## Vineet Loomba is teaching live on Unacademy Plus

IIT-JEE CRASH COURSE FOR SURE SHOT SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) IIT-JEE MATHS REVISION COURSE PREPARED BY: ER. VINEET LOOMBA IITian | IIT-JEE MENtOR Search vineet loomba unacademy" on GOOGLE

REVISION COURSE FOR JEE MAIN AND ADVANCED (IIT-JEE) ABOUT ME B.Tech. From IIT Roorkee IIT-JEE Mentor Since 2010. Doubts/Feedback in Comment Section. Comment other topics you want to revise. Follow me @ https://unacademy.com/user/vineetloomba to get updates. Share among your peers as SHARING is CARING !!

TRIGONOMETRY #PROPERTIES AND SOLUTION OF TRIANGLE IIT-IEE MATHS REVISION COURSE PREPARED BY: ER. VINEET LOOMBA IITiAN | IIT-JEE MENTOR FoR SURE SHoT SUcCESS IN JEE MAIN AND ADVANCED (IIT-JEE)

REVISION COURSE FOR JEE MAIN AND ADVANCED (IIT-JEE) Sine Rule In any triangle ABC abc 2A sin A sin B sin C where R is circumradius and is area of triangle. MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)

REVISION COURSE FOR JEE MAIN AND ADVANCED (IIT-JEE) A -B a+b Example : In any AABC, prove that- sin A -B 2 a b Solution.We have to prove Sin_ From sine rule, we know that sinA sin B sin C A+B A B A -B Sin |- cOS coS a+b k(sin A +sin B) k sin C L.H.S. - Sin R.H.S Proved MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)

REVISION COURSE FOR JEE MAIN AND ADVANCED (IIT-JEE) Example Angles of a triangle are in 4 11ratio. The ratio between its greatest side and perimeter is 3 N3 2+43 2-3 2+3 Solution Angles are in ratio 4 1 1. angles are 120 , 30 , 30 If sides opposite to these angles are a, b, c respectively, then a will be the greatest side Now from sine formula s 3/2 1/2 1/2 sin 120 sin 30 sin 30 =1=1=K (Say) then a=13k, permeter = (2+43)k ..r required ratio Ans. (B) (2+V3)k 2+V3 MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)

REVISION COURSE FOR JEE MAIN AND ADVANCED (IIT-JEE) Cosine Rule b2 +c2-a2 (a) cos A = 2 2 (b) cos Beta"- 2ca 2 = a +b2-c2 (c)cos C 2ab MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)

REVISION COURSE FOR JEE MAIN AND ADVANCED (IIT-JEE) Example : In a triangle ABC, if B -30 and c3 b, then A can be equal to (A) 45 (B) 60 (C) 90 (D) 120 Solution = S2+a2-b2 We have cOS B a2-3ab + 2b2 = 0 Either a = b a=2b-> (a-2b) (a-b) = 0 A = 30 or a2 = 4b2-b2 + c2-A-90 MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)

REVISION COURSE FOR JEE MAIN AND ADVANCED (IIT-JEE) PROJECTION FORMULAE (a) b cos C + c cos B = a (b) c cos A+ a cos C = b (c) acos B + b cos A = c th MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)

REVISION COURSE FOR JEE MAIN AND ADVANCED (IIT-JEE) In a ABC, c cos2-acos2 C 3b =- Example: , then show a, b, c are in A.P. 2 3b 2 3b Solution Here, (1 + cos A) + 2 (1+cos C) = 2 a+c+(c cos A + a cos C) fusing projection formula) a + c = 2b which shows a, b, c are in A.P. MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)