MAT129 Fall 2025 Notes
mathnstuff.com/math/precalc/MAT129fa25/p.22.htm
© 2025   A2, Agnes Azzolino
Law of Sines; Law of Cosines, Text 10.1, 10.2, PG. 762, 776

Welcome to Precalc !
Sine Law & Cosine Law
Hi,
    With these two tools, ANY triangle may be solved -- even triangles which are not right triangles.
   
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10.1  - Non-right Triangles: Law of Sines, pg. 762,  Learning Objectives
	* Use the Law of Sines to solve oblique triangles.
	* Find the area of an oblique triangle using the sine function.
	* Solve applied problems using the Law of Sines.
Law of Sines
sin (A) / a = sin(B) / b = sin(C) / c
a / sin (A) = b / sin (B) = c / sin (C)		 Law of Sines
a = (c*sin(A) / sin(C)
a = (b*sin(A) / sin (B)
b = (a*sin(B)) / sin (A)
b = (c*sin(B)) / sin (C), etc.
  
10.2  - Non-right Triangles: Law of Cosines, pg. 776,  Learning Objectives
	* Use the Law of Cosines to solve oblique triangles.
	* Solve applied problems using the Law of Cosines.
	* Use Heron’s formula to find the area of a triangle.
 Law of Cosines
  a2 = b2 + c2 - 2bccos(A)
  b2 = a2 + c2 - 2accos (B)
  c2 = a2 + b2 - 2abcos(C) 		cos(A) = ( b2 + c2 - a2) / ( 2(b)(c) )
cos(B) = ( a2 + c2 - c2) / ( 2(a)(c) )
cos(C) = ( a2 + b2 - c2) / ( 2(a)(b) )
 
  A = cos-1 [ b2 + c2 - a2) / (2(b)(c) ) ]
  B = cos-1 [ a2 + c2 - b2) / (2(a)(c) ) ]
  C = cos-1 [ a2 + b2 - c2) / (2(a)(b) ) ]
Solving a Triangle -- Easy to Harder
Use Pythagorean Theorem, arithmetic, and basic trig with angles in degrees.
  Input leg a and leg b then press
each button to compute A, B, and
c to 3 decimal places.
is ° done a is  
is ° done b is  
C is 90° is done
A is sin-1(a/c)*180/(pi) °
B is 90-A °
C is 90°
c is sqrt(a2 + b2)
  Input hypotenuse c and leg a then
press ach button to compute A, B,
and b to 3 decimal places.
is ° done a is
is ° done is done
C is 90° c is
A is sin-1(a/c)*180/(pi) °
B is cos-1(a/c)*180/(pi) °
C is 90°
b is sqrt(c2 - a2)
  Input angle A and leg a then press
each button to compute B, b and
c to 3 decimal places.
A is ° a is
is ° done is done
C is 90° is done
b is a / (tan(A * (pi/180) )
B is 90 - A °
C is 90°
c is sqrt( a2 + b2)

Use the Sine Law
If a side and the opposite angle, a pair, a side and corresponding angle, are given.
Input angles A , B, side a. Seek two sides and an angle.
Input angle A , side a, side b. Seek two angles and a side.

Input sides a & b & angle A. Seek B, C, c.
A is ° done a is   done
is ° working ** b is   done
is ° working is
B = sin-1(b*sin(A) / a)
C = sin-1(c*sin(A) / a)
c = a*sin(C) /sin(A)
Input sides a & b & angle A.
Seek angle B, b, and angle C.
Ambiguous case -- ASS or SSA
A is a is
B is is
C compute c is compute
a / sin(A) = b / sin(B) = c / sin(C)
sin(A) / a = sin(B) / b = sin(C) / c
b = (a*sin(B)) / sin (A)
c = (a*sin(C)) / sin (A)
C = sin-1(c*sin(A) / a)

Use the Law of Cosines
Input sides a, b, c. Seek each angle.
Input angle A , sides b,c. Seek no solution, 1 solution, or 2 solutions.
Input sides a & b & angle A. Seek side c.
Seek: no solution, 1 solution, or 2 solutions.
is ° a is  
is ° b is  
C is   also working is 
Law of Cosines
a2 = b2 + c2 -  2bccos(A)
b2 = a2 + c2 -  2accos (B)
c2 = a2 + b2 -  2abcos(C)
Input sides a, b, c. Seek each angle.
is ° done a is
is ° done b is
is ° done c is
cos(A) = ( b2 + c2 - a2) / ( 2(b)(c) )
cos(B) = ( a2 + c2 - c2) / ( 2(a)(c) )
cos(C) = ( a2 + b2 - c2) / ( 2(a)(b) )
 
  A = cos-1 [ b2 + c2 - a2) / (2(b)(c) ) ]
  B = cos-1 [ a2 + c2 - b2) / (2(a)(c) ) ]
  C = cos-1 [ a2 + b2 - c2) / (2(a)(b) ) ]