1. Xtra Edge February 2012 - Career Point

sin A
a
sin B
=
b
=
sin C
c
1
=
2R
Where R is the radius of circumcircle of triangle
ABC.
2. Cosine formulae :
2 2 2
2 2 2
b + c − a a + c − b
cos A =
, cos B =
,
2bc
2ac
2 2 2
a + b − c
cos C =
2ab
It should be remembered that, in a triangle ABC
If ∠A = 60º, then b 2 + c 2 – a 2 = bc
If ∠B = 60º, then a 2 + c 2 – b 2 = ac
If ∠C = 60º, then a 2 + b 2 – c 2 = ab
3. Projection formulae :
a = b cos C + c cos B, b = c cos A + a cos C
c = a cos B + b cos A
Trigonometrical Ratios of the Half Angles of a Triangle:
a + b + c
If s = in triangle ABC, where a, b and c are
2
the lengths of sides of ∆ABC, then
A s ( s − a)
B s ( s − b)
(a) cos = , cos = ,
2 bc 2 ac
C
cos =
2
A
(b) sin =
2
C
sin =
2
A
(c) tan =
2
s ( s − c)
ab
( s − b)(
s − c)
bc
( s − a)(
s − b)
ab
( s − b)(
s − c)
,
s(
s − a)
B
, sin =
2
( s − a)(
s − c)
,
ac
B ( s − a)(
s − c)
C ( s − a)(
s − b)
tan =
, tan
2 s(
s − b)
2 s(
s − c)
Napier's Analogy :
B − C b − c A C − A c − a B
tan = cot , tan = cot
2 b + c 2 2 c + a 2
A − B a − b C
tan = cot
2 a + b 2
Area of Triangle :
1 1 1
∆ = bc sin A= ca sin B = ab sin C
2 2 2
∆ =
a sin B sin C b sinC
sin A c sin Asin
B
=
=
2 sin( B + C)
2 sin( C + A)
2 sin( A + B)
1 2
1 2
1 2
2
2 ∆
sin A = s( s − a)(
s − b)(
s − c)
=
bc
bc
2 ∆
2 ∆
Similarly sin B = & sin C =
ca
ab
Some Important Results :
A B
1. tan tan =
2 2
s − c A B
∴ cot cot =
s 2 2
A B c C c
2. tan + tan = cot = (s – c)
2 2 s 2 ∆
A B a − b
3. tan – tan = (s – c)
2 2 ∆
A B
4. cot + cot =
2 2
A B
tan + tan
2 2
A B
tan tan
2 2
XtraEdge for IIT-JEE 46 FEBRUARY 2012
=
s
s − c
c
s − c
cot C
2
5. Also note the following identities :
Σ(p – q) = (p – q) + (q – r) + (r – p) = 0
Σp(q – r) = p(q – r) + q(r – p) + r(p – q) = 0
Σ(p + a)(q – r) = Σp(q – r) + aΣ(q – r) = 0
Solution of Triangles :
1. Introduction : In a triangle, there are six
elements viz. three sides and three angles. In
plane geometry we have done that if three of the
elements are given, at least one of which must be
a side, then the other three elements can be
uniquely determined. The procedure of
determining unknown elements from the known
elements is called solving a triangle.
2. Solution of a right angled triangle :
Case I. When two sides are given : Let the
triangle be right angled at C. Then we can
determine the remaining elements as given in the
following table.
Given Required
(i) a, b
(ii) a, c
a
tanA = , B = 90º – A, c =
b
a
sin A
a
sinA = , b = c cos A, B = 90º – A
c
Case II. When a side and an acute angle are given –
In this case, we can determine
Given Required
(i) a, A
B = 90º – A, b = a cot A, c =
a
sin A
(ii) c, A B = 90º – A, a = c sin A, b = c cos A