Solution of Triangles
Basic Rules of Triangle
In a ∆ABC, the angles are denoted by capital letters A, B and C and
the lengths of the sides opposite to these angles are denoted by small
letters a , b and c, respectively. Area and perimeter of a triangle are
denoted by ∆ and 2s respectively.
A
b
c
B C
a
a+ b+ c
Semi-perimeter of the triangle is written as s = .
2
sin A sin B sin C 1
(i) Sine Rule = = = , where R is radius of the
a b c 2R
circumcircle of ∆ABC.
b2 + c2 − a 2 a 2 + c2 − b2
(ii) Cosine Rule cos A = , cos B =
2bc 2ac
a 2 + b2 − c2
and cosC =
2ab
(iii) Projection Rule a = b cos C + c cos B, b = c cos A + a cos C
and c = a cos B + b cos A
B−C b− c A
(iv) Napier’s Analogy tan = cot ,
2 b+ c 2
C− A c−a B A− B a− b C
tan = cot and tan = cot
2 c+a 2 2 a+b 2
, Trigonometrical Ratios of
Half of the Angles of Triangle
A ( s − b)( s − c) B ( s − c)( s − a )
(i) sin = , sin = ,
2 bc 2 ac
C ( s − a )( s − b)
sin =
2 ab
A s( s − a ) B s( s − b) C s( s − c)
(ii) cos = , cos = , cos =
2 bc 2 ac 2 ab
A ( s − b)( s − c) B ( s − a )( s − c)
(iii) tan = , tan =
2 s( s − a ) 2 s( s − b)
C ( s − a )( s − b)
tan =
2 s( s − c)
Area of a Triangle
Consider a triangle of side a , b and c.
1 1 1
(i) ∆ = bc sin A = ca sin B = ab sin C
2 2 2
c2 sin A sin B a 2 sin B sin C b2 sin C sin A
(ii) ∆ = = =
2 sin C 2 sin A 2 sin B
(iii) ∆ = s ( s − a ) ( s − b) ( s − c), its known as Heron’s formula.
a+ b+ c
where, s = [semi-perimeter of triangle]
2
abc
(iv) ∆ = = rs, where R and r are radii of the circumcircle and the
4R
incircle of ∆ABC, respectively.
Solutions of a Triangle
Elements of a Triangle
There are six elements of a triangle, in which three are its sides and
other three are its angle. If three elements of a triangle are given,
atleast one of which is its side, then other elements can be uniquely
calculated. This is called solving the triangle.
Basic Rules of Triangle
In a ∆ABC, the angles are denoted by capital letters A, B and C and
the lengths of the sides opposite to these angles are denoted by small
letters a , b and c, respectively. Area and perimeter of a triangle are
denoted by ∆ and 2s respectively.
A
b
c
B C
a
a+ b+ c
Semi-perimeter of the triangle is written as s = .
2
sin A sin B sin C 1
(i) Sine Rule = = = , where R is radius of the
a b c 2R
circumcircle of ∆ABC.
b2 + c2 − a 2 a 2 + c2 − b2
(ii) Cosine Rule cos A = , cos B =
2bc 2ac
a 2 + b2 − c2
and cosC =
2ab
(iii) Projection Rule a = b cos C + c cos B, b = c cos A + a cos C
and c = a cos B + b cos A
B−C b− c A
(iv) Napier’s Analogy tan = cot ,
2 b+ c 2
C− A c−a B A− B a− b C
tan = cot and tan = cot
2 c+a 2 2 a+b 2
, Trigonometrical Ratios of
Half of the Angles of Triangle
A ( s − b)( s − c) B ( s − c)( s − a )
(i) sin = , sin = ,
2 bc 2 ac
C ( s − a )( s − b)
sin =
2 ab
A s( s − a ) B s( s − b) C s( s − c)
(ii) cos = , cos = , cos =
2 bc 2 ac 2 ab
A ( s − b)( s − c) B ( s − a )( s − c)
(iii) tan = , tan =
2 s( s − a ) 2 s( s − b)
C ( s − a )( s − b)
tan =
2 s( s − c)
Area of a Triangle
Consider a triangle of side a , b and c.
1 1 1
(i) ∆ = bc sin A = ca sin B = ab sin C
2 2 2
c2 sin A sin B a 2 sin B sin C b2 sin C sin A
(ii) ∆ = = =
2 sin C 2 sin A 2 sin B
(iii) ∆ = s ( s − a ) ( s − b) ( s − c), its known as Heron’s formula.
a+ b+ c
where, s = [semi-perimeter of triangle]
2
abc
(iv) ∆ = = rs, where R and r are radii of the circumcircle and the
4R
incircle of ∆ABC, respectively.
Solutions of a Triangle
Elements of a Triangle
There are six elements of a triangle, in which three are its sides and
other three are its angle. If three elements of a triangle are given,
atleast one of which is its side, then other elements can be uniquely
calculated. This is called solving the triangle.
