RD Chapter 10 Sine and Cosine Formulae and Their Applications Ex 10.2 Solutions
Question - 1 : - In any ∆ABC, prove the following:In a ∆ABC,if a = 5, b = 6 and C = 60o, show that its area is (15√3)/2 sq.units.
Answer - 1 : -
Given:
In a ∆ABC, a = 5, b = 6 and C = 60o
By using the formula,
Area of ∆ABC = 1/2 ab sin θ where, a and b are thelengths of the sides of a triangle and θ is the angle between sides.
So,
Area of ∆ABC = 1/2 ab sin θ
= 1/2 × 5 × 6 × sin 60o
= 30/2 × √3/2
= (15√3)/2 sq. units
Hence proved.
Question - 2 : - In a ∆ABC, if a = √2, b = √3 and c = √5 show that its area is1/2 √6 sq. units.
Answer - 2 : -
Given:
In a ∆ABC, a = √2, b = √3 and c = √5
By using the formulas,
We know, cos A = (b2 + c2 –a2)/2bc
By substituting the values we get,
= [(√3)2 + (√5)2 –(√2)2] / [2 × √3 × √5]
= 3/√15
We know, Area of ∆ABC = 1/2 bc sin A
To find sin A:
Sin A = √(1 – cos2 A) [by usingtrigonometric identity]
= √(1 – (3/√15)2)
= √(1- (9/15))
= √(6/15)
Now,
Area of ∆ABC = 1/2 bc sin A
= 1/2 × √3 × √5 × √(6/15)
= 1/2 √6 sq. units
Hence proved.
Question - 3 : - The sides of a triangle are a = 4, b = 6 and c = 8, show that: 8 cos A + 16 cos B + 4 cos C = 17.
Answer - 3 : -
Given:
Sides of a triangle are a = 4, b = 6 and c = 8
By using the formulas,
Cos A = (b2 + c2 – a2)/2bc
Cos B = (a2 + c2 – b2)/2ac
Cos C = (a2 + b2 – c2)/2ab
So now let us substitute the values of a, b and c weget,
Cos A = (b2 + c2 – a2)/2bc
= (62 + 82 – 42)/2×6×8
= (36 + 64 – 16)/96
= 84/96
= 7/8
Cos B = (a2 + c2 – b2)/2ac
= (42 + 82 – 62)/2×4×8
= (16 + 64 – 36)/64
= 44/64
Cos C = (a2 + b2 – c2)/2ab
= (42 + 62 – 82)/2×4×6
= (16 + 36 – 64)/48
= -12/48
= -1/4
Now considering LHS:
8 cos A + 16 cos B + 4 cos C = 8 × 7/8 + 16 × 44/64 +4 × (-1/4)
= 7 + 11 – 1
= 17
Hence proved.
Question - 4 : - In a ∆ABC, if a = 18, b = 24, c = 30, find cos A, cos B and cos C
Answer - 4 : -
Given:
Sides of a triangle are a = 18, b = 24 and c = 30
By using the formulas,
Cos A = (b2 + c2 – a2)/2bc
Cos B = (a2 + c2 – b2)/2ac
Cos C = (a2 + b2 – c2)/2ab
So now let us substitute the values of a, b and c weget,
Cos A = (b2 + c2 – a2)/2bc
= (242 + 302 – 182)/2×24×30
= 1152/1440
= 4/5
Cos B = (a2 + c2 – b2)/2ac
= (182 + 302 – 242)/2×18×30
= 648/1080
= 3/5
Cos C = (a2 + b2 – c2)/2ab
= (182 + 242 – 302)/2×18×24
= 0/864
= 0
∴ cos A = 4/5, cos B =3/5, cos C = 0
Question - 5 : - For anyΔABC, show that b (c cos A – a cos C) = c2 – a2
Answer - 5 : -
Let us consider LHS:
b (c cos A – a cos C)
As LHS contain bc cos A and ab cos C which can beobtained from cosine formulae.
From cosine formula we have:
Cos A = (b2 + c2 – a2)/2bc
bc cos A = (b2 + c2 –a2)/2 … (i)
Cos C = (a2 + b2 – c2)/2ab
ab cos C = (a2 + b2 –c2)/2 … (ii)
Now let us subtract equation (i) and (ii) we get,
bc cos A – ab cos C = (b2 + c2 –a2)/2 – (a2 + b2 – c2)/2
= c2 – a2
∴ b (c cos A – a cos C) =c2 – a2
Hence proved.
Question - 6 : - For any ΔABC show that c (a cos B – b cos A) = a2 – b2
Answer - 6 : -
Let us consider LHS:
c (a cos B – b cos A)
As LHS contain ca cos B and cb cos A which can beobtained from cosine formulae.
From cosine formula we have:
Cos A = (b2 + c2 – a2)/2bc
bc cos A = (b2 + c2 –a2)/2 … (i)
Cos B = (a2 + c2 – b2)/2ac
ac cos B = (a2 + c2 –b2)/2 … (ii)
Now let us subtract equation (ii) from (i) we get,
ac cos B – bc cos A = (a2 + c2 –b2)/2 – (b2 + c2 – a2)/2
= a2 – b2
∴ c (a cos B – b cos A) =a2 – b2
Hence proved.
Question - 7 : - For any ΔABC show that
2 (bc cos A+ ca cos B + ab cos C) = a2 + b2 + c2
Answer - 7 : -
Let us consider LHS:
2 (bc cos A + ca cos B + ab cos C)
As LHS contain 2ca cos B, 2ab cos C and 2cb cos A,which can be obtained from cosine formulae.
From cosine formula we have:
Cos A = (b2 + c2 – a2)/2bc
2bc cos A = (b2 + c2 –a2) … (i)
Cos B = (a2 + c2 – b2)/2ac
2ac cos B = (a2 + c2 –b2)… (ii)
Cos C = (a2 + b2 – c2)/2ab
2ab cos C = (a2 + b2 –c2) … (iii)
Now let us add equation (i), (ii) and (ii) we get,
2bc cos A + 2ac cos B + 2ab cos C = (b2 +c2 – a2) + (a2 + c2 –b2) + (a2 + b2 – c2)
Upon simplification we get,
= c2 + b2 + a2
2 (bc cos A + ac cos B + ab cos C) = a2 +b2 + c2
Hence proved.
Question - 8 : - For any ΔABC show that
(c2 –a2 + b2) tan A = (a2 – b2 +c2) tan B = (b2 – c2 + a2)tan C
Answer - 8 : -
Let us consider LHS:
(c2 – a2 + b2),(a2 – b2 + c2), (b2 –c2 + a2)
We know sine rule in Δ ABC
As LHS contain (c2 – a2 +b2), (a2 – b2 + c2)and (b2 – c2 + a2), which canbe obtained from cosine formulae.
From cosine formula we have:
Cos A = (b2 + c2 – a2)/2bc
2bc cos A = (b2 + c2 –a2)
Let us multiply both the sides by tan A we get,
2bc cos A tan A = (b2 + c2 –a2) tan A
2bc cos A (sin A/cos A) = (b2 + c2 –a2) tan A
2bc sin A = (b2 + c2 –a2) tan A … (i)
Cos B = (a2 + c2 – b2)/2ac
2ac cos B = (a2 + c2 –b2)
Let us multiply both the sides by tan B we get,
2ac cos B tan B = (a2 + c2 –b2) tan B
2ac cos B (sin B/cos B) = (a2 + c2 –b2) tan B
2ac sin B = (a2 + c2 –b2) tan B … (ii)
Cos C = (a2 + b2 – c2)/2ab
2ab cos C = (a2 + b2 –c2)
Let us multiply both the sides by tan C we get,
2ab cos C tan C = (a2 + b2 –c2) tan C
2ab cos C (sin C/cos C) = (a2 + b2 –c2) tan C
2ab sin C = (a2 + b2 –c2) tan C … (iii)
As we are observing that sin terms are being involvedso let’s use sine formula.
From sine formula we have,
Let us multiply abc to each of the expression we get,
bc sin A = ac sin B = ab sin C
2bc sin A = 2ac sin B = 2ab sin C
∴ From equation(i), (ii) and (iii) we have,
(c2 – a2 + b2)tan A = (a2 – b2 + c2) tan B = (b2 –c2 + a2) tan C
Hence proved.
Question - 9 : - For any ΔABC show that:
Answer - 9 : -
Let us consider LHS:
We can observe that we can get terms c – b cos A and b– c cos A from projection formula
From projection formula we get,
c = a cos B + b cos A
c – b cos A = a cos B …. (i)
And,
b = c cos A + a cos C
b – c cos A = a cos C …. (ii)
Dividing equation (i) by (ii), we get,
= RHS
Hence proved.