Falling Behind in Studies? Join Our Performance Improvement Batch. Enroll For Free
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Chapter 10: Sine and Cosine Formulae and Their Applications – Exercise 10.2 Sine and Cosine Formulae and Their Applications – Exercise 10.2 – Q.1 The area of a triangle ABC is given by Sine and Cosine Formulae and Their Applications – Exercise 10.2 – Q.2 The area of a triangle ABC is given by Therefore, Sine and Cosine Formulae and Their Applications – Exercise 10.2 – Q.3 We have, a = 4, b = 6 and c = 8 Sine and Cosine Formulae and Their Applications – Exercise 10.2 – Q.4 In any ∆ABC, we have we have, a = 18, b = 24, c = 30 Therefore, Sine and Cosine Formulae and Their Applications – Exercise 10.2 – Q.5 b(c cos A – a cos C) = c2 - a2 RHS = c2 - a2 = k2 sin2C - k2 sin2 A = k2(sin2 C - sin2 A) = k2sin(C + A). sin(C - A) = k2 sin(π - B).sin(C - A) = k2 sin B.sin(C - A) = k sin B.k sin(C - A) = bk sin (C-A) = bk(sin C.cos A - sin A.con C) = b(k sin C.cos A - k sin A. cos C) = b(b cos A - a cos C) = LHS Sine and Cosine Formulae and Their Applications – Exercise 10.2 – Q.6 c(a cos B - b cos A) = ac.cos B - bc cos A Sine and Cosine Formulae and Their Applications – Exercise 10.2 – Q.7 2(bc cos A + ca cos B + ab cos C) = a2 + b2 + c2 LHS = 2bc cos A + 2ca cos B + 2ab cos C = b2 + c2 - a2 + a2 + c2 - b2 + a2 + b2 - c2 = a2 + b2 + c2 = RHS Sine and Cosine Formulae and Their Applications – Exercise 10.2 – Q.8 For any ∆ABC, We have therefore, Also, Now, Hence proved. Sine and Cosine Formulae and Their Applications – Exercise 10.2 – Q.10 In any ∆ABC, we have a = b cos C +c cos B b = c cos A + a cos C c = a cos B + b cos A Therefore, L.H.S = a(cos B + cos C - 1) + b(cos C + cos A - 1) + c(cos A + cos B - 1) = a cos B + a cos C - a + b cos C + b cos A - b + c cos A + c cos B - c = c - b cos A + a cos C - a + a - c cos B + b cos A - b + b - a cos C + c cos B - c = 0 = RHS Hence proved. Sine and Cosine Formulae and Their Applications – Exercise 10.2 – Q.11 By sine rule we have K sin A = a, K sin B = b, k sin C = c a cos A + b cos B + cos C = k sin A cos A + k sin B cos B + k sin C cos C =k[sin(A+B) cos (A – B) + sin C cos C] =k [sin C cos (A – B) – sin C cos (A + B)] =k [sin C(cos (A – B) – cos(A + B))] = k sin C [2 sin A sin B] = 2 sin C (k sin A) sin B Sine and Cosine Formulae and Their Applications – Exercise 10.2 – Q.12 We know that by cosine rule ⇒2bc cos A = b2 + c2 - a2 ⇒ a2 = b2+c2 - 2bc cosA Sine and Cosine Formulae and Their Applications – Exercise 10.2 – Q.13 LHS, = 2bc – b2 - c2 + a2 + 2ca – a2 - c2 + b2 + 2ab – b2 - a2 + c2 = a2 + b2 + c2+ 2ab + 2bc + 2ca = (a + b + c)2 = RHS Sine and Cosine Formulae and Their Applications – Exercise 10.2 – Q.14 sin3 A cos(B - C) + sin3B.cos(C - A) + sin3C.cos(A - B) = sin2A sin A cos (B - C) + sin2B.sin B cos(C - A) + sin2 C. sin C. cos (A - B) = sin2A sin (π -(B + C))cos(B - C) + sin2B. sin(π - (A + C)).cos(C - A) + sin2 C. sin (π - (A + B)). cos (A - B) = sin2 A sin(B + C) cos(B - C) + sin2B. sin(C + A). cos(C - A) + sin2 C. sin(A + B).cos(A - B) = sin2A. (sin2B + sin2C) + sin2B. (sin2C + sin2A) + sin2C.(sin2A + sin2B) = sin2A.(2sinB.cosB + 2sinCcosC) + 2sin2B.(2sin C cos C + 2 sin A cos A) + sin2C. (2 sin A cos A + 2 sin B cos B) = sin2A. (2sin B cos B + 2 sin C cos C) + sin2B. (2sin C cos C + 2 sin A cos A) + sin2C. (2sin A cos A + 2 sin B cos B) = sin2A. 2 sin B cos B + sin2A. 2 sin C cos C + sin2B. 2 sin C cos S + sin2B. 2sin A cos A + sin2C. 2 sin A cos A + sin2 C. 2sin B cos B = k2a2 2kb cos B + k2a2.2kc cos C + k2b2.2ka cos C + k2b2.2 ka cos A + k2c2.2ka cos A + k2c2.2kb cos B = k3ab(a cos B + b cos A) + k3ac (a cos C + c cos A) + k3bc(c cos B + b cos C) = k3abc + k3acb + k3bca = k33abc = 3(k sin A k sin B k sin C) = 3abc = RHS Sine and Cosine Formulae and Their Applications – Exercise 10.2 – Q.15 b + c = 12λ, c + a = 13λ, a + b = 15λ (b + c + c + a + a + b) = 12λ + 13λ + 15λ 2(a + b + c) = 40λ a + b + c = 20λ b + c = 12λ and a + b + c = 20λ ⟹ a = 8λ c + a = 13λ and a + b + c = 20λ ⟹ b = 7λ a + b = 15λ and a + b + c = 20λ ⟹ c = 5λ
The area of a triangle ABC is given by
Therefore,
We have, a = 4, b = 6 and c = 8
In any ∆ABC, we have
we have,
a = 18, b = 24, c = 30
b(c cos A – a cos C) = c2 - a2
RHS
= c2 - a2
= k2 sin2C - k2 sin2 A
= k2(sin2 C - sin2 A)
= k2sin(C + A). sin(C - A)
= k2 sin(π - B).sin(C - A)
= k2 sin B.sin(C - A)
= k sin B.k sin(C - A)
= bk sin (C-A)
= bk(sin C.cos A - sin A.con C)
= b(k sin C.cos A - k sin A. cos C)
= b(b cos A - a cos C) = LHS
c(a cos B - b cos A)
= ac.cos B - bc cos A
2(bc cos A + ca cos B + ab cos C) = a2 + b2 + c2
LHS
= 2bc cos A + 2ca cos B + 2ab cos C
= b2 + c2 - a2 + a2 + c2 - b2 + a2 + b2 - c2
= a2 + b2 + c2 = RHS
For any ∆ABC, We have
therefore,
Also,
Now,
Hence proved.
a = b cos C +c cos B
b = c cos A + a cos C
c = a cos B + b cos A
L.H.S = a(cos B + cos C - 1) + b(cos C + cos A - 1) + c(cos A + cos B - 1)
= a cos B + a cos C - a + b cos C + b cos A - b + c cos A + c cos B - c
= c - b cos A + a cos C - a + a - c cos B + b cos A - b + b - a cos C + c cos B - c
= 0
= RHS
By sine rule we have
K sin A = a, K sin B = b, k sin C = c
a cos A + b cos B + cos C = k sin A cos A + k sin B cos B + k sin C cos C
=k[sin(A+B) cos (A – B) + sin C cos C]
=k [sin C cos (A – B) – sin C cos (A + B)]
=k [sin C(cos (A – B) – cos(A + B))]
= k sin C [2 sin A sin B]
= 2 sin C (k sin A) sin B
We know that by cosine rule
⇒2bc cos A = b2 + c2 - a2
⇒ a2 = b2+c2 - 2bc cosA
LHS,
= 2bc – b2 - c2 + a2 + 2ca – a2 - c2 + b2 + 2ab – b2 - a2 + c2
= a2 + b2 + c2+ 2ab + 2bc + 2ca
= (a + b + c)2 = RHS
sin3 A cos(B - C) + sin3B.cos(C - A) + sin3C.cos(A - B)
= sin2A sin A cos (B - C) + sin2B.sin B cos(C - A) + sin2 C. sin C. cos (A - B)
= sin2A sin (π -(B + C))cos(B - C) + sin2B. sin(π - (A + C)).cos(C - A)
+ sin2 C. sin (π - (A + B)). cos (A - B)
= sin2 A sin(B + C) cos(B - C) + sin2B. sin(C + A). cos(C - A)
+ sin2 C. sin(A + B).cos(A - B)
= sin2A. (sin2B + sin2C) + sin2B. (sin2C + sin2A) + sin2C.(sin2A + sin2B)
= sin2A.(2sinB.cosB + 2sinCcosC) + 2sin2B.(2sin C cos C + 2 sin A cos A)
+ sin2C. (2 sin A cos A + 2 sin B cos B)
= sin2A. (2sin B cos B + 2 sin C cos C) + sin2B. (2sin C cos C + 2 sin A cos A)
+ sin2C. (2sin A cos A + 2 sin B cos B)
= sin2A. 2 sin B cos B + sin2A. 2 sin C cos C + sin2B. 2 sin C cos S
+ sin2B. 2sin A cos A + sin2C. 2 sin A cos A + sin2 C. 2sin B cos B
= k2a2 2kb cos B + k2a2.2kc cos C + k2b2.2ka cos C
+ k2b2.2 ka cos A + k2c2.2ka cos A + k2c2.2kb cos B
= k3ab(a cos B + b cos A) + k3ac (a cos C + c cos A) + k3bc(c cos B + b cos C)
= k3abc + k3acb + k3bca
= k33abc
= 3(k sin A k sin B k sin C)
= 3abc = RHS
b + c = 12λ, c + a = 13λ, a + b = 15λ
(b + c + c + a + a + b) = 12λ + 13λ + 15λ
2(a + b + c) = 40λ
a + b + c = 20λ
b + c = 12λ and a + b + c = 20λ ⟹ a = 8λ
c + a = 13λ and a + b + c = 20λ ⟹ b = 7λ
a + b = 15λ and a + b + c = 20λ ⟹ c = 5λ
Signing up with Facebook allows you to connect with friends and classmates already using askIItians. It’s an easier way as well. “Relax, we won’t flood your facebook news feed!”
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
Chapter 10: Sine and Cosine Formulae and Their...
