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# RD Chapter 10- Sine and Cosine Formulae and Their Applications Ex-10.2 Interview Questions Answers

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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.

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

= (153)/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.

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.

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

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

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

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

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

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:

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.

krishan