}{-34} = cos(180-D)$$

$$\frac{AC^2 - 82}{-34} = cos(180)cos(D) + sin(180)sin(D)$$

$$\frac{AC^2 - 82}{-34} = -1cos(D) + 0sin(D)$$$$\frac{AC^2 - 82}{-34} = -1cos(D)$$$$\frac{AC^2 - 82}{34} = cos(D)$$

now I am stuck

 

[tiny-tim]

(try using the X² tag just above the Reply box )

erm … AC² = 6² + 9² - … ?

 

[um0123]

i can't assume its a right triangle, can i? i was using the law of cosines because the Pythagorean theorem is only for right triangles, unless i can prove its a right triangle i don't think I am allowed to use that.

BTW what does the X^2 tag do if i put it in the tag box?

 

[Mark44]

No, it's not a right triangle. tiny-tim was not saying you shouldn't use the Law of Cosines, but he was saying that you should use it correctly, and was trying to guide you in that direction.

 

[tiny-tim]

I meant there are two triangles with AC …

you've only used one of them!

 

[um0123]

oh, sorry, i didn't notice the minus sign after your previous post tiny-tim (my eyesight is less than acceptable).

so i continue with:$$AC^2 = 6^2 + 9^2 - 4(6)(9)cos(D)$$

$$AC^2 = 36 + 81 - 216cos(D)$$

$$AC^2 = 117 - 216cos(D)$$

$$AC^2 -117 = -216cos(D)$$

$$\frac{AC^2 -117}{-216} = cos(D)$$

but i still get stuck...i must be overlooking something really obvious.

 

[tiny-tim]

Forget the fractions …

you now have two expressions for AC² (one for each triangle),

so put them equal, and that gives you an equation in cosD.

 

[willem2]

You can use the cosine rule in the triangle ADC as well, giving another expression for AC and cos(D).
You put a 4 in the cosine rule where a 2 was wanted

 

[um0123]

So if i understand you correctly, tiny-tim, i should have:
$$117 - 216Cos(D) = 82 - 34Cos(180-D)$$

$$35 - 216Cos(D) = -34Cos(180-D)$$

$$\frac{35 - 216Cos(D)}{-34} = -1Cos(180-D)$$

$$\frac{35 - 216Cos(D)}{34} = Cos(180-D)$$

$$\frac{35 - 216Cos(D)}{34} = Cos(180)Cos(D) + Sin(180)Sin(D)$$

$$\frac{35 - 216Cos(D)}{34} = -1Cos(D) + 0Sin(D)$$

$$\frac{35}{34} = \frac{216Cos(D)}{34} - Cos(D)$$

now i really can't go anywhere

 

[tiny-tim]

(I haven't checked the arithmetic, but …)

that should give you an equation of the form PcosD = Q.

 

[um0123]

im sorry, i don't think i have learned that equation, if i have then i am not recognizing it.

btw, i just wanted to say thanks for helping me through this, i know I am probably really hard to handle.

 

[willem2]

$$\frac{35}{34} = \frac{216Cos(D)}{34} - Cos(D)$$

now i really can't go anywhere

You're almost there. You just have to give the 2 fractions on the right the same denominator so you can add them.

You also need to use the correct cosine rule: a^2 = b^2 + c^2 -2 bc cos(A)

 

[um0123]

oh, wow, i thought it was 4bc cos(a)

also i did a mathematical error when i calculated 4(1)(9) as 34, it sohuld be 36,

so redoing the calculations i get:

$$\frac{35}{18} = \frac{108cos(D)}{18} - cos(D)$$

$$\frac{35}{18} = 6cos(D) - cos(D)$$

$$\frac{35}{18} = 5cos(D)$$

$$35 = 90cos(D)$$

$$\frac{35}{90} = cos(D)$$

please tell me I am right!?

 

[um0123]

form there all i have to do is cos^-1(35/90) to ge tthe angle and subtract from 180. I hope this is correct.Thanks so much for you help, tiny-tim and willem2!

 

[willem2]

So if i understand you correctly, tiny-tim, i should have:

$$35 - 216Cos(D) = -34Cos(180-D)$$

$$\frac{35 - 216Cos(D)}{-34} = -1Cos(180-D)$$

Unfortunately there's a sign error here, that's still present in your final answer.

 

[icystrike]

Express AC² in terms of cosB and cosD.
Now recall cosD=cos(180-B)=-cosB
Try to get cos B