“Solutions of
Triangles”
THE AREA RULE
THE SINE RULE
THE COSINE RULE
The formulae can be used in any triangle. No restrictions to right-angled triangles as
previously
Capital letters are used to indicate the angles and small letters for the lengths of sides
Round your answers off to 2 decimal places (unless otherwise indicated)
, THE AREA RULE
PROOF OF THE AREA RULE
B (c cosA; c sinA)
a
c
h
A (0; 0) D C (b; 0)
b
Given : △ABC
Required to prove :
1
Area △ABC = bc sin A
2
Proof : Place A in standard position and drop a perpendicular line from B to the x-
axis, the coordinate of B are (c cosA; c sinA)
If we consider AC to be the base, the y-coordinate of B is the height
∴ h = c sin A
1
∴ Area △ABC = 2 base x height
1
= 2 b x c sin A
Similarly, by placing B in standard position one can prove that
1
Area △ABC = 2 ac sin B
by placing B in standard position one can prove that
1
Area △ABC = 2 ab sin C
1 1 1
∴ Area △ABC = bc sin A = ac sin B = ab sin C
2 2 2
Triangles”
THE AREA RULE
THE SINE RULE
THE COSINE RULE
The formulae can be used in any triangle. No restrictions to right-angled triangles as
previously
Capital letters are used to indicate the angles and small letters for the lengths of sides
Round your answers off to 2 decimal places (unless otherwise indicated)
, THE AREA RULE
PROOF OF THE AREA RULE
B (c cosA; c sinA)
a
c
h
A (0; 0) D C (b; 0)
b
Given : △ABC
Required to prove :
1
Area △ABC = bc sin A
2
Proof : Place A in standard position and drop a perpendicular line from B to the x-
axis, the coordinate of B are (c cosA; c sinA)
If we consider AC to be the base, the y-coordinate of B is the height
∴ h = c sin A
1
∴ Area △ABC = 2 base x height
1
= 2 b x c sin A
Similarly, by placing B in standard position one can prove that
1
Area △ABC = 2 ac sin B
by placing B in standard position one can prove that
1
Area △ABC = 2 ab sin C
1 1 1
∴ Area △ABC = bc sin A = ac sin B = ab sin C
2 2 2