Relation between Sides and Angles:
A triangle has six components, three sides and three angles. The three angles of a ΔABC are denoted by letters A, B, C and the sides opposite to these angles by letters a, b and c respectively.
A + B + C = 180º = (or π)
a + b > c, b + c > a, c + a > b
|a – b| < c, |b – c| < a, |c – a| < b
The Law of sines or sine Rule:
The sides of a triangle are proportional to the sines of the angles opposite to them i.e., (a/sin A) = (b/sin B) = (c/sin C) = k, (say)
More generally, if R be the radius of the circumcircle of the triangle ABC, (a/sin A) = (b/sin B) = (c/sin C) = 2R.
The Law of cosines or cosine Rule:
For a triangle ABC,
a2 = b2 + c2 – 2bc cos A ⇒ cos A = (b2 + c2 – a2 )/2bc
b2 = c2 + a2 – 2ca cos b ⇒ cos B = (c2 + a2 – b2)/2ca
c2 = a2 + b2 – 2ab cos C ⇒ cos C = (a2 + b2 – c2)/2ab
Combining with sin A = (a/2R), sin B = (b/2R), sin C = (c/2R)
We have by division,
where, R be the radius of the circum-circle of the triangle ABC.
Projection Formulae:
In any triangle ABC
Theorem of the medians: (Apollonius theorem):
In every triangle the sum of the squares of any two sides is equal to twice the square on half the third side together with twice the square on the median that bisects the third side.
For any triangle ABC, b2 + c2 = 2(h2 + m2) = 2{h2 + m2} = 2{m2 + (a/2)2} by use of cosine rule.
If Δ be right angled, the mid point of hypotenuse is equidistant from the three vertices so that DA = DB = DC.
Napier's analogy (Law of tangents):
For any triangle ABC,
Mollweide's Formula: For any triangle,
Area of Triangle
Let three angles of ΔABC are denoted by A, B, C and the sides opposite to these angles by letters a, b, c respectively.
(1) When two sides and the included angle be given:
The area of triangle ABC is given by,
i.e., Δ = (1/2) (Product of two sides) × sine of included angle
(2) When three sides are given:
Area of ΔABC = Δ = √(s(s – a)(s – b) (s – c))
where semiperimeter of triangle s = (a + b + c)/2
(3) When three sides and the circum-radius be given:
Area of triangle Δ = (abc)/4R, where R be the circum-radius of the triangle.
(4) When two angles and included side be given:
Half Angle Formulae
If 2s shows the perimeter of a triangle ABC then, i.e., 2s = a + b + c, then
Formulae:
(1) Circumcircle of a triangle and its radius:
(i) Circumcircle: The circle which passes through the angular points of a triangle is called its circumcircle. The centre of this circle is the point of intersection of perpendicular bisectors of the sides and is called the circumcentre. Its radius is always denoted by R. The circumcentre may lie within, outside or upon one of the sides of the triangle.
(ii) Circum-radius: The circum-radius of a ΔABC is given by
(a)
(b) R = (abc/4Δ) [Δ = area of ΔABC]
(2) Inscribed circle or in-circle of a triangle and its radius:
(i) In-circle or inscribed circle: The circle which can be inscribed within a triangle so as to touch each of its sides is called its inscribed circle or in circle. The centre of this circle is the point of intersection of the bisectors of the angles of the triangle. The radius of this circle is always denoted by r and is equal to the length of the perpendicular from its centre to any one of the sides of triangle.
(ii) In-radius: The radius r of the inscribed circle of a triangle ABC is given by
(a) r = Δ/s
(b)
(c) r = (s – a) tan(A/2),
(d)
(e) cos A + cos B + cos C = 1 + (r/R)
(3) Escribed circles of a triangle and their radii:
(i) Escribed Circle: The circle which touches the side BC and two sides AB and AC produced of a triangle ABC is called the escribed circle opposite to the angle A. Its radius is denoted by r1. Similarly, r2 and r3 denote the radii of the escribed circles opposite to the angles B and C respectively.
The centres of the escribed circles are called the ex-centres. The centre of the escribed circle opposite to the angle A is the point of intersection of the external bisectors of angles B and C. The internal bisectors of angle A also passes through the same point. The centre is generally denoted by I1.
(ii) Radii of ex-circles In any ΔABC, we have
(a) r1 = (Δ/(s–a)), r2 = (Δ/(s–b)), r3 = (Δ/(s–c))
(b) r1 = s (tan (A/2)), r2 = s(tan B/2) , r3 = s(tan(C/2))
(c)
(d) r1 + r2 + r3 – r = 4R
(e)
(f)
(g)
(h) r1r2 + r2r3 + r3r1 = s2
(i) Δ = 2R2 sin A. sin B. sin C = 4R cos(A/2). cos(B/2). cis(C/2)
(j) r1 = 4R sin(A/2) cos (B/2) cos (C/2); r2 = 4R cos (A/2). sin (B/2). cos (C/2)
(4) Centroid (G):
Common point of intersection of medians of a triangle. Divides every median in the ratio 2:1. Always lies inside the triangle.
(5) Orthocentre of a triangle:
The point of intersection of perpendicular drawn from the vertices on the opposite sides of a triangle is called its orthocentre.
► Let the perpendiculars AD, BE and CF from the vertices A, B and C on the opposite sides BC, CA and AB of ΔABC respectively, meet at O. Then O is the orthocentre of the ΔABC. The triangle DEF is called the pedal triangle of the ΔABC.
► Othocentre of the triangle is the incentre of the pedal triangle.
► If O is the orthocentre and DEF the pedal triangle of the ΔABC, where AD, BE, CF are the perpendiculars drawn from A, B, C on the opposite sides BC, CA, AB respectively, then
OA = 2R cos A, OB = 2R cs B and OC = 2R cos C
OD = 2R cos B cos C, OE = 2R cos C cos A and OF = 2R cos A cos B
(1) Sides and angles of a pedal triangle:
The angles of pedal triangle DEF are: 180 – 2A, 180 – 2B, 180 – 2C and sides of pedal triangle are: EF = a cos A or R sin 2A; FD = b cos B or R sin 2B; DE = c cos C or R sin 2C
If given ΔABC is obtuse, then angles are represented by 2A, 2B, 2C – 180º and the sides are a cos A, b cos B, –c cos C.
(2) Area and circum-radius and in-radius of pedal triangle:
Area of pedal triangle = 1/2 (product of the sides) × (sine of included angle)
Δ = (1/2)R2. sin 2A. sin 2B. sin 2C
Circum-radius of pedal triangle
In-radius of pedal triangle
= 2R cos A. cos B. cos C
Ex-central Triangle
► Let ABC be a triangle and I be the centre of incircle. Let I1, I2, and I3 be the centres of the escribed circles which are opposite to A, B, C respectively then I1I2I3 is called the Ex-central triangle of ΔABC
► I1I2I3 is a triangle, thus the triangle ABC is the pedal triangle of its ex-central triangle I1I2I3. The angles of ex-central triangle I1I2I3 are 90º – (A/2), 90º – (B/2), 90º – (C/2) and sides are I1I2 = 4R cos (B/2); I1I2 = 4R cos (C/2); I1I2 = 4R cos (A/2)
Area and circum-radius of the ex-central triangle:
Area of triangle
= 1/2 (Product of two sides) × (sine of included angles)
Circum-radius
Cyclic Quadrilateral
► A quadrilateral PQRS is said to be cyclic quadrilateral if there exists a circle passing through all its four vertices P, Q, R and S.
► Let a cyclic quadrilateral be such that PQ = a, QR = b, RS = c and SP = d.
► Then ∠Q + ∠S = 180º, ∠A + ∠C = 180º
► Let 2s = a + b + c + d
► Area of cyclic quadrilateral = 1/2(ab + cd) sin Q
► Also, area of cyclic quadrilateral = √(s–a) (s–b) (s–c) (s–d), where 2s = a + b + c + d and cos Q = (a2 + b2 – c2 – d2)/2(ab + cd).
(1) Circumradius of cyclic quadrilateral: Circum circle of quadrilateral PQRS is also the circumcircle of ΔPQR.
(2) Ptolemy's theorem: In a cyclic quadrilateral PQRS, the product of diagonals is equal to the sum of the products of the length of the opposite sides i.e., According to Ptolemy's theorem, for a cyclic quadrilateral PQRS PR.QS = PQ.RS + PQ.PS
► A regular polygon is a polygon which has all its sides equal and all its angles equal.
(1) Each interior angle of a regular polygon of n sides is ((2n – 4)/n) × right angles = [(2n–4)/n] × (π/2) = radians.
(2) The circle passing through all the vertices of a regular polygon is called its circumscribed circle.
► If a is the length of each side of a regular polygon of n sides, then the radius R of the circumscribed circle, is given by R = (a/2). cosec(π/n)
(3) The circle which can be inscribed within the regular polygon so as to touch all its sides is called its inscribed circle.
► Again if a is the length of each side of a regular polygon of n sides, then the radius r of the inscribed circle is given by r = (a/2). cot(π/n)
(4) The area of a regular polygon is given by Δ = n × area of triangle OAB
= (1/4)na2 cot(π/n), (in terms of side)
= nr2. tan(π/n), (in terms of in-radius)
= (n/2).R2 sin(2π/n), (in terms of circum-radius)
Point at the Glance
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►
►
►
► Circum-centre, Centroid and Orthocentre are collinear.
► In any right angled triangle, the orthocentre coincides with the vertex containing the right angled.
► The mid-point of the hypotenuse of a right angled triangle is equidistant from the three vertices of the triangle.
► The mid-point of the hypotenuse of a right angled triangle is the circumcentre of the triangle.
► The length of the medians AD, BE, CF of ΔABC are given by AD = (1/2)√2b2 + 2c2 – a2, = (1/2)√b2 + c2 – 2bc cos A
►
►
► The distance between the circumcentre O and centroid G of ΔABC is given by
► , where H is the orthocentre of ΔABC.
► The distance between the orthocentre H and centroid G of ΔABC is given by .
► The distance between the circumcentre O and the incentre I of ΔABC given by
► If I1 is the centre of the escribed circle opposite to the angle B, then
Similarly,
► sin A + sin B + sin C is maximum, when A = B = C.
► cos A + cos B + cos C is maximum, when A = B = C.
► tan A + tan B + tan C is minimum, when A = B = C.
► cot A + cot B + cot C is minimum, when A = B = C
► If cos A + cos B + cos C = (3/2), then the triangle is equilateral.
► If sin A + sin B + sin C = ((3√/3)/2), then the triangle is equilateral.
► If tan A + tan B + tan C = 3√3, then the triangle is equilateral.
► If cot A + cot B + cot C = √3, then the triangle is equilateral.
► If cos2 A + cos2 B + cos2 C = 1, then the triangle is right angled.
► Circle circumscribing the pedal triangle of a given triangle bisects the sides of the given triangle and also the lines joining the vertices of the given triangle to the orthocentre of the given triangle. This circle is known as "Nine point circle".