Cud u give me the formula of Law of Cosine on how to solve a triangle w/ 3 sides

given?

Im kind of confused with this problem:

a = 5, b = 6, c = 9

find all 3 angles A,B & C

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- Jul 29th 2006, 12:31 PM^_^Engineer_Adam^_^Law of cosine Help!!
Cud u give me the formula of Law of Cosine on how to solve a triangle w/ 3 sides

given?

Im kind of confused with this problem:

a = 5, b = 6, c = 9

find all 3 angles A,B & C - Jul 29th 2006, 01:10 PMtopsquarkQuote:

Originally Posted by**^_^Engineer_Adam^_^**

Where $\displaystyle \gamma$ is the angle across from side c.

Thus $\displaystyle cos( \gamma ) = \frac{a^2 + b^2 - c^2}{2ab}$.

The other formulae simply permute the values of a, b, and c and need not be given.

So for example, to find the angle across from side c we have:

$\displaystyle cos( \gamma ) = \frac{5^2 + 6^2 - 9^2}{2\cdot 5 \cdot 6} = -\frac{1}{3}$

Thus $\displaystyle \gamma$ is second quadrant and $\displaystyle \gamma \approx 109.5^o $

To find the angle across from side a, use a = 6, b = 9, c = 5. etc.

-Dan - Jul 29th 2006, 05:16 PMThePerfectHacker
You can also find the angles by using the fact that.

$\displaystyle \frac{1}{2}ab\sin \gamma =A$

where, $\displaystyle A=\mbox{ area }$

And you can calculate area by*Heron's Formula*.

$\displaystyle A=\sqrt{s(s-a)(s-b)(s-c)}$

$\displaystyle s=\mbox{ semi-perimeter }$ - Jul 30th 2006, 04:05 AMSoroban
Hello, Adam!

Quote:

Cud u give me the formula of Law of Cosine on how to solve a triangle w/ 3 sides given?

Im kind of confused with this problem:

a = 5, b = 6, c = 9

find all 3 angles A,B & C

No,confused . . .*I'm*

You're familiar with the Law of Cosines

. . but you'vesolved for an angle . . .__never__*ever?*

Okay, just this once . . .

I assume you know that: .$\displaystyle a^2\:=\:b^2+c^2 - 2bc\cos A$

Rearrange the terms: .$\displaystyle 2bc\cos A\:=\:b^2 + c^2 - a^2$

Divide by $\displaystyle 2bc:\;\;\boxed{\cos A\:=\:\frac{b^2 + c^2 - a^2}{2bc}}$ . . . a formula for finding $\displaystyle \angle A.$

Similarly, we can derive formulas for the other two angles:

. . $\displaystyle \boxed{\cos B\:=\:\frac{a^2+c^2-b^2}{2ac}}\qquad\boxed{\cos C\:=\:\frac{a^2+b^2-c^2}{2ab}} $

You should memorize these or be able to derive them when needed.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Your problem has: .$\displaystyle a = 5,\lb = 6,\;c = 9$

We have: .$\displaystyle \cos A\:=\:\frac{6^2 + 9^2 - 5^2}{2(6)(9)}\:=\:\frac{92}{108}$

. . Therefore: .$\displaystyle A\:=\:\cos^{-1}\left(\frac{92}{108}\right)\:=\:31.5863381\quad \Rightarrow\quad \boxed{A\:\approx\:31.6^o}$

We have: .$\displaystyle \cos B\:=\:\frac{5^2 + 9^2 - 6^2}{2(5)(9)}\:=\:\frac{70}{90}$

. . Therefore: .$\displaystyle B\:=\:\cos^{-1}\left(\frac{7}{9}\right)\:=\:38.94244127\quad \Rightarrow\quad \boxed{B\:\approx\:38.9^o}$

We have: .$\displaystyle \cos C\:=\:\frac{5^2+6^2-9^2}{2(5)(6)}\:=\:\frac{-20}{60}$

. . Therefore: .$\displaystyle C\:=\:\cos^{-1}\left(-\frac{1}{3}\right)\:=\:109.4712206\quad \Rightarrow\quad \boxed{C\:\approx\:109.5^o}$

Check: .$\displaystyle A + B + C\:=\:31.6^o + 38.9^o + 109.5^o \:=\:180^o$*. . . Yay!*

- Jul 30th 2006, 05:20 AM^_^Engineer_Adam^_^
Yea ...

But the only confusing thing is that after solving the law of cosine to get the 1st angle A which is 32 degrees, i solve the angle using the law of sine... so sin32 / 5 = sin C / 9 and it gave the C an angle of 72.5 degrees...

How come?

Btw

Thanks topsquart, ThePerfectHacker and thanks again Soroban!! - Jul 30th 2006, 06:03 AMSoroban
Hello again, Adam!

Quote:

After solving the Law of Cosine to get $\displaystyle A = 32^o$

i solved the angle using the Law of Sines.

So $\displaystyle \frac{\sin32}{5} = \frac{\sin C}{9}$ and it gave $\displaystyle C = 72.5^o.$

How come?

You fell for a very common "trap" in these problems.

Recall that an inverse sine can have__two__possible values.

. . For example: .$\displaystyle \sin^{-1}(0.5)\,=\,30^o$**or**$\displaystyle 150^o$

And your calculator gives you only the smaller value.

It is up to__you__to determine which angle is appropriate.

You got: .$\displaystyle C = 72.5^o$, but $\displaystyle 107.5^o$ is also possible.

I've explained this to my students:

"The Law of Sines is much easier to use for determining angles.

But the Law of Sines (and your calculator) canto you.*lie*

. . (It says the angle is $\displaystyle 60^o$, but it's really $\displaystyle 120^o.)$

Hence, I recommend that you use the*Law of Cosines*to find angles.

. . (It doesn't lie.)"

- Jul 30th 2006, 06:11 AM^_^Engineer_Adam^_^
I c so u hav ta use cosine all the time involving 3 sides

.... thanks master soroban!! i finally found the answer to my frustration... coz i got a 20 / 50 in a seatwork with that problem...

really appreciated!