Trigonometric - Properties of Triangles (Lecture-04)

Properties and Solutions of Triangles

Notation :

Vertices A,B,C

Sides a,b,c

Centroid G (Point of intersection of medians )

Orthocentre O (Point of intersection of altitudes)

Circumcentre S (Point of intersection of perpendicular bisectors of the sides)

Incentre I (Point of intersection of internal bisectors of the angles)

Excentres I1,I2,I3 (Point of intersection of internal bisector of an angle \& external bisectors of the other two angles).

Circumradius R (radius of circle with centre S and passing through the vertices)

Inradius r (radius of circle with centre I and touching the sides)

Exradii r1,r2,r3 ( radii of circles with centres I1,I2,I3 respectively and touching the sides)

Semiperimeter s=a+b+c2

Area of triangle Δ

Concepts and Formula

1. Sine law

In any triangle ABC,asinA=bsinB=csinC

2. Cosine law

In any triangle ABC

  • cosA=b2+c2a22bc or a2=b2+c22bccosA
  • cosB=c2+a2b22ac or b2=c2+a22cacosB
  • cosC=a2+b2c22ab or c2=a2+b22abcosC
3. Projection formula
  • a=bcosC+ccosB
  • b=cosA+acosC
  • c=acosB+bcosA
4. Napier’s analogy (Tangent rule)
  • tanBC2=bcb+ccotA2

  • tanCA2=cac+acotB2

  • tanAB2=aba+bcotC2

5. Auxiliary formulae

(Trigonometrical ratios of half angles of a triangle)

  • sinA2=(sb)(sc)bc;sinB2=(sc)(sa)ca;sinC2=(sa)(sb)ab;

  • cosA2=s(sa)bc;cosB2=s(sb)ca;cosC2=s(sc)ab;

  • tanA2=(sb)(sc)s(sa);tanB2=(sc)(sa)s(sb);tanC2=(sa)(sb)s(sc);

(=Δs(sa))(=Δs(sb))(=Δs(sc))

6. Area of Triangle
  • Area of triangle ABCΔ=s(sa)(sb)(sc) (Hero’s formula)

  • Δ=12absinC=12bcsinA=12casinB

  • Δ=12a2sinBsinCsin(B+C)=12b2sinCsinAsin(C+A)=12c2sinAsinBsin(A+B)

7.
  • sinA=2Δbc=2bcs(sa)(sb)(sc)

  • sinB=2Δca=2cas(sa)(sb)(sc)

  • sinC=2Δab=2abs(sa)(sb)(sc)

Also sinAa=sinBb=sinCc=2Δabc

8. mn Rule

In any triangle,

(m+n)cotθ=mcotαncotβ

=ncotBmcotC

9. Circumcircle of a triangle.

Circle passing through the angular point of a ABC is called its circumcircle. Its radius is denoted by R. The circum centre may lie within, outside or upon one of the sides of the

triangle. In a right angled triangle the circum centre is the mid-point of hypotenuse.

  • R=a2sinA=b2sinB=c2sinC

  • R=abc4Δ

10. Incircle of a triangle

The circle which touches the sides is called inscribed circle or incircle. Its radius is denoted by r.

  • r=Δs

  • r=(sa)tanA2=(sb)tanB2=(sc)tanC2

  • r=asinB2sinC2cosA2=bsinA2sinC2cosB2=csinB2sinA2cosC2

  • r=4RsinA2sinB2sinC2

11. Escribed circles of a triangle.

The circle which touches side BC and two sides AB&AC produced of ABC is called escribed circle opposite to the angle A. Its radius is denoted by r1. Similarly r2&r3 denote the radii of escribed circles opposite to angles B \& C.

  • r1=Δsa;r2=Δsb;r3=Δsc

  • r1=stanA2;r2=stanB2;r3=stanC2

  • r1=acosB2cosC2cosA2;r2=bcosA2cosC2cosB2r3=ccosB2cosA2cosC2

  • r1=4RsinA2cosB2cosC2;r2=4RcosA2sinB2cosC2;r3=4RcosA2cosB2sinC2

12. Orthocentre and Pedal triangle
  • The triangle MNP formed by joining the feet of the altitudes is called the pedal triangle

  • The distance of orthocentre O from vertices A,B,C are 2RcosA,2RcosB and 2RcosC

  • Distance of O from sides are 2RcosBcosC,2RcosCcosA and 2RcosAcosB.

  • InMNP,

M=π2C;MN=acosAN=π2 B;NP=cosCP=π2 A;MP=bcosB

i.e. sides of pedal triangles are a cosA(Rsin2 A), bcosB(Rsin2 B) and cosC(Rsin2C).

  • Circumradii of the triangles OBC,OCA&OAB are equal.
13. Note :
  • Orthocentre of ABC is the incentre of pedal triangle MNP.

  • Incentre I of ΔABC is the orthocentre of ΔΔI1,I2I3.

  • Centroid of ABC lies on the line joining the circumcentre to the orthocentre and divides it in the ratio 1:2

  • Circumcentre of the pedal triangle bisects the line joining the circumcentre of the triangle to the orthocentre.

14. Excentral Triangle.

The triangle formed by joining the three excentres I1,I2,&I3 of ABC is called the excentral or excentric triangle.

 Here AI1I2BI2I1&CI2I2

So clearly ΔABC is pedal triangle of excentral triangle Δ1I1I2I3

  • Sides of excentral trangle are 4RcosA2,4RcosB2 and 4RcosC2 and its angles are π2A2,π2B2,π2C2

  • II1=4RsinA2;II2=4RsinB2;II3=4RsinC2

15. Nine point circle

Circle circumscribing the pedal triangle of a given triangle bisects the sides of the given triangle and also the line joining the vertices of the given triangle to the orthocentre of the given triangle. This circle is known as nine point circle.

i.e. Nine point circle passes through the mid point of the sides, feet of the perpendiculars and the mid points of the line joining the orthocentre to the angular points.

16. Distance between special points
  • Distance between excentre and circumcentre

OI1=R1+8sinA2cosB2cosC2

OI2=R1+8cosA2sinB2cosC2OI3=R1+8cosA2cosB2sinC2

  • Distance between cirumcentre \& orthocentre is 18cosAcosBcosC

  • Distance between cirumcentre \& incentre is R22Rr

  • Distance between incentre \& orthocentre is 2r24R2cosAcosBcosC

  • Perimeter (P) and area (A) of a regular polygon of n sides inscribed in a circle of radius r are given by P=2nsinπn and A=12nr2sin2πn.

  • Perimeter (P) and area (A) of a regular polygon of n sides circumscribed about a given circle of radius r are given by P=2nrtanπn and A=12nr2tan2πn

17. Length of medians (Apollonious rule)

m12=b2+c22a24

m22=c2+a22b24

m32=a2+b22c24

Also m12+m22+m32=34(a2+b2+c2)

If A=90, then m22+m32=5 m12

Area =43m(mm1)(mm2)(mm3) where 2m=m1+m2+m3.

18. Altitudes

h1=2Δa,h2=2Δb,h3=2Δc,

Area is given by 1Δ=41 h(1 h1 h1)(1 h1 h2)(1 h1 h3) where 2 h=1 h1+1 h2+1 h3

19. Length of internal bisectors of angles A,B,C are given by 2bcb+ccosA2;2cac+acosB2;

2aba+bcosC2 respectively

20. Some useful results.
  • r1+r2=4Rcos2C2

  • r3r=4Rsin2C2.

  • r1+r2+r3r=4R

  • rr1r2r3=Δ2

  • 1r1+1r2+1r3=1r

  • 1r2+1r12+1r22+1r32=a2+b2+c2Δ2

  • s=r1r2+r2r3+r3r1=r1r2

  • Δ=r1r2r3 s=r1r2r3r1r2

  • r=r1r2r3r1r2

  • R=(r1+r2)(r2+r3)(r3+r1)4r1r2 a=r1(r2+r3)r1r2;b=r2(r3+r1)r1r2;c=r3(r3+r2)r1r2

21. Solution of triangles

(a) Solution of a general triangle.

(b) Solution of a right angled triangle

Let C=90

Given To find Formulae used
a,b A, B, c c2=a2+b2
(two sides) tanA=ab( or tanB=ba)
B=90A( or A=90B)
c,a A, B, b b2=c2a2
(hypotenuse & one side) SinA=ac
b=cosA or b=acotA
B=90A
a, A B, b, c B=90A
(Side and angle) b=acotA
c=asinA
c,A B, a, b B=90A
(hypotenuse & angle) a=csinA
b=ccosA

Solved Examples

1. If in ABC,a=6,b=3 and cos(AB)=45 then its area square units is

(a) 8

(b) 9

(c) 6

(d) None of these

Show Answer

Solution: Using Napier’s analogy,

tanAB2=aba+bcotC2 and tan(AB2)=1cos(AB)1+cos(AB)

13=636+3cotC2 tan(AB2)=1451+45=13

cotC2=1

C=90

A=12absinC=1263sin90=9 square units

Answer (b)

2. If the angles A,B,C are the solutions of the equation tan3x3ktan2x3tanx+k=0, then the ABC is

(a) isosceles

(b) equilateral

(c) acute angled

(d) None of these

Show Answer

Solution: tanA,tanB,tanC are roots of the given equation

tanA+tanB+tanC=3k,tanAtanB+tanBtanC+tanCtanA=3 and tanAtanBtanC=k

But tanA+tanB+tanC=tanAtanBtanC

3k=k4k=0 gives k=0

tanAtanBtanC=0

tanA=0 or tanB=0 or tanC=0 which is not possible

Answer (d)

3. In ABC

sin4 A+sin4 B+sin4C=sin2 Bsin2C+2sin2Csin2 A+2sin2 Asin2 B, then A=

(a) π6,5π6

(b) π6,5π3

(c) 5π6,π3

(d) None of these

Show Answer

Solution: Given equation is,

a4+b4+c4=b2c2+2c2a2+2a2 b2 (using sine rule)

a4+b4+c4+2 b2c22c2a22a2 b2=3 b2c2

(b2+c2a2)2=3 b2c2

(b2+c2a2)24 b2c2=34

(b2+c2a22bc)2=(32)2

cos2 A=34cosA=±32

A=π6,5π6

Answer (a)

4. If the median of a triangle through A is perpendicular to AB, then

(a) 2tanA+tanB=0

(b) 2tanAtanB=0

(c) tanA2tanB=0

(d) tanA+2tanB=0

Show Answer

Solution:

tanA=tan(180A)=2xc=2(xc)=2tanBtanA+2tanB=0

Answer (d)

5. In ABC, medians AD and BE are drawn. If AD=4,DAB=π6 and ABE=π3, then the area of ABC is (in square units)

(a) 6433

(b) 833

(c) 1633

(d) 3233

Show Answer

Solution:

InΔABGcosπ6=AGAB32=23ADAB32=2×43ABAB=1633

Area of ABD=12ABADsinπ6

=121633412=1633 units 2

Area of ABC=2 area of ABD=3233 units 2

Answer (d)

6. In ABC,a2+b2+c2=ac+3 ab, then the triangle is

(a) equilateral

(b) isosceles

(c) right angled

(d) None of these

Show Answer

Solution:

a2+b2+c2ac3ab=0

(a24+c2ac)+(3a24+b23ab)=0

(a2c)2+(3a2b)2=0

a2=c&32a=b

a=2c&a=2 b3

a=2c=2 b3=λ

a=λ,c=λ2, b=3λ2

a2=b2+c2

right angled

Answer (c)

7. In ABC, if cotA+cotB+cosC=3, then show that the triangle is equilateral

Show Answer

Solution:

In ABC,tanA+tanB+tanC=tanAtanBtanC

cotBcotC+cotCcotA+cotBcotA=1……………………..(1)

Let cotA=x,cotB=y,cotC=z

given that x+y+z=3

Squaring,

x2+y2+z2+2xy+2yz+2zx=3×1

x2+y2+z2+2xy+2yz+2zx=3(xy+yz+zx) using (1) 

i.e. x2+y2+z2xyyzzx=0

2x2+2y2+2z22xy2yz2zx=0

(xy)2+(yz)2+(zx)2=0

x=y&y=z&z=x

x=y=z

cotA=cotB=cotC

A=B=C

The triange is equilateral.

Exercise

1. In ABC,B=π3,C=π4, Let D divides BC internally in the ratio 1:3, then sinBADsinCAD is equal to

(a) 16

(b) 13

(c) 13

(d) 23

Show Answer Answer: a

2. If in PQR,sinP,sinQ,sinR are in A.P., then

(a) The altitudes are in A.P

(b) The medians are in G.P

(c) The altitudes are in H.P

(b) The medians are in A.P

Show Answer Answer: c

3. In PQR,R=π2, if tan(P2) and tan(Q2) are the roots of ax2+bx+c=0(a0), then

(a) a+b=c

(b) b+c=a

(c) c+a=b

(d) b=c

Show Answer Answer: a

4. In ABC,2acsin(AB+C)2=

(a) a2+b2c2

(b) c2+a2b2

(c) b2c2a2

(d) c2a2b2

Show Answer Answer: b

5. In ABC, let C=π2. If r is the inradius and R is the circumradius of the triangle, then 2(r+R) is equal to

(a) a+b

(b) b+c

(c) c+a

(d) a+b+c

Show Answer Answer: a

6. Inradius of a circle which is inscribed in an isosceles triangle one of whose angle is 2π3 is 3, then area of triangle is

(a) 43

(b) 1273

(c) 12+73

(d) None of these

Show Answer Answer: c

7. If the angles A,B&C of a triangle are in A.P and if a,b and c denote the lengths of the sides opposite to A,B&C respectively, then the value of the expression acsin2C+casin2 A is

(a) 12

(b) 32

(c) 1

(d) 3

Show Answer Answer: d

8. Read the following passage and answer the questions.

Consider the circle x2+y2=9 and the parabola y2=8x. They intersect at P&Q in the first and fourth quadrants, respectively. Tangents to the circle at P&Q intersect the x-axis at R and tangents to the parabola at P&Q intersect the x axis at S.

(i) The ratio of the areas of ΔPQS and ΔPQR is

(a) 1:2

(b) 1:2

(c) 1:40

(d) 1:8

(ii) The radius of the circumcircle of the ΔPRS is

(a) 5

(b) 33

(c) 32

(d) 23

(iii) The radius of the incircle of the triangle PQR is

(a) 4

(b) 3

(c) 83

(d) 2

Show Answer Answer: (i) c (ii) b (ii) d

9.* Internal bisector A of ABC meets side BC at D. A line drawn through D perpendicular to AD intersects the side AC at E& side AB at F. If a,b,c represent sides of ABC, then

(a) AE is the HM of b \& c

(b) AD=2bcb+ccosA2

(c) EF=4bcb+csinA2

(d) AEF is isosceles

Show Answer Answer: a,b,c,d

10.* A straight line through the vertex P of a ΔPQR intersects the side QR at the point S and the circumcircle of the triangle PQR at the point T. If S is not the centre of the circumcircle, then

(a) 1PS+1ST<2QS×SR

(b) 1PS+1ST>2QS×SR

(c) 1PS+1ST<4QR

(d) 1PS+1ST>4QR

Show Answer Answer: b,d

11.* In a ABC with fixed base BC, the vertex A moves such that cosB+cosC=4sin2A2. If a,b and c denote the lengths of the sides of the triangle opposite to the angles A,B and C respectively, then

(a) b+c=4a

(b) b+c=2a

(c) locus of point A is an ellipse

(d) locus of point A is a pair of straight line

Show Answer Answer: b,c

12. Match the following:

Column I Column II
(a) In ABC, if cosAa=cosBb=cosCc and the side a=2, then area of the triangle is (p) 23+6
(b) In ABC,abc, if a3+b3+c3sin3 A+sin3 B+sin3C=7, then the maximum possible value of a is (q) (q) 16
(c) Two sides of a triangle are given by the roots of the equation x223x+2=0. the angle between the sides is π3.The perimeter of the triangle is (r) 73
(s) 3
Show Answer Answer: as;br;cp

13. In ABC if cosA+CosB+cosC=74, then Rr=

(a) 43

(b) 34

(c) 23

(d) 32

Show Answer Answer: a

14. In ABC, sides a,b,c are in A.P and 21!9!+23!7!+15!5!=8a(2 b)! then the maximum value of tanAtanB is

(a) 12

(b) 13

(c) 14

(d) 15

Show Answer Answer: b

15. If A1 A2 A3.An be a regular polygon of n sides and 1 A1 A2=1 A1 A3+1 A1 A4, then

(a) n=5

(b) n=6

(c) n=7

(d) None of these

Show Answer Answer: c