Law of cosine Help!!

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July 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
July 29th 2006, 01:10 PM
topsquark

Quote:

Originally Posted by ^_^Engineer_Adam^_^

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

$c^2 = a^2 + b^2 - 2ab \cdot cos( \gamma )$

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

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

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

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

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

-Dan
July 29th 2006, 05:16 PM
ThePerfectHacker

You can also find the angles by using the fact that.
$\frac{1}{2}ab\sin \gamma =A$
where, $A=\mbox{ area }$
And you can calculate area by Heron's Formula.
$A=\sqrt{s(s-a)(s-b)(s-c)}$
$s=\mbox{ semi-perimeter }$
July 30th 2006, 04:05 AM
Soroban

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, I'm confused . . .

You're familiar with the Law of Cosines
. . but you've never solved for an angle . . . ever?

Okay, just this once . . .

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Check: . $A + B + C\:=\:31.6^o + 38.9^o + 109.5^o \:=\:180^o$ . . . Yay!
July 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!!
July 30th 2006, 06:03 AM
Soroban

Hello again, Adam!

Quote:

After solving the Law of Cosine to get $A = 32^o$
i solved the angle using the Law of Sines.

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

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

Recall that an inverse sine can have two possible values.
. . For example: . $\sin^{-1}(0.5)\,=\,30^o$ or $150^o$
And your calculator gives you only the smaller value.
It is up to you to determine which angle is appropriate.

You got: . $C = 72.5^o$ , but $107.5^o$ 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) can lie to you.
. . (It says the angle is $60^o$ , but it's really $120^o.)$
Hence, I recommend that you use the Law of Cosines to find angles.
. . (It doesn't lie.)"
July 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!