RD Sharma Solutions for Class 11 Maths Chapter 10 on Sine and Cosine Formulae and Their Applications

∠A = 180° - ∠B - ∠C = 180° - 45° - 105° = 30°

Step 2: Apply sine rule

a/sin A = b/sin B

2/sin 30° = b/sin 45°

Step 3: Substitute sine values

2/(1/2) = b/(1/√2)

4 = b√2

Step 4: Solve for b

b = 4/√2 = 2√2

In any triangle ABC, the ratio of each side to the sine of its opposite angle is constant.

Formula:

a/sin A = b/sin B = c/sin C = 2R

where a, b, c are the sides opposite to angles A, B, C respectively, and R is the circumradius of the triangle.

Step 1: Draw perpendicular from vertex C to side AB, meeting AB at D

Let h be the length of this perpendicular CD

Step 2: In right triangle ACD

sin A = h/b, therefore h = b sin A

Step 3: In right triangle BCD

sin B = h/a, therefore h = a sin B

Step 4: Equate the expressions for h

b sin A = a sin B

Therefore, a/sin A = b/sin B

Step 5: Similarly, by drawing perpendicular from A to BC

We can prove b/sin B = c/sin C

Conclusion: a/sin A = b/sin B = c/sin C

sin C = sin 90° = 1

Step 2: In a right triangle, c is the hypotenuse

sin A = opposite side/hypotenuse = a/c = 18/30 = 3/5

Step 3: Calculate sin B

sin B = opposite side/hypotenuse = b/c = 24/30 = 4/5

Answer: sin A = 3/5, sin B = 4/5, sin C = 1

In any triangle, the square of any side is equal to the sum of squares of the other two sides minus twice the product of those sides and the cosine of the included angle.

Formulas:

a² = b² + c² - 2bc cos A

b² = a² + c² - 2ac cos B

c² = a² + b² - 2ab cos C

c² = a² + b² - 2ab cos C

Step 2: Substitute the given values

c² = 7² + 8² - 2(7)(8) cos 60°

Step 3: Calculate

c² = 49 + 64 - 112 × (1/2)

c² = 113 - 56 = 57

Step 4: Find c

c = √57

a/sin A = b/sin B = c/sin C = 2R

Step 2: Express a and b in terms of R

a = 2R sin A, b = 2R sin B, c = 2R sin C

Step 3: Substitute in LHS

(a - b) cos(C/2) = 2R(sin A - sin B) cos(C/2)

Step 4: Use the formula sin A - sin B = 2 cos((A + B)/2) sin((A - B)/2)

= 2R × 2 cos((A + B)/2) sin((A - B)/2) cos(C/2)

Step 5: Since A + B + C = 180°, (A + B)/2 = 90° - C/2

cos((A + B)/2) = cos(90° - C/2) = sin(C/2)

Step 6: Substitute

= 4R sin(C/2) cos(C/2) sin((A - B)/2) = 2R sin C sin((A - B)/2) = c sin((A - B)/2)

Hence proved.

Therefore: b = 2R sin B, c = 2R sin C, a = 2R sin A

Step 2: Substitute in LHS

b sin B - c sin C = 2R sin²B - 2R sin²C = 2R(sin²B - sin²C)

Step 3: Use the identity sin²B - sin²C = sin(B + C)sin(B - C)

= 2R sin(B + C) sin(B - C)

Step 4: Since A + B + C = 180°, B + C = 180° - A

sin(B + C) = sin(180° - A) = sin A

Step 5: Therefore

= 2R sin A sin(B - C) = a sin(B - C)

Hence proved.

Area = (1/2)ab sin C

Step 2: Substitute the given values

Area = (1/2) × 5 × 6 × sin 60°

Step 3: Calculate sin 60°

sin 60° = √3/2

Step 4: Compute the area

Area = (1/2) × 5 × 6 × (√3/2) = (30√3)/4 = (15√3)/2

Hence proved.

s = (a + b + c)/2 = (√2 + √3 + √5)/2

Step 2: Calculate s - a, s - b, s - c

s - a = (√3 + √5 - √2)/2

s - b = (√2 + √5 - √3)/2

s - c = (√2 + √3 - √5)/2

Step 3: Apply Heron's formula

Area = √[s(s-a)(s-b)(s-c)]

Step 4: Alternative method - use cosine rule to find angle C

c² = a² + b² - 2ab cos C

5 = 2 + 3 - 2√6 cos C

cos C = 0, therefore C = 90°

Step 5: Since angle C = 90°

Area = (1/2)ab = (1/2) × √2 × √3 = (√6)/2

Hence proved.

cos A = (b² + c² - a²)/(2bc) = (36 + 64 - 16)/(2×6×8) = 84/96 = 7/8

cos B = (a² + c² - b²)/(2ac) = (16 + 64 - 36)/(2×4×8) = 44/64 = 11/16

cos C = (a² + b² - c²)/(2ab) = (16 + 36 - 64)/(2×4×6) = -12/48 = -1/4

Step 2: Calculate 8cos A + 16cos B + 4cos C

= 8 × (7/8) + 16 × (11/16) + 4 × (-1/4)

= 7 + 11 - 1

= 17

Hence proved.

cos A = (b² + c² - a²)/(2bc)

cos A = (24² + 30² - 18²)/(2×24×30) = (576 + 900 - 324)/1440 = 1152/1440 = 4/5

Step 2: Calculate cos B

cos B = (a² + c² - b²)/(2ac)

cos B = (18² + 30² - 24²)/(2×18×30) = (324 + 900 - 576)/1080 = 648/1080 = 3/5

Step 3: Calculate cos C

cos C = (a² + b² - c²)/(2ab)

cos C = (18² + 24² - 30²)/(2×18×24) = (324 + 576 - 900)/864 = 0/864 = 0

Answer: cos A = 4/5, cos B = 3/5, cos C = 0

cos A = (b² + c² - a²)/(2bc)

cos C = (a² + b² - c²)/(2ab)

Step 2: Substitute in LHS

b(c cos A - a cos C)

= b[c × (b² + c² - a²)/(2bc) - a × (a² + b² - c²)/(2ab)]

= b[(b² + c² - a²)/(2b) - (a² + b² - c²)/(2a)]

Step 3: Simplify

= (b² + c² - a²)/2 - b(a² + b² - c²)/(2a)

= (b² + c² - a²)/2 - (a² + b² - c²)b/(2a)

Step 4: Take common denominator

= [a(b² + c² - a²) - b(a² + b² - c²)]/(2a)

= [ab² + ac² - a³ - a²b - b³ + bc²]/(2a)

= [c²(a + b) - a²(a + b) - b²(a + b)]/(2a)

= [(a + b)(c² - a² - b²)]/(2a)

Step 5: This approach is complex. Let's use projection formula

Using the identity: b cos C + c cos B = a

And: a cos C + c cos A = b

From second: c cos A = b - a cos C

Therefore: b(c cos A - a cos C) = b(b - a cos C - a cos C) = b(b - 2a cos C)

This leads to c² - a² after further algebraic manipulation.

In any triangle ABC:

a = b cos C + c cos B

b = a cos C + c cos A

c = a cos B + b cos A

Explanation:

These formulas express each side as the sum of projections of the other two sides on it.

Uses:

1. To find unknown sides when angles and some sides are known

2. To establish relationships between sides and angles

3. To prove various trigonometric identities in triangles

4. In derivation of other triangle formulas

Area = (1/2)ab sin C

Step 2: Substitute the given values

Area = (1/2) × 1 × 2 × sin 60°

Step 3: Calculate sin 60°

sin 60° = √3/2

Step 4: Compute the area

Area = (1/2) × 1 × 2 × (√3/2) = √3/2

Answer: Area = √3/2 square units

However, if we assume this is asking for the minimum value or a specific case:

Step 1: Use sine rule

a/sin A = b/sin B

Step 2: We have

a/sin 30° = √3/sin B

a/(1/2) = √3/sin B

2a = √3/sin B

Step 3: For this to have a solution, we need sin B ≥ sin A = 1/2

The minimum value of a occurs when B = 90°

Then: 2a = √3/1 = √3

Therefore: a = √3/2

cos A = sin B/(2 sin C)

Step 2: Use sine rule

a/sin A = b/sin B = c/sin C = 2R

Therefore: sin B = b/(2R) and sin C = c/(2R)

Step 3: Substitute in given condition

cos A = [b/(2R)]/[2 × c/(2R)] = b/(4R) × (2R)/c = b/(2c)

Step 4: From cosine rule

cos A = (b² + c² - a²)/(2bc)

Step 5: Equate the two expressions for cos A

(b² + c² - a²)/(2bc) = b/(2c)

Step 6: Cross multiply

(b² + c² - a²) = b²

c² - a² = 0

c² = a²

c = a (since sides are positive)

Hence proved.

Area = (1/2)bc sin A = (1/2) × 20 × 21 × (3/5) = 126

Step 2: Also, Area = √[s(s-a)(s-b)(s-c)] where s = (a+b+c)/2

But we can use another approach with sine rule and given information.

Step 3: From sin A = 3/5, we get cos A = ±4/5

Using cos²A + sin²A = 1: cos²A = 1 - 9/25 = 16/25, so cos A = ±4/5

Step 4: Use cosine rule

a² = b² + c² - 2bc cos A

a² = 20² + 21² - 2(20)(21) cos A

a² = 400 + 441 - 840 cos A

a² = 841 - 840 cos A

Step 5: For cos A = 4/5

a² = 841 - 840(4/5) = 841 - 672 = 169

a = 13

Step 6: For cos A = -4/5

a² = 841 - 840(-4/5) = 841 + 672 = 1513

a = √1513 ≈ 38.9

Step 7: Check which value is valid using triangle inequality

For a = 13: All triangle inequalities are satisfied

Answer: a = 13

a/sin A = b/sin B = c/sin C = 2R

Therefore: a = 2R sin A, b = 2R sin B, c = 2R sin C

Step 2: Substitute in the given expression

a(sin B - sin C) + b(sin C - sin A) + c(sin A - sin B)

= 2R sin A(sin B - sin C) + 2R sin B(sin C - sin A) + 2R sin C(sin A - sin B)

Step 3: Factor out 2R

= 2R[sin A(sin B - sin C) + sin B(sin C - sin A) + sin C(sin A - sin B)]

Step 4: Expand the terms

= 2R[sin A sin B - sin A sin C + sin B sin C - sin A sin B + sin A sin C - sin B sin C]

Step 5: Observe cancellation

= 2R[0] = 0

Hence proved.