Expert: Paul Klarreich - 11/23/2008

Question
Instructions: Use either Law of Sines or Law of Cosines as appropriate to solve the triangle.

1. measure of angle C = 150
side b = 7
side c = 5

What I tried doing:
Step 1. Law of Cosines
c^2 = a^2 + b^2 - 2abcos(C)
5^2 = a^2 + 7^2 - 2(a)(7)cos(150)
25 = a^2 + 49 - 14acos(150)
a^2 = 24 - 14acos(150)
((a^2)/(-14a))= 24 + cos(150)
((a)/(14))= 23.134
a = 323.876

(So, I think to myself that is a REALLY big number; but regardless, I continue)

Step 2. Law of Sines
(sinA)/323.876 = (sin(150))/5
5(sinA) = 161.938
sinA = 32.3876
A = arcsin(32.3876)

And, I get an error:domain message. :S

Answer
Questioner:   Laura
Category:  Advanced Math
Private:  No
 
Subject:  Calculus: Law of Sines and Cosines
Question:  Instructions: Use either Law of Sines or Law of Cosines as appropriate to solve the triangle.

1. measure of angle C = 150
side b = 7
side c = 5

What I tried doing:
Step 1. Law of Cosines

>> Not usually a good idea.  Here is the clue:

If you have both a side and its opposite angle GIVEN, use law of sines.

If you have SSS, or SAS given, (remember your basic geometry and congruent triangles?) use law of cosines.

In this case you have C = 150, c = 5, so you DO have a side and opposite angle.

>> HOWEVER, Here is another big clue:

*****************************************************************
In any triangle, the largest angle is opposite the largest side.
*****************************************************************

Since C is clearly biggest, so is c.  Therefore  a = 323 is impossible.  IN FACT, b = 7 is also impossible.  No wonder you are having some trouble.

Why not look over the example again?


P.S. This belongs in Advanced Math, not Calculus.