Area of a triangle

1
Area of a triangle
B
a
c
h
C
A
b
In the triangle ABC a perpendicular is drawn from
B to meet AC at point X. Using the sine ratio
for the triangle on the right
sin C h / a or h a sin C But the
area ABC ½ base x height ½ b h By
substitution area ABC ½ b x a sin C or ½ a b
sin C Area of a triangle ½ ab sin C
2
The Cosine Rule
B
a
c
Using Pythagoras Theorem in triangle BPA c2
h2 (b x) 2 which expanding gives c2
h2 x2 b2 - 2bx
h
A
C
x P b x
b

• Using Pythagoras Theorem again but in triangle
  BPC
• a2 h2 x2
• Now for some substitution into the 1st equation
• c2 a2 b2 - 2bx
• But using the cos ratio on triangle BPC cos C
  x / a or x a cos C
• so c2 a2
  b2 2ab cos C

3
The Sine Rule
C
a
b
h
B
A
X

• The diagram shows a general triangle ABC with
  side lengths of a,b and c and a perpendicular
  height h.
• Using the sine ratio in triangle AXC
• Sin A h / b rearranging h b sin A
• also Sin B h / a or h a
  sin B
• Hence b sin A a sin B
• a b
  c
• sin A sin B
  sin C