Expert: Sherry Wallin - 5/8/2011

Question
QUESTION: Hi, I'm currently covering the law of sines and cosines. I'm understanding the algebra portin so far, but when it comes to sketching and labeling the given parts I'm a bit lost. Although, I know that angles A,B & C are notated counter clockwise, and sides a,b & c are opposite the angles, I just can't draw it.

For instance: when given A = 32.0 dgr, B= 81.8 dgr, and a = 42.9 cm, It's been recommended that I draw a sketch before solving, but I'm just not seeing what the triangle looks like and where the hypotenuse is before I do so. How can I make this easier.

To be honest, I'm even having trouble sketching after I solve... This is really holding me up!

ANSWER: Gene~

Just from your statement "...where the hypotenuse is..." tells me you don't understand what is going on because the Law of Sines and Law of Cosines are for triangles that are NOT right triangles, so there is NO HYPOTENUSE, just another side. You could use either of those rules if you have the 'appropriate' information on a right triangle, but why? Pythagorus or regular trig using SOHCAHTOA is sufficient. So now down to the business of labeling the triangle. You are right that you have three angles and they can be labeled A,B, and C but you can call them anything you want in any order. It is standard if you use the alphabet to use capital letters to name the angles and lower case letters to name the sides opposite the angle in question.

         A = 32 deg
         /\
         b /   \
         /      \ c
         C        \
         \
      a = 42.9 cm   \        
         \
         B = 81.8 deg

Where you will need to assume there is a line between B and C that we call 'a'.

The Law of Sines says that the ratio between the sine of an angle and it side opposite will be equal to the sine of any other angle and it's opposite side. What this whittles down to is that in order to be able to use the Law of Sines you need either a pair of angles and one of it's opposite sides or a pair of sides with one of the opposite angles and set up the appropriate ratio and solve for the unknown side or angle respectively.

The Law of Cosines works when all you have is either all three angles or all three sides or a pair of angles and no opposite side or a pair of sides and not opposite angle.

Incidentally, it isn't necessary, and it is hardly ever true, that a triangle actually 'looks' like it is. As long as you have the measurements of the sides and angles you can use either of the Laws.

So in your example you have a pair of angle measures, >A, >B,  and you have an opposite side measure a, so you would use the Law of Sines in this way:

First you want to note that you can find the m<A by 180 - 81.8 - 32 => m<A = 66.2 deg
You also want to note what it is that you are searching for. Since you have angle B find 'b' first

a/sin A = b/sin B => 42.9/sin 32 = b/sin 81.8, so using algebra you get b = 42.9*sin 81.8/sin 32 so b = 80.1

You may also find c in a similar way: c/sin C = a/sin A => c/sin 66.2 = 42.9/sin 32 => c = 42.9*sin 66.2/sin 32 => c = 74.1

Hopefully I have helped you Gene. Feel free to ask further questions.

Math Prof

---------- FOLLOW-UP ----------

QUESTION: Wow thanks... you cleared alot up

Would you clarify the "ambigiuos case of the law of sines" and using the inverse function to see if there's a second triangle ( If I'm explaining it right)? My text provides a description, but its not completely registering.  

And

Answer
Gene~

The ambiguous case comes from looking at any triangle and having know information of SSA, that is two consecutive sides and the angle after them.
Picture a triangle labeling it A, B, and C. If you know a and b (two consecutive sides) and you know angle A (the angle after them in that order) then you have SSA. If you know b and a then the angle after them is angle B. This is the ambiguous case because depending on the length of the sides you might have no triangle (side a isn't long enough to close the triangle), one triangle where either a = the height of the triangle so a is just long enough to close the triangle hence you would have a right triangle or the side a is as long or longer than side b which makes there be only one triangle that is obtuse. The other scenario is that you could get two different triangles depending on if angle C is acute or obtuse. Note if angle C is acute then angle B is obtuse and if angle C is obtuse then angle B is acute. This is a slightly different way of explaining than in most textbooks, see if this is easier to understand and see.

The other more common way of explaining the ambiguous cases is to have you examine the side length a and the height h of the triangle which could be side a if it is a right triangle. Just using the right triangle assuming a is the height then sin A = h/b => h = b*sin A. If  = b*sin A and a is less than b then there are two possible triangles where either B is acute or C is acute (like I talked above). If a is greater than or equal to b then you get one triangle that is not a right triangle. If a = h then we have a single right triangle. The only other possibility is if a is less than b*sin A then the triangle will not close, This is primarily because the triangle inequality that says the sum of any two sides must be greater than the third side, or in other words, any one side must not be bigger than the sum of the other two sides.

Math Prof