Page : 1 of 21 PROPRETIES OF TRIANGLE
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STUDY PACKAGE
Subject : Mathematics
Topic : Properties of Triangle
FREE Download Study Package from website: www.tekoclasses.com
98930 58881 , BHOPAL
R
TEKO CLASSES, H.O.D. MATHS : SUHAG R. KARIYA (S. R. K. Sir) PH: 0 903 903 7779,
Index
1. Theory
2. Short Revision
3. Exercise (Ex. 3 + 2 = 5)
4. Assertion & Reason
5. Que. from Compt. Exams
6. 39 Yrs. Que. from IIT-JEE
7. 15 Yrs. Que. from AIEEE
Student’s Name :______________________
Class :______________________
Roll No. :______________________
Address : Plot No. 27, III- Floor, Near Patidar Studio,
Above Bond Classes, Zone-2, M.P. NAGAR, Bhopal
: 0 903 903 7779, 98930 58881, WhatsApp 9009 260 559
www.TekoClasses.com www.MathsBySuhag.com
, Page : 2 of 21 PROPRETIES OF TRIANGLE
Properties & Solution of Triangle
1. Sine Rule:
In any triangle ABC, the sines of the angles are proportional to the opposite sides i.e.
a b c
= = .
sin A sin B sin C
A −B
cos
a+b 2
Example : In any ∆ABC, prove that = C .
c sin
2
−B
A
cos
a+b 2
Solution. ∵ We have to prove = C .
c sin
2
FREE Download Study Package from website: www.tekoclasses.com
∵ From sine rule, we know that
a b c
98930 58881 , BHOPAL
= = = k (let)
sin A sin B sin C
⇒ a = k sinA, b = k sinB and c = k sinC
a+b
∵ L.H.S. =
c
A +B A −B
sin cos
k(sin A + sin B) 2 2
= = C C
k sin C sin cos
2 2
C A −B A −B
cos cos
TEKO CLASSES, H.O.D. MATHS : SUHAG R. KARIYA (S. R. K. Sir) PH: 0 903 903 7779,
cos
2 2 2
= C C = C
sin cos sin
2 2 2
= R.H.S.
Hence L.H.S. = R.H.S. Proved
Example : In any ∆ABC, prove that
(b 2 – c 2 ) cot A + (c 2 – a 2 ) cot B + (a 2 – b 2 ) cot C = 0
Solution. ∵ We have to prove that
(b 2 – c 2 ) cot A + (c 2 – a 2 ) cot B + (a2 – b 2 ) cot C = 0
∵ from sine rule, we know that
a = k sinA, b = k sinB and c = k sinC
∴ (b 2 – c 2 ) cot A = k 2 (sin 2 B – sin 2 C) cot A
∵ sin 2 B – sin 2 C = sin (B + C) sin (B – C)
∴ (b 2 – c 2 ) cot A = k 2 sin (B + C) sin (B – C) cotA ∵ B+C=π–A
cos A
∴ (b 2 – c 2 ) cot A = k 2 sin A sin (B – C) ∵ cosA = – cos(B + C)
sin A
= – k 2 sin (B – C) cos (B + C)
k2
=– [2sin (B – C) cos (B + C)]
2
k2
⇒ (b 2 – c 2 ) cot A = – [sin 2B – sin 2C] ..........(i)
2
k2
Similarly (c 2 – a 2 ) cot B = – [sin 2C – sin 2A] ..........(ii)
2
k2
and (a 2 – b 2 ) cot C = – [sin 2A – sin 2B] ..........(iii)
2
adding equations (i), (ii) and (iii), we get
(b 2 – c 2 ) cot A + (c 2 – a 2 ) cot B + (a2 – b 2 ) cot C = 0 Hence Proved
Self Practice Problems
In any ∆ABC, prove that
A A
1. a sin + B = (b + c) sin .
2 2
A B
tan + tan
a 2 sin(B − C) b 2 sin(C − A ) c 2 sin( A − B) c 2 2
2. + + =0 3. = .
sin B + sin C sin C + sin A sin A + sin B a−b A B
tan − tan
2 2
fo/u fopkjr Hkh# tu] ugha vkjEHks dke] foifr ns[k NksM+s rqjra e/;e eu dj ';keA
iq#"k flag ladYi dj] lgrs foifr vusd] ^cuk^ u NksM+s /;s; dks] j?kqcj jk[ks VsdAA
jfpr% ekuo /keZ iz.ksrk
ln~ xq# Jh j.kNksM +nklth egkjkt
STUDY PACKAGE
Subject : Mathematics
Topic : Properties of Triangle
FREE Download Study Package from website: www.tekoclasses.com
98930 58881 , BHOPAL
R
TEKO CLASSES, H.O.D. MATHS : SUHAG R. KARIYA (S. R. K. Sir) PH: 0 903 903 7779,
Index
1. Theory
2. Short Revision
3. Exercise (Ex. 3 + 2 = 5)
4. Assertion & Reason
5. Que. from Compt. Exams
6. 39 Yrs. Que. from IIT-JEE
7. 15 Yrs. Que. from AIEEE
Student’s Name :______________________
Class :______________________
Roll No. :______________________
Address : Plot No. 27, III- Floor, Near Patidar Studio,
Above Bond Classes, Zone-2, M.P. NAGAR, Bhopal
: 0 903 903 7779, 98930 58881, WhatsApp 9009 260 559
www.TekoClasses.com www.MathsBySuhag.com
, Page : 2 of 21 PROPRETIES OF TRIANGLE
Properties & Solution of Triangle
1. Sine Rule:
In any triangle ABC, the sines of the angles are proportional to the opposite sides i.e.
a b c
= = .
sin A sin B sin C
A −B
cos
a+b 2
Example : In any ∆ABC, prove that = C .
c sin
2
−B
A
cos
a+b 2
Solution. ∵ We have to prove = C .
c sin
2
FREE Download Study Package from website: www.tekoclasses.com
∵ From sine rule, we know that
a b c
98930 58881 , BHOPAL
= = = k (let)
sin A sin B sin C
⇒ a = k sinA, b = k sinB and c = k sinC
a+b
∵ L.H.S. =
c
A +B A −B
sin cos
k(sin A + sin B) 2 2
= = C C
k sin C sin cos
2 2
C A −B A −B
cos cos
TEKO CLASSES, H.O.D. MATHS : SUHAG R. KARIYA (S. R. K. Sir) PH: 0 903 903 7779,
cos
2 2 2
= C C = C
sin cos sin
2 2 2
= R.H.S.
Hence L.H.S. = R.H.S. Proved
Example : In any ∆ABC, prove that
(b 2 – c 2 ) cot A + (c 2 – a 2 ) cot B + (a 2 – b 2 ) cot C = 0
Solution. ∵ We have to prove that
(b 2 – c 2 ) cot A + (c 2 – a 2 ) cot B + (a2 – b 2 ) cot C = 0
∵ from sine rule, we know that
a = k sinA, b = k sinB and c = k sinC
∴ (b 2 – c 2 ) cot A = k 2 (sin 2 B – sin 2 C) cot A
∵ sin 2 B – sin 2 C = sin (B + C) sin (B – C)
∴ (b 2 – c 2 ) cot A = k 2 sin (B + C) sin (B – C) cotA ∵ B+C=π–A
cos A
∴ (b 2 – c 2 ) cot A = k 2 sin A sin (B – C) ∵ cosA = – cos(B + C)
sin A
= – k 2 sin (B – C) cos (B + C)
k2
=– [2sin (B – C) cos (B + C)]
2
k2
⇒ (b 2 – c 2 ) cot A = – [sin 2B – sin 2C] ..........(i)
2
k2
Similarly (c 2 – a 2 ) cot B = – [sin 2C – sin 2A] ..........(ii)
2
k2
and (a 2 – b 2 ) cot C = – [sin 2A – sin 2B] ..........(iii)
2
adding equations (i), (ii) and (iii), we get
(b 2 – c 2 ) cot A + (c 2 – a 2 ) cot B + (a2 – b 2 ) cot C = 0 Hence Proved
Self Practice Problems
In any ∆ABC, prove that
A A
1. a sin + B = (b + c) sin .
2 2
A B
tan + tan
a 2 sin(B − C) b 2 sin(C − A ) c 2 sin( A − B) c 2 2
2. + + =0 3. = .
sin B + sin C sin C + sin A sin A + sin B a−b A B
tan − tan
2 2