cos sin tan

I need help with this please!
IT says that I need to (bascally) find all the measurements for this triangle, by using the law of cosines. Given: a=51, b=61, m of angle B=19 (lower case letters=sides Upper=angles). I really need help thanx!

Originally Posted by Wikipedia.org

In trigonometry, the law of cosines (also known as the cosine formula) is a statement about a general triangle which relates the lengths of its sides to the cosine of a known angle. It generalizes the Pythagorean theorem, which applies only to right triangles, by correcting it with a term proportional to the cosine of the known angle. Let a, b, and c be the lengths of the sides of the triangle and let C be the angle opposite c. Then,


This formula is useful for computing the third side of a triangle when two sides and their enclosed angle are known, and in computing the angles of a triangle if all three sides are known.

The law of cosines also shows that

if and only if
The statement cos C = 0 implies that C is a right angle, since a and b are positive. In other words, this is the Pythagorean theorem and its converse. Although the law of cosines is a broader statement of the Pythagorean theorem, it isn't a proof of the Pythagorean theorem, because the law of cosines derivation given below depends on the Pythagorean theorem.

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Thats what wikipedia says. So yeah hope that helps.

No, but thanx though. I know what it is, but Im in a tight spot lol.
So far I have: c^2=61^2+51^2-(2(61)(51))cos?
Thats what I have (^2=squared cos?= I dont know what to put here).

I hope you have a calculator with sin, cosin, and tangent!

Sin= Opposite over Hypoteneuse

Cos= Adjacent over Hypoteneuse

Tan= Opposite over Adjacent

If I remember right, all you have to do is divide the number of the side adjacent to B by the Hypoteneuse.

If I saw the triangle I could write out the formula for you.


Sorry if I'm not helping...

Simple.

Cosine - Opposite over hypotenuse

cos19=51/x
cos19 is approxamatley - .9455

.9455 = 51
------- ----
1-------x

.9455x=51
x=53.93

Now that we have solved the 3rd side using cosines, we need to solve for the other acute angle. Simple - Add 19+90=109
180-109=71

Tada.

As taht is right, thats not what im trying to do. The law of cosines state:
a^2= b^2+c^2-2bc cos A
b^2= a^2+c^2-2ac cos B
c^2= a^2+b^2-2ab cos C
Thats the law of cosines. lower case= sides upper=angles.

here is what the triangle is supposed to look like:

(note: if you cant see anything dont worry bout it then. the top number (on top of the triangle) is 19 which is the top angle.)

the formulas (a better name is equation) are:

c^2 = a^2 + b^2 - 2*a*b*cosC

a^2 = c^2 + b^2 - 2*c*b*cosA

b^2 = a^2 + c^2 - 2*a*c*cosB

(^2 means square)


now:

you cannot use the law of sines, right? (it would be easier).... so, use the third one:

61^2 = 51^2 + c^2 - 51*2*c*cos(19)

this is: c^2 + 102cos(19)*c - 1120 = 0

using Vietta, or any other method for cuadratic ecuations you get this:

c = - ( squareroot(2061*(cos(19))^2 + 1120) - 51 cos (19) ) or
squareroot(2061*(cos(19))^2 +1120) + 51 cos(19)

aproximately:

c = 110.943 or -10.095

don't use the the negative one for obvious reasons, so...

c = 110.943

use this to get the other ones...

EDIT: oops.. 19 is the external angle... that changes it

Rock lee: Please be mindful that using the edit button is preferable to using the 'post reply' button whenever it would result in a double post. I edited your post to include your double post, and deleted the double post.-- Bobslob

no 19 is on the inside, thats why i put an "arrow" there, even thought its hard to see.

Edit: Im just going to ask my teach tom. about it. So request lockage plz!

oops again, I think I was using radians, sorry

c = 106.918....

and I`m not entirely sure about loosing the negative one...
oh you'll ask a more experienced person, that's the best, good luck

Edit: ok, now I'm sure, this answer c = 106.918 is right (checked), and the negative one is useless....