Mathematics

Session

Properties of Triangle - 2

Session Objectives

Session Objective

Solution of Right-angled Triangle

Solution of a Oblique Triangle

(a) When three sides are given

(b) When three angles are given

(c) When two sides and the included angle between them are given

(d) Two angles and one of the corresponding sides are given

(e) When two sides and an angle opposite to one of them is given

Introduction

A triangle has three sides and three angles. If three parts of a triangle, at least one of which is side, are given then other three parts can be uniquely determined.

Finding other unknown parts, when three parts are known is called ‘solution of triangle’.

Solution of Right-angled Triangle

(a) Given a side and an acute angle

Let the angle A (acute) and side c of a right angle at C be given

or b = c sinB, a = c cosB

B 90 A, b c cos A, a c sinA

c

b

a

C A

B

Solution of Right-angled Triangle

(b) Given two sides

22

atanA;B90A;

b

cab,etc.

c

a

b

C B

A

Let a and b are the sides of .C is theright angle. Then we can find the remaining angles and sides by the following way

ABC

Solution of a Oblique Triangle

(a) When three sides are given

(i) If the given data is in sine, use the following formulae

s – b s – c s – c s – aA Bsin , sin , etc.

2 bc 2 ca

(ii) If the given data is in cosine, use the following formulae.

s s a s s bA Bcos , cos , etc

2 bc 2 ac

Solution of a Oblique Triangle

s b s c s c s aA B

tan , tan ' etc2 s s a 2 s s b

(iii) If the given data is in tangent, then we use

(iv) If the lengths of the sides a, b and c are small, the angle of triangle can also be obtained by cosine rule.

(v) For logarithmic computation, we define the following.

Lcos 10 logcos , Lsin 10 logsin ,etc.

Solution of a Oblique Triangle

(b) When three angles are given:

In this case, the sides cannot be determined uniquely. Only ratio of the sides can be determined by sine rule and hence there will be infinite number of such triangles.

Solution of a Oblique Triangle

(c)When two sides and the included angle between them

are given:

If two sides b and c and the included angle A are given, then (B – C) can be found by using the following formula:

B C b c A

tan cot if b c2 b c 2

C B c b A

tan cot2 c b 2

If b < c, then we use

Solution of a Oblique Triangle

(d)Two angles and one of the corresponding sides are given:

The value of the other side and remaining angle can be found by

oa b cand A B C 180

sinA sinB sinC

Solution of a Oblique Triangle

(e) When two sides and an angle opposite to one of them is given:

In this case, either no triangle or one triangle or two triangles are possible depending on the given parts.

Therefore, this case is known as ambiguous case.

Let a, b and the angle A are given.

2 2 2b c acos A

2bc 2 2 2c 2bcos A c b a 0

Solution of a Oblique Triangle

This is a quadratic equation in c.

Let c1 and c2 be two values of c

2 2 2 22bcos A 4b cos A 4 b ac

2

2 2 2 2bcos A b cos A b a

2 2 2bcos A a b 1 cos A

2 2 2c bcos A a b sin A

Solution of a Oblique Triangle

2 2 21c bcos A a b sin A .....(i)

2 2 22and c bcos A a b sin A .....(ii)

Case I: When a < b sinA

2 2 2if a bsinA a b sin A

2 2 2or a b sin A 0

Hence, from (i) and (ii), become imaginary.

1 2c and c

No triangle is possible.

Solution of a Oblique Triangle

Case II: When a = b sinA

2 2 2If a bsinA a b sin A

2 2 2a b sin A 0

From (i) and (ii),

1 2c c bcos A

But will be positive when A is acute angle. In this case, only one triangle is possible provided is acute.

1 2c and c

A

Solution of a Oblique Triangle

a k sinA

b k sinBsinA sinA

sinB 1 B 90

The triangle is right-angled in this case.

Case III: When a > b sinA

2 2 2 2 2 2If a bsinA a b sin A a b sin A 0

From (i) and (ii), are real.1 2c and c

Also a = b sinA

Solution of a Oblique Triangle

But triangle is possible only when are positive. For this we have to consider the following cases:

1 2c and c

(a) When a > b

2 2a b a b

2 2 2 2a b cos A sin A

2 2 2 2 2a b sin A b cos A

2 2 2a b sin A bcos A

2 2 2bcos A a b sin A 0

1 2c 0 and c 0,

Hence only one triangle is possible.

Solution of a Oblique Triangle

(b) When a = b

2 2a b a b

2 2 2 2a b cos A sin A

2 2 2 2 2a b sin A b cos A

2 2 2a b sin A bcos A

1 2c 0 and c 0

Hence, only one triangle is possible.

Solution of a Oblique Triangle

(c) When a < b

2 2a b a b

2 2 2 2a b cos A sin A

2 2 2 2 2a b sin A b cos A

2 2 2a b sin A bcos A

1 2c 0 and c 0

Two triangles are possible. (Ambiguous case)

Thus, when a, b and are given, two triangles are possible when a > b sinA and a < b [For ambiguous case]

A

Class Test

Class Exercise - 1

If the sides of a triangle are

Prove that its largest angle is 120°.

2 2x x 1, 2x 1 and x 1.

Solution

2 2Let a x x 1, b 2x 1 and c x 1

a, b, c are the sides of the triangle,

a > 0, b > 0, c > 0

2 2 1 1 1a x x 1 x 2.x. 1

2 4 4

221 3x 0 for all x.

2 2

1

b 2x 1 0 x2

Solution Cont.

2c x 1 0 x 1 x 1 0

x 1 or x 1

Hence, a, b, c are positive, when x > 1Now

2 2x x 1 2x 1 x x x x 1 0 as x > 1

a – c = x2 + x + 1 – x2 + 1 = x + 2 > 0 as x > 1

a is the largest side A is the largest angle.

2 2 2b c acos A

2bc

Solution Cont.

2 22 2 2

2

2x 1 x 1 x x 1

2 2x 1 x 1

2 3

2

1 2x x 2x

2 2x 1 x 1

22

2 2

1 2x 1 x1 2x x 1 2x

2 2x 1 x 1 2 1 2x x 1

1A 120 (Pr oved)

2

Class Exercise - 2

The angles of a triangle are in the ratio 1 : 2 : 7. Show that the ratio of the greatest side to the least sides is 5 1 : 5 1 .

Solution

Let the angles of the triangle be x°, 2x°, 7x°.

x + 2x + 7x = 180° x = 18°

A = 18°, B = 36°, C = 126°

Least side is a and the greatest side is c.

a c c sinC

sinA sinC a sinA

sin 180 54c sin126 sin54

a sin18 sin18 sin18

Solution Cont.

5 1

c cos36 5 14a sin18 5 15 1

4Proved.

Class Exercise - 3

Let The number of triangle such that log b + 10 = log c + L sinB is

Lsin 10 logsin .

(a) one (b) two

(c) infinite (d) None of these

Solution

log b + 10 = log c + L sin B

log b + 10 = log c + 10 + log sinB

log b = log (c sinB)

b = c sinB

Only one triangle is possible

Class Exercise - 4

In the ambiguous case, if the remaining angles of the triangles formed with a, b and A be

1 21 1 2 2

1 2

sinC sinCB , C and B , C , then is

sinB sinB

(a) 2 cosA (b) cosA (c) 2 sinA (d) sinA

Solution

The two triangles formed are

1 2AB C and AB C.

1 1Let AB c

2 2AB c

A

a

C 1

C 2

B 1 L

C

B 2

b

1In AB C1

1 1

c bsinC sinB

1 1

1

sinC c............(1)

sinB b

2In AB C,

Solution Cont.

2 2 2

2 2 2

c sinC cb......(ii)

sinC sinB sinB b

1 2 1 2

1 2

sinC sinC c cFrom (i) and (ii) , ....(iii)

sinB sinB b

Triangles are formed with a, b and A,

2 2 2b c acos A

2bc

2 2 2c 2bcos A c b a 0

Solution Cont.

Here two values of c are .1 2c and c

1 2c c 2b cos A (Sum of roots)

1 2

1 2

From (iii),

sinC sinC 2bcos A2cos A

sinB sinB b

Hence answer is (a).

Class Exercise - 5

In ambiguous case, where b, c, B are given and prove that

23b c 1 8sin B.

2 1a 2a ,

Solution

b, c and B are given,

2 2 2c a bcosB

2ca

2 2 2a 2c cosB a c b 0

This is quadratic equation in a.

2 2 2 22ccosB 4c cos B 4 c ba

2

2 2 2 2c cosB c cos B c b

Solution Cont.

2 2 2a c cosB b c sin B

2 2 21a ccosB b c sin B

2 2 22and a c cosB b c sin B

It is given that 2 1a 2a

2 2 2c cosB b c sin B 2 2 22c cosB 2 b c sin B

Solution Cont.

2 2 2 2 2c cos B 9 b c sin B

2 2 2 2 2c 1 sin B 9b 9c sin B

2 2 2 29b c 1 8sin B 3b c 1 8sin B (Pr oved)

(Negative sign is neglected as ‘b’ is the length of side of triangle).

Thank you