Trigonometry

⇒ sin θ =

cos θ =

tan θ =

cosec θ =

sec θ =

cot θ =

⇒ sin θ =

cos θ =

tan θ =

cosec θ =

sec θ =

cot θ =

sin θ =

cos θ =

tan θ =

cosec θ =

sec θ =

cot θ =

sin θ =

cos θ =

tan θ =

cosec θ =

sec θ =

cot θ =

sin θ =

cos θ =

tan θ =

cosec θ =

sec θ =

cot θ =

sin θ =

cos θ =

tan θ =

cosec θ =

sec θ =

cot θ =

sin θ =

cos θ =

tan θ =

cosec θ =

sec θ =

cot θ =

sin θ =

cos θ =

tan θ =

cosec θ =

sec θ =

cot θ =

Let the third side be p,

By Pythagoras theorem,

⇒ 9² + p² = 15²

⇒ 81 + p² = 225

⇒ p² = 144

⇒ p = 12

Therefore, the other angles are:

cos A =

tan A =

cosec A =

sec A =

cot A =

Let the third side be p,

By Pythagoras theorem,

⇒ 9² + p² = 15²

⇒ 81 + p² = 225

⇒ p² = 144

⇒ p = 12

Therefore, the other angles are:

cos A =

tan A =

cosec A =

sec A =

cot A =

Let the third side be p,

By Pythagoras theorem,

⇒ 15² + p² = 17²

⇒ 225 + p² = 289

⇒ p² = 64

⇒ p = 8

Therefore, the other angles are:

sin P =

tan P =

cosec P =

sec P =

cot P =

Let the third side be p,

By Pythagoras theorem,

⇒ 15² + p² = 17²

⇒ 225 + p² = 289

⇒ p² = 64

⇒ p = 8

Therefore, the other angles are:

sin P =

tan P =

cosec P =

sec P =

cot P =

Let the third side be p,

By Pythagoras theorem,

⇒ 5² + 12² = P²

⇒ 25 + 144 = p²

⇒ p² = 169

⇒ p = 13

Therefore, the other angles are:

sin P =

cos P =

cosec P =

sec P =

cot P =

Let the third side be p,

By Pythagoras theorem,

⇒ 5² + 12² = P²

⇒ 25 + 144 = p²

⇒ p² = 169

⇒ p = 13

Therefore, the other angles are:

sin P =

cos P =

cosec P =

sec P =

cot P =

Let the third side be p,

By Pythagoras theorem,

⇒ 15² + p² = 17²

⇒ 64 + p² = 289

⇒ p² = 225

⇒ p = 15

Therefore, the other angles are:

sin θ =

cos θ =

tan θ =

cosec θ =

cot θ =

Let the third side be p,

By Pythagoras theorem,

⇒ 15² + p² = 17²

⇒ 64 + p² = 289

⇒ p² = 225

⇒ p = 15

Therefore, the other angles are:

sin θ =

cos θ =

tan θ =

cosec θ =

cot θ =

Let the third side be p,

By Pythagoras theorem,

⇒ 60² + p² = 61²

⇒ 3600 + p² = 3721

⇒ p² = 121

⇒ p = 11

Therefore, the other angles are:

sin θ =

cos θ =

tan θ =

sec θ =

cot θ =

Let the third side be p,

By Pythagoras theorem,

⇒ 60² + p² = 61²

⇒ 3600 + p² = 3721

⇒ p² = 121

⇒ p = 11

Therefore, the other angles are:

sin θ =

cos θ =

tan θ =

sec θ =

cot θ =

Let the third side be p,

By Pythagoras theorem,

⇒ x² + p² = y²

⇒ p² = y² – x²

⇒ p = √y² – x²

Therefore, the other angles are:

cos θ =

tan θ =

cosec θ =

sec θ =

cot θ =

Let the third side be p,

By Pythagoras theorem,

⇒ x² + p² = y²

⇒ p² = y² – x²

⇒ p = √y² – x²

Therefore, the other angles are:

cos θ =

tan θ =

cosec θ =

sec θ =

cot θ =

(i)

θ = 45° (From the table)

(ii)

θ = 0° (From the table)

(iii)

θ = 60° (From the table)

(iv)

θ = 30° (From the table)

(i)

θ = 45° (From the table)

(ii)

θ = 0° (From the table)

(iii)

θ = 60° (From the table)

(iv)

θ = 30° (From the table)

For ∠A:

sin A =

cos A =

tan A =

cosec A =

sec A =

cot A =

For ∠C:

sin C =

cos C =

tan C =

cosec C =

sec C =

cot C =

For ∠A:

sin A =

cos A =

tan A =

cosec A =

sec A =

cot A =

For ∠C:

sin C =

cos C =

tan C =

cosec C =

sec C =

cot C =

5cosθ –12sinθ = 0

⇒ 5cosθ = 12sinθ

⇒ tanθ = 5/12

Let the third side be p,

By Pythagoras theorem,

⇒ 5² + 12² = P²

⇒ 25 + 144 = p²

⇒ p² = 169

⇒ p = 13

Therefore, the other angles are:

sin θ =

cos θ =

Putting values in the function as:

⇒

⇒

⇒

5cosθ –12sinθ = 0

⇒ 5cosθ = 12sinθ

⇒ tanθ = 5/12

Let the third side be p,

By Pythagoras theorem,

⇒ 5² + 12² = P²

⇒ 25 + 144 = p²

⇒ p² = 169

⇒ p = 13

Therefore, the other angles are:

sin θ =

cos θ =

Putting values in the function as:

⇒

⇒

⇒

29cosθ = 20

⇒ cosθ = 20/29

Let the third side be p,

By Pythagoras theorem,

⇒ 20² + p² = 29²

⇒ 400 + p² = 841

⇒ p² = 441

⇒ p = 21

Therefore, the other angles are:

sec θ =

tan θ =

Putting the values in function:

⇒ sec² θ – tan² θ

⇒

⇒

⇒

29cosθ = 20

⇒ cosθ = 20/29

Let the third side be p,

By Pythagoras theorem,

⇒ 20² + p² = 29²

⇒ 400 + p² = 841

⇒ p² = 441

⇒ p = 21

Therefore, the other angles are:

sec θ =

tan θ =

Putting the values in function:

⇒ sec² θ – tan² θ

⇒

⇒

⇒

∠BAC = θ

Let the third side be p,

By Pythagoras theorem,

⇒ 10² + p² = 26²

⇒ 100 + p² = 676

⇒ p² = 576

⇒ p = 24

Therefore, the other angles are:

sin θ =

cos θ =

Putting values in the function as:

⇒

⇒

⇒

∠BAC = θ

Let the third side be p,

By Pythagoras theorem,

⇒ 10² + p² = 26²

⇒ 100 + p² = 676

⇒ p² = 576

⇒ p = 24

Therefore, the other angles are:

sin θ =

cos θ =

Putting values in the function as:

⇒

⇒

⇒

Let the third side be p,

By Pythagoras theorem,

⇒ a² + b² = p²

⇒ p = √a² + b²

Therefore, the other angles are:

sin θ =

cos θ =

Putting the values in function:

⇒ sin² θ + cos² θ

⇒

⇒

⇒

Let the third side be p,

By Pythagoras theorem,

⇒ a² + b² = p²

⇒ p = √a² + b²

Therefore, the other angles are:

sin θ =

cos θ =

Putting the values in function:

⇒ sin² θ + cos² θ

⇒

⇒

⇒

Simplifying the function:

⇒

⇒

Let the third side be p,

By Pythagoras theorem,

⇒ 8² + 15² = P²

⇒ 64 + 225 = p²

⇒ p² = 289

⇒ p = 17

Therefore, the other angles are:

sin θ =

cos θ =

Putting these values in the simplied function:
⇒

⇒

⇒

⇒

Simplifying the function:

⇒

⇒

Let the third side be p,

By Pythagoras theorem,

⇒ 8² + 15² = P²

⇒ 64 + 225 = p²

⇒ p² = 289

⇒ p = 17

Therefore, the other angles are:

sin θ =

cos θ =

Putting these values in the simplied function:
⇒

⇒

⇒

⇒

Let the third side be p,

By Pythagoras theorem,

⇒ 1² + (√3)² = P²

⇒ 1 + 3 = p²

⇒ p² = 4

⇒ p = 2

Therefore, the other angles are:

sin P =

cos P =

sin R =

cos R =

(i) Putting the values in the expression:

sin P cos R + cos P sin R

⇒

⇒

⇒ 1

(ii) Putting the values in the expression:

cos P cos R – sin P sin R.

⇒

⇒ 0

Let the third side be p,

By Pythagoras theorem,

⇒ 1² + (√3)² = P²

⇒ 1 + 3 = p²

⇒ p² = 4

⇒ p = 2

Therefore, the other angles are:

sin P =

cos P =

sin R =

cos R =

(i) Putting the values in the expression:

sin P cos R + cos P sin R

⇒

⇒

⇒ 1

(ii) Putting the values in the expression:

cos P cos R – sin P sin R.

⇒

⇒ 0

Let the third side be p,

By Pythagoras theorem,

⇒ 5² + p² = 13²

⇒ 25 + p² = 169

⇒ p² = 144

⇒ p = 12

Therefore, the other angles are:

sin θ =

cos θ =

Putting these values in the left hand side:
⇒

⇒

⇒

⇒ 3

Which is equal to right hand side.

Hence proved.

Let the third side be p,

By Pythagoras theorem,

⇒ 5² + p² = 13²

⇒ 25 + p² = 169

⇒ p² = 144

⇒ p = 12

Therefore, the other angles are:

sin θ =

cos θ =

Putting these values in the left hand side:
⇒

⇒

⇒

⇒ 3

Which is equal to right hand side.

Hence proved.

Let the third side be p,

By Pythagoras theorem,

⇒ 8² + p² = 17²

⇒ 64 + p² = 289

⇒ p² = 225

⇒ p = 15

Therefore, the other angles are:

sin θ =

cos θ =

Putting these values in the left hand side:
⇒ 1–2×

⇒ 1–2×

⇒ 1–

⇒

Putting values in the right hand side:

⇒ 2×

⇒ 2× –1

⇒

⇒

Therefore, Left hand side and right hand side are equal.

Hence proved.

Let the third side be p,

By Pythagoras theorem,

⇒ 8² + p² = 17²

⇒ 64 + p² = 289

⇒ p² = 225

⇒ p = 15

Therefore, the other angles are:

sin θ =

cos θ =

Putting these values in the left hand side:
⇒ 1–2×

⇒ 1–2×

⇒ 1–

⇒

Putting values in the right hand side:

⇒ 2×

⇒ 2× –1

⇒

⇒

Therefore, Left hand side and right hand side are equal.

Hence proved.

(i) We know sin 45° = (1/√2) and cos 45° = (1/√2) (From the table)

⇒

⇒

(ii) We know sin 30° = 1/2 and tan 45° = 1 (From the table)

⇒

(iii) We know tan 30° = (1/√3) (From the table)

tan 45° = 1

tan 60° = √3

⇒

⇒

⇒

(iv) We know sin 30° = cos 60° = 1/2 (From the table)

tan 30° = cot 60° = 1/√3

⇒

⇒

⇒

⇒

(v) We know that cos 90° = 0 , sin 90° = 1 and tan 45° = 1 (From the table)

⇒ 6(0)² + 3(1)² + 4(1)²

⇒ 3 + 4

⇒ 7

(vi) We know cot 60° = (1/√3) (From the table)

sec 30° = 2/√3

sin 45° = 1/√2

sin 60° = √3/2

cos 45° = 1/√2

⇒

⇒

⇒

⇒

(vii) We know tan 60° = √3 (From the table)

cos 45° = 1/√2

sec 30° = 2/√3

cos 90° = 0

cosec 30° = 2

sec 60° = 2

cot 30° = √3

⇒

⇒

⇒ 9

(viii) We know

sin 30° = cos 60° = 1/2 (From the table)

cos 45° = 1/√2

sin 90° = 1

⇒ 4(

⇒ 4(

⇒ 4(

⇒ 2

(i) We know sin 45° = (1/√2) and cos 45° = (1/√2) (From the table)

⇒

⇒

(ii) We know sin 30° = 1/2 and tan 45° = 1 (From the table)

⇒

(iii) We know tan 30° = (1/√3) (From the table)

tan 45° = 1

tan 60° = √3

⇒

⇒

⇒

(iv) We know sin 30° = cos 60° = 1/2 (From the table)

tan 30° = cot 60° = 1/√3

⇒

⇒

⇒

⇒

(v) We know that cos 90° = 0 , sin 90° = 1 and tan 45° = 1 (From the table)

⇒ 6(0)² + 3(1)² + 4(1)²

⇒ 3 + 4

⇒ 7

(vi) We know cot 60° = (1/√3) (From the table)

sec 30° = 2/√3

sin 45° = 1/√2

sin 60° = √3/2

cos 45° = 1/√2

⇒

⇒

⇒

⇒

(vii) We know tan 60° = √3 (From the table)

cos 45° = 1/√2

sec 30° = 2/√3

cos 90° = 0

cosec 30° = 2

sec 60° = 2

cot 30° = √3

⇒

⇒

⇒ 9

(viii) We know

sin 30° = cos 60° = 1/2 (From the table)

cos 45° = 1/√2

sin 90° = 1

⇒ 4(

⇒ 4(

⇒ 4(

⇒ 2

(i) We know sin 30° = (1/2) and cos 30° = (√3/2) (From the table)

Putting the values in the left hand side:

⇒

⇒

⇒ 1

Which is equal to the right hand side.

Hence verified.

(ii) We know tan 45° = 1 and sec 45° = √2 (From the table)

Putting the values in the left hand side:

⇒ 1 + (1)²

⇒ 1 + 1

⇒ 2

Putting the values in the right hand side:

⇒ (√2)²

⇒ 2

∴ LHS = RHS

Hence verified.

(iii) We know sin 30° = cos 60° = (1/2) and cos 30° = (√3/2) (From the table)

So the leftmost function = cos60° = (1/2)

Putting values in the middle function:
⇒ 1 – 2×

⇒

⇒ 1/2

Putting values in the rightmost function:
⇒ 2×

⇒

⇒ 1/2

Therefore, all simplify to 1/2 and are equal.

Hence verified.

(iv) We know sin 45° = cos 45° = (1/√ 2) and cos 90° = 0 (From the table)

So the leftmost function = cos90° = 0

Putting values in the middle function:
⇒ 1 – 2×

⇒

⇒ 0

Putting values in the rightmost function:
⇒ 2×

⇒

⇒ 0

Therefore, all simplify to 0 and are equal.

Hence verified.

(v) We know sin 60° = (√3/2) (From the table)

cos 60° = (1/2)

tan 60° = √3

and sec 30° = 2

Putting values in the left hand side:
⇒

⇒

⇒

Putting values in the right hand side:
⇒

Therefore, left hand side and right hand side are equal.

Hence verified.

(vi) We know cos 60° = (1/2) (From the table)

tan 60° = √3

Putting values in the left hand side:
⇒

⇒

⇒

Putting values in the right hand side:
⇒

⇒

⇒

⇒

Therefore, left hand side and right hand side are equal.

Hence verified.

(vii) We know sec 60° = (2/√3) (From the table)

tan 30° = (1/√3)

sin 30° = (1/2)

Putting values in the left hand side:
⇒

⇒

⇒ 3

Putting values in the right hand side:
⇒

⇒

⇒ 3

Therefore, left hand side and right hand side are equal.

Hence verified.

(viii) We know sin 30° = (1/2) (From the table)

cot 30° = √3

cosec 45° = √2

tan 60° = √3

and tan 45° = 1

Putting the values in the left hand side:

⇒

⇒

⇒ 0

Which is equal to the right hand side.

Hence verified.

(ix) We know cos 60° = (1/2) (From the table)

sin 60° = (√3/2)

sec 60° = 2

and cot 45° = 1

Putting the values in the left hand side:

⇒

⇒

⇒ 1

Which is equal to the right hand side.

Hence verified.

(x) We know sin 30° = cos 60° = (1/2) (From the table)

Cos 30° = sin 60° = (√3/2)

Sin 90° = 1

So the right hand side = sin 90° = 1

Putting the values in the left hand side:

⇒

⇒

⇒ 1

Which is equal to the right hand side.

Hence verified.

(i) We know sin 30° = (1/2) and cos 30° = (√3/2) (From the table)

Putting the values in the left hand side:

⇒

⇒

⇒ 1

Which is equal to the right hand side.

Hence verified.

(ii) We know tan 45° = 1 and sec 45° = √2 (From the table)

Putting the values in the left hand side:

⇒ 1 + (1)²

⇒ 1 + 1

⇒ 2

Putting the values in the right hand side:

⇒ (√2)²

⇒ 2

∴ LHS = RHS

Hence verified.

(iii) We know sin 30° = cos 60° = (1/2) and cos 30° = (√3/2) (From the table)

So the leftmost function = cos60° = (1/2)

Putting values in the middle function:
⇒ 1 – 2×

⇒

⇒ 1/2

Putting values in the rightmost function:
⇒ 2×

⇒

⇒ 1/2

Therefore, all simplify to 1/2 and are equal.

Hence verified.

(iv) We know sin 45° = cos 45° = (1/√ 2) and cos 90° = 0 (From the table)

So the leftmost function = cos90° = 0

Putting values in the middle function:
⇒ 1 – 2×

⇒

⇒ 0

Putting values in the rightmost function:
⇒ 2×

⇒

⇒ 0

Therefore, all simplify to 0 and are equal.

Hence verified.

(v) We know sin 60° = (√3/2) (From the table)

cos 60° = (1/2)

tan 60° = √3

and sec 30° = 2

Putting values in the left hand side:
⇒

⇒

⇒

Putting values in the right hand side:
⇒

Therefore, left hand side and right hand side are equal.

Hence verified.

(vi) We know cos 60° = (1/2) (From the table)

tan 60° = √3

Putting values in the left hand side:
⇒

⇒

⇒

Putting values in the right hand side:
⇒

⇒

⇒

⇒

Therefore, left hand side and right hand side are equal.

Hence verified.

(vii) We know sec 60° = (2/√3) (From the table)

tan 30° = (1/√3)

sin 30° = (1/2)

Putting values in the left hand side:
⇒

⇒

⇒ 3

Putting values in the right hand side:
⇒

⇒

⇒ 3

Therefore, left hand side and right hand side are equal.

Hence verified.

(viii) We know sin 30° = (1/2) (From the table)

cot 30° = √3

cosec 45° = √2

tan 60° = √3

and tan 45° = 1

Putting the values in the left hand side:

⇒

⇒

⇒ 0

Which is equal to the right hand side.

Hence verified.

(ix) We know cos 60° = (1/2) (From the table)

sin 60° = (√3/2)

sec 60° = 2

and cot 45° = 1

Putting the values in the left hand side:

⇒

⇒

⇒ 1

Which is equal to the right hand side.

Hence verified.

(x) We know sin 30° = cos 60° = (1/2) (From the table)

Cos 30° = sin 60° = (√3/2)

Sin 90° = 1

So the right hand side = sin 90° = 1

Putting the values in the left hand side:

⇒

⇒

⇒ 1

Which is equal to the right hand side.

Hence verified.

⇒

⇒

⇒

⇒ 1

⇒

⇒

⇒

⇒ 1

⇒

⇒

⇒

⇒ 1

⇒

⇒

⇒

⇒ 1

⇒ sin θ sec (90–θ)

⇒ sin θ cosec θ

⇒ sin θ × (1/sin θ )

⇒ 1

⇒ sin θ sec (90–θ)

⇒ sin θ cosec θ

⇒ sin θ × (1/sin θ )

⇒ 1

⇒

⇒

⇒

⇒ 1

⇒

⇒

⇒

⇒ 1

⇒

⇒

⇒

⇒ 1

⇒

⇒

⇒

⇒ 1

⇒

⇒

⇒

⇒ 1

⇒

⇒

⇒

⇒ 1

⇒ cos 38° cos 52° – sin 38° sin 52°

⇒ cos 38° sin (90–52)° – sin 38° cos (90–52)°

⇒ cos 38° sin 38° – sin 38° cos 38°

⇒ 0

⇒ cos 38° cos 52° – sin 38° sin 52°

⇒ cos 38° sin (90–52)° – sin 38° cos (90–52)°

⇒ cos 38° sin 38° – sin 38° cos 38°

⇒ 0

⇒

⇒

⇒ 1 + 1 = 2

⇒

⇒

⇒ 1 + 1 = 2

⇒

⇒

⇒ 1 – 1 = 0

⇒

⇒

⇒ 1 – 1 = 0

⇒

⇒

⇒ (3×1) + (1/2)×1 + (5/2)×1

⇒ 3 + 3 = 6

⇒

⇒

⇒ (3×1) + (1/2)×1 + (5/2)×1

⇒ 3 + 3 = 6

⇒

⇒

⇒ 1 × 1 = 1

⇒

⇒

⇒ 1 × 1 = 1

⇒

⇒ 2 × 1 + 7 × 1

⇒ 2 + 7 = 9

⇒

⇒ 2 × 1 + 7 × 1

⇒ 2 + 7 = 9

⇒

⇒

⇒

⇒

⇒ sec θ – sec θ

⇒ 0

⇒

⇒

⇒

⇒

⇒ sec θ – sec θ

⇒ 0

⇒

⇒

⇒ 1 + 1 –(1/2)

⇒ 3/2

⇒

⇒

⇒ 1 + 1 –(1/2)

⇒ 3/2

⇒ cot 12° cot 38° tan (90–52)° tan (90–78)°

⇒ cot 12° cot 38° tan (38)° tan (12)°

⇒ cot 12° × 1 × tan (12)°

⇒ 1

⇒ cot 12° cot 38° tan (90–52)° tan (90–78)°

⇒ cot 12° cot 38° tan (38)° tan (12)°

⇒ cot 12° × 1 × tan (12)°

⇒ 1

(i) sin A = sin (90–30)°

⇒ sin A = sin 60°

⇒ A = 60

(ii) cot(90–49) = cot A

⇒ cot 41 = cot A

⇒ A = 41

(iii) tan A = 1/ tan35°

⇒ tan A = cot 35°

⇒ tan A = tan (90–35)

⇒ tan A = tan 55

⇒ A = 55

(iv)cosec(90–35) = cosec A

⇒ cosec 55 = cosec A

⇒ A = 55

(v) cosec A = 1/ cos 43°

⇒ cosec A = sec 43°

⇒ cosec A = cosec (90–43)

⇒ cosec A = cosec 47

⇒ A = 47

(vi) sin 20° tan A cosec(90–70) = √3

⇒ sin 20° tan A cosec 20° = √3

⇒ tan A = √3

⇒ tan A = tan 60°

⇒ A = 60°

(i) sin A = sin (90–30)°

⇒ sin A = sin 60°

⇒ A = 60

(ii) cot(90–49) = cot A

⇒ cot 41 = cot A

⇒ A = 41

(iii) tan A = 1/ tan35°

⇒ tan A = cot 35°

⇒ tan A = tan (90–35)

⇒ tan A = tan 55

⇒ A = 55

(iv)cosec(90–35) = cosec A

⇒ cosec 55 = cosec A

⇒ A = 55

(v) cosec A = 1/ cos 43°

⇒ cosec A = sec 43°

⇒ cosec A = cosec (90–43)

⇒ cosec A = cosec 47

⇒ A = 47

(vi) sin 20° tan A cosec(90–70) = √3

⇒ sin 20° tan A cosec 20° = √3

⇒ tan A = √3

⇒ tan A = tan 60°

⇒ A = 60°

Simplifying the left hand side:

⇒ cos 48° – cos(90–42)

⇒ cos 48° – cos(48)°

⇒ 0

Which is equal to right hand side.

Hence proved.

Simplifying the left hand side:

⇒ cos 48° – cos(90–42)

⇒ cos 48° – cos(48)°

⇒ 0

Which is equal to right hand side.

Hence proved.

Simplifying the left hand side:

⇒ cos 20° cos 70° – cos(90–70)° cos (90–20)°

⇒ cos 20° cos 70° – cos 20° cos 70°

⇒ 0

Which is equal to right hand side.

Hence proved.

Simplifying the left hand side:

⇒ cos 20° cos 70° – cos(90–70)° cos (90–20)°

⇒ cos 20° cos 70° – cos 20° cos 70°

⇒ 0

Which is equal to right hand side.

Hence proved.

Simplifying the left hand side:

⇒ cos θ tan θ

⇒ cos θ (sin θ/cos θ)

⇒ sin θ

Which is equal to right hand side.

Hence proved.

Simplifying the left hand side:

⇒ cos θ tan θ

⇒ cos θ (sin θ/cos θ)

⇒ sin θ

Which is equal to right hand side.

Hence proved.

Simplifying the left hand side:

⇒

⇒

⇒ 1

Which is equal to right hand side.

Hence proved.

Simplifying the left hand side:

⇒

⇒

⇒ 1

Which is equal to right hand side.

Hence proved.

The relevant part of the sine table is given below.

From the table,

We have sin 26° = 0.43837 ≈ 0.4384

The relevant part of the sine table is given below.

From the table,

We have sin 26° = 0.43837 ≈ 0.4384

The relevant part of the cosine table is given below.

From the table,

We have cos 72° = 0.3090

The relevant part of the cosine table is given below.

From the table,

We have cos 72° = 0.3090

The relevant part of the tangent table is given below.

From the table, we have

We have tan 35° = 0.7002

The relevant part of the tangent table is given below.

From the table, we have

We have tan 35° = 0.7002

We write sin 75° 15’ = sin 75° 12’ + 3’.

The relevant part of the sine table is given below.

From the table, we have

Sin 75° 12’ = 0.9668

Mean difference for 3’ = 0.0002

We know that mean difference is to be added in the case of sine.

∴ sin75° 15’ = 0.9668 + 0.0002 = 0.9670

We write sin 75° 15’ = sin 75° 12’ + 3’.

The relevant part of the sine table is given below.

From the table, we have

Sin 75° 12’ = 0.9668

Mean difference for 3’ = 0.0002

We know that mean difference is to be added in the case of sine.

∴ sin75° 15’ = 0.9668 + 0.0002 = 0.9670

The relevant part of the sine table is given below.

From the table, we have

Sin 12° 12’ = 0.2113

The relevant part of the sine table is given below.

From the table, we have

Sin 12° 12’ = 0.2113

We write cos 12° 35’ = cos 12° 30’ + 5’.

The relevant part of the cosine table is given below.

From the table, we have

cos 12° 30’ = 0.9763

Mean difference for 5’ = 0.0003

We know that mean difference is to be subtracted in the case of cosine.

∴ cos 12° 35’ = 0.9763 - 0.0003 = 0.9760

We write cos 12° 35’ = cos 12° 30’ + 5’.

The relevant part of the cosine table is given below.

From the table, we have

cos 12° 30’ = 0.9763

Mean difference for 5’ = 0.0003

We know that mean difference is to be subtracted in the case of cosine.

∴ cos 12° 35’ = 0.9763 - 0.0003 = 0.9760

We write cos 40° 20’ = cos 40° 18’ + 2’.

The relevant part of the cosine table is given below.

From the table, we have

cos 40° 18’ = 0.7627

Mean difference for 2’ = 0.0004

We know that mean difference is to be subtracted in the case of cosine.

∴ cos 40° 20’ = 0.7627 - 0.0004 = 0.7623

We write cos 40° 20’ = cos 40° 18’ + 2’.

The relevant part of the cosine table is given below.

From the table, we have

cos 40° 18’ = 0.7627

Mean difference for 2’ = 0.0004

We know that mean difference is to be subtracted in the case of cosine.

∴ cos 40° 20’ = 0.7627 - 0.0004 = 0.7623

We write tan 10° 26’ = tan 10° 24’ + 2’.

The relevant part of the tangent table is given below.

From the table, we have

tan 10° 24’ = 0.1835

Mean difference for 2’ = 0.0006

We know that mean difference is to be added in the case of tangent.

∴ tan 10° 26’ = 0.1835 + 0.0006 = 0.1841

We write tan 10° 26’ = tan 10° 24’ + 2’.

The relevant part of the tangent table is given below.

From the table, we have

tan 10° 24’ = 0.1835

Mean difference for 2’ = 0.0006

We know that mean difference is to be added in the case of tangent.

∴ tan 10° 26’ = 0.1835 + 0.0006 = 0.1841

The relevant part of the cotangent table is given below.

From the table, we have

cot 20° = 2.7475

The relevant part of the cotangent table is given below.

From the table, we have

cot 20° = 2.7475

The relevant part of the cotangent table is given below.

From the table, we have

cot 40° 20’ = 1.17777 ≈ 1.1778

The relevant part of the cotangent table is given below.

From the table, we have

cot 40° 20’ = 1.17777 ≈ 1.1778

(i) From the sine table, we find 0.7009 is corresponding to sin 44° 30’.

⇒ sin 44° 30’ = 0.7009

∴ θ = 44° 30’

(ii) From the cosine table, we find 0.9664 is corresponding to cos 14° 54’.

⇒ cos 14° 54’ = 0.9664

∴ θ = 14° 54’

(iii) From the tangent table, we find 0.3679 is corresponding to tan 20° 12’.

⇒ tan 20° 12’ = 0.3679

∴ θ = 20° 12’

(iv) From the cotangent table, we find 0.2334 is corresponding to cot 76° 30’.

⇒ cot 76° 30’ = 0.2334

∴ θ = 76° 30’

(v) From the tangent table, we find 63.6567 is corresponding to tan 89° 6’.

⇒ tan 89° 6’ = 63.6567

∴ θ = 89° 6’

(i) From the sine table, we find 0.7009 is corresponding to sin 44° 30’.

⇒ sin 44° 30’ = 0.7009

∴ θ = 44° 30’

(ii) From the cosine table, we find 0.9664 is corresponding to cos 14° 54’.

⇒ cos 14° 54’ = 0.9664

∴ θ = 14° 54’

(iii) From the tangent table, we find 0.3679 is corresponding to tan 20° 12’.

⇒ tan 20° 12’ = 0.3679

∴ θ = 20° 12’

(iv) From the cotangent table, we find 0.2334 is corresponding to cot 76° 30’.

⇒ cot 76° 30’ = 0.2334

∴ θ = 76° 30’

(v) From the tangent table, we find 63.6567 is corresponding to tan 89° 6’.

⇒ tan 89° 6’ = 63.6567

∴ θ = 89° 6’

First consider sin 30° 30’,

The relevant part of the sine table is given below.

From the table, we have

Sin 30° 30’ = 0.5075

Then consider cos 40° 20’,

We write cos 40° 20’ = cos 40° 18’ + 2’.

The relevant part of the cosine table is given below.

From the table, we have

cos 40° 18’ = 0.7627

Mean difference for 2’ = 0.0004

We know that mean difference is to be subtracted in the case of cosine.

∴ cos 40° 20’ = 0.7627 - 0.0004 = 0.7623

∴ sin 30° 30’ + cos 40° 20’ = 0.5075 + 0.7623 = 1.2698

First consider sin 30° 30’,

The relevant part of the sine table is given below.

From the table, we have

Sin 30° 30’ = 0.5075

Then consider cos 40° 20’,

We write cos 40° 20’ = cos 40° 18’ + 2’.

The relevant part of the cosine table is given below.

From the table, we have

cos 40° 18’ = 0.7627

Mean difference for 2’ = 0.0004

We know that mean difference is to be subtracted in the case of cosine.

∴ cos 40° 20’ = 0.7627 - 0.0004 = 0.7623

∴ sin 30° 30’ + cos 40° 20’ = 0.5075 + 0.7623 = 1.2698

First we consider tan 45° 27’.

We write tan 45° 27’ = tan 45° 24’ + 3’.

The relevant part of the tangent table is given below.

From the table, we have

tan 45° 24’ = 1.0141

Mean difference for 3’ = 0.0018

We know that mean difference is to be added in the case of tangent.

∴ tan 45° 27’ = 1.0141 + 0.0018 = 1.0159

Then, consider sin 20°,

The relevant part of the sine table is given below.

From the table, we have

Sin 20° = 0.3420

∴ tan 45° 27’ + sin 20° = 1.0159 + 0.3240 = 1.3399

First we consider tan 45° 27’.

We write tan 45° 27’ = tan 45° 24’ + 3’.

The relevant part of the tangent table is given below.

From the table, we have

tan 45° 24’ = 1.0141

Mean difference for 3’ = 0.0018

We know that mean difference is to be added in the case of tangent.

∴ tan 45° 27’ = 1.0141 + 0.0018 = 1.0159

Then, consider sin 20°,

The relevant part of the sine table is given below.

From the table, we have

Sin 20° = 0.3420

∴ tan 45° 27’ + sin 20° = 1.0159 + 0.3240 = 1.3399

First we consider tan 63° 12’,

The relevant part of the tangent table is given below.

From the table, we have

tan 63° 12’ = 1.9797

Then consider cos 12° 42’,

The relevant part of the cosine table is given below.

From the table, we have

cos 12° 42’ = 0.9755

∴ tan 63° 12’ – cos 12° 42’ = 1.9797 – 0.9755 = 1.0042

First we consider tan 63° 12’,

The relevant part of the tangent table is given below.

From the table, we have

tan 63° 12’ = 1.9797

Then consider cos 12° 42’,

The relevant part of the cosine table is given below.

From the table, we have

cos 12° 42’ = 0.9755

∴ tan 63° 12’ – cos 12° 42’ = 1.9797 – 0.9755 = 1.0042

First, we consider sin 50° 26’,

We write sin 50° 26’ = sin 50° 24’ + 2’.

The relevant part of the sine table is given below.

From the table, we have

Sin 50° 24’ = 0.7705

Mean difference for 2’ = 0.0004

We know that mean difference is to be added in the case of sine.

∴ sin50° 26’ = 0.7705 + 0.0004 = 0.7709

Now consider cos 18°,

The relevant part of the cosine table is given below.

From the table, we have

cos 18° = 0.9511

Then consider tan 70° 12’,

The relevant part of the tangent table is given below.

From the table, we have

tan 70° 12’ = 2.7776

∴ sin 50° 26’ + cos 18° + tan 70° 12’ = 0.7709 + 0.9511 + 2.7776 = 4.4996

First, we consider sin 50° 26’,

We write sin 50° 26’ = sin 50° 24’ + 2’.

The relevant part of the sine table is given below.

From the table, we have

Sin 50° 24’ = 0.7705

Mean difference for 2’ = 0.0004

We know that mean difference is to be added in the case of sine.

∴ sin50° 26’ = 0.7705 + 0.0004 = 0.7709

Now consider cos 18°,

The relevant part of the cosine table is given below.

From the table, we have

cos 18° = 0.9511

Then consider tan 70° 12’,

The relevant part of the tangent table is given below.

From the table, we have

tan 70° 12’ = 2.7776

∴ sin 50° 26’ + cos 18° + tan 70° 12’ = 0.7709 + 0.9511 + 2.7776 = 4.4996

First we consider tan 72°,

The relevant part of the tangent table is given below.

From the table, we have

tan 72° = 3.0777

Then consider cot 30°,

The relevant part of the cotangent table is given below.

From the table, we have

cot 30° = 1.73205

∴ tan 72° + cot 30° = 3.0777 + 1.73205 = 4.80975 ≈ 4.8098

First we consider tan 72°,

The relevant part of the tangent table is given below.

From the table, we have

tan 72° = 3.0777

Then consider cot 30°,

The relevant part of the cotangent table is given below.

From the table, we have

cot 30° = 1.73205

∴ tan 72° + cot 30° = 3.0777 + 1.73205 = 4.80975 ≈ 4.8098

From the above figure,

⇒ sin θ =

⇒ sin 48° =

From the sine table, sin 48° = 0.7431.

∴ 0.7431 =

∴ AB = 0.7431 × 20 = 14.862 cm

Then cos θ =

⇒ cos 48° =

From the cosine table, cos 48° = 0.6691

∴ 0.6691 =

∴ BC = 0.6691 × 20 = 13.382 cm

We know that Area of right angled triangle = 1/2 bh.

∴ Area of ΔABC = 1/2 × 13.382 × 14.862

= 99.441642

∴ Area of the triangle is 99.441 cm².

From the above figure,

⇒ sin θ =

⇒ sin 48° =

From the sine table, sin 48° = 0.7431.

∴ 0.7431 =

∴ AB = 0.7431 × 20 = 14.862 cm

Then cos θ =

⇒ cos 48° =

From the cosine table, cos 48° = 0.6691

∴ 0.6691 =

∴ BC = 0.6691 × 20 = 13.382 cm

We know that Area of right angled triangle = 1/2 bh.

∴ Area of ΔABC = 1/2 × 13.382 × 14.862

= 99.441642

∴ Area of the triangle is 99.441 cm².

From the above figure,

⇒ sin θ =

⇒ sin 57° =

From the sine table, sin 57° = 0.8387.

∴ 0.8387 =

∴ AB = 0.8387 × 8 = 6.7096 cm

Then cos θ =

⇒ cos 57° =

From the cosine table, cos 57° = 0.5446

∴ 0.5446 =

∴ BC = 0.5446 × 8 = 4.3568 cm

We know that Area of right angled triangle = 1/2 bh.

∴ Area of ΔABC = 1/2 × 4.3568 × 6.7096

= 14.616192

∴ Area of the triangle is 14.62 cm² (Approximately).

From the above figure,

⇒ sin θ =

⇒ sin 57° =

From the sine table, sin 57° = 0.8387.

∴ 0.8387 =

∴ AB = 0.8387 × 8 = 6.7096 cm

Then cos θ =

⇒ cos 57° =

From the cosine table, cos 57° = 0.5446

∴ 0.5446 =

∴ BC = 0.5446 × 8 = 4.3568 cm

We know that Area of right angled triangle = 1/2 bh.

∴ Area of ΔABC = 1/2 × 4.3568 × 6.7096

= 14.616192

∴ Area of the triangle is 14.62 cm² (Approximately).

Draw CD perpendicular to AB.

∴ D is the midpoint of AB and ∠ACB = 60° 40’

Then ∠ACD = = 30° 20’ = ∠BCD

In right angled triangle ACD,

⇒ tan 30° 20’ =

⇒ tan 30° 20’ =

From tangent table, tan 30° 18’ = 0.5844 and Mean difference of 2’ = 0.0008.

∴ tan 30° 20’ = 0.5844 + 0.0008 = 0.5852

⇒ CD = 8 ÷ 0.5852 = 13.672 cm

We know that Area of right angled triangle = 1/2 bh.

⇒ Area of ΔACD = 1/2 × 8 × 13.672 = 54.688 cm²

Since ΔABC is isosceles, area of ΔACD = area of ΔBCD = 54.688 cm².

∴ Area of ΔABC = Area of (ΔACD + ΔBCD)

= 54.688 + 54.688

= 109.376 cm²

∴ The area of given isosceles triangle = 109.376 cm².

Draw CD perpendicular to AB.

∴ D is the midpoint of AB and ∠ACB = 60° 40’

Then ∠ACD = = 30° 20’ = ∠BCD

In right angled triangle ACD,

⇒ tan 30° 20’ =

⇒ tan 30° 20’ =

From tangent table, tan 30° 18’ = 0.5844 and Mean difference of 2’ = 0.0008.

∴ tan 30° 20’ = 0.5844 + 0.0008 = 0.5852

⇒ CD = 8 ÷ 0.5852 = 13.672 cm

We know that Area of right angled triangle = 1/2 bh.

⇒ Area of ΔACD = 1/2 × 8 × 13.672 = 54.688 cm²

Since ΔABC is isosceles, area of ΔACD = area of ΔBCD = 54.688 cm².

∴ Area of ΔABC = Area of (ΔACD + ΔBCD)

= 54.688 + 54.688

= 109.376 cm²

∴ The area of given isosceles triangle = 109.376 cm².

Draw CD perpendicular to AB.

∴ D is the midpoint of AB and ∠ACB = 80°

Then ∠ACD = = 40° = ∠BCD

In right angled triangle ACD,

⇒ tan 40° =

⇒ tan 40° =

From tangent table, tan 40° = 0.8391.

⇒ CD = 7.5 ÷ 0.8391 = 8.938 cm

We know that Area of right angled triangle = 1/2 bh.

⇒ Area of ΔACD = 1/2 × 7.5 × 8.938 = 33.5175 cm²

Since ΔABC is isosceles, area of ΔACD = area of ΔBCD = 33.5175 cm².

∴ Area of ΔABC = Area of (ΔACD + ΔBCD)

= 33.5175 + 33.5715

= 67.035 cm²

∴ The area of given isosceles triangle = 67.035 cm².

Draw CD perpendicular to AB.

∴ D is the midpoint of AB and ∠ACB = 80°

Then ∠ACD = = 40° = ∠BCD

In right angled triangle ACD,

⇒ tan 40° =

⇒ tan 40° =

From tangent table, tan 40° = 0.8391.

⇒ CD = 7.5 ÷ 0.8391 = 8.938 cm

We know that Area of right angled triangle = 1/2 bh.

⇒ Area of ΔACD = 1/2 × 7.5 × 8.938 = 33.5175 cm²

Since ΔABC is isosceles, area of ΔACD = area of ΔBCD = 33.5175 cm².

∴ Area of ΔABC = Area of (ΔACD + ΔBCD)

= 33.5175 + 33.5715

= 67.035 cm²

∴ The area of given isosceles triangle = 67.035 cm².

Let AC be the length of the ladder.

AB is the distance between the foot of the ladder and the wall = 12m

In ΔABC,

⇒ cos 30° =

⇒ cos 30° =

From cosine table, cos 30° = 0.8660

⇒ 0.8660 =

⇒ AC = = 13.856 m

∴ The length of the ladder is 13.856 m.

Let AC be the length of the ladder.

AB is the distance between the foot of the ladder and the wall = 12m

In ΔABC,

⇒ cos 30° =

⇒ cos 30° =

From cosine table, cos 30° = 0.8660

⇒ 0.8660 =

⇒ AC = = 13.856 m

∴ The length of the ladder is 13.856 m.

Let θ be the angle made by the ladder with the ground.

AB is the length of the ladder = 4m

AC is the distance between end of ladder and the wall = 2m

In ΔABC,

⇒ cos θ =

⇒ cos θ = = 1/2 = 0.5

From cosine table, we find 0.5 is corresponding to cos 60°.

⇒ cos θ = cos 60°

∴ θ = 60°

∴ The angle made by the ladder with the ground is 60°.

Let θ be the angle made by the ladder with the ground.

AB is the length of the ladder = 4m

AC is the distance between end of ladder and the wall = 2m

In ΔABC,

⇒ cos θ =

⇒ cos θ = = 1/2 = 0.5

From cosine table, we find 0.5 is corresponding to cos 60°.

⇒ cos θ = cos 60°

∴ θ = 60°

∴ The angle made by the ladder with the ground is 60°.

Let AB be the chord of a circle of radius 5 cm with O as centre.

Draw OC perpendicular to AB.

∴ C is the midpoint of AB and ∠AOB = 108°

Then ∠AOC = = 54°

In right angled triangle OCA,

⇒ sin 54° =

⇒ sin 54° =

⇒ AC = sin 54° × 5

= 0.8090 × 5

= 4.045 cm

∴ Length of the chord AB = AC × 2 = 4.045 × 2 = 8.90 cm

Let AB be the chord of a circle of radius 5 cm with O as centre.

Draw OC perpendicular to AB.

∴ C is the midpoint of AB and ∠AOB = 108°

Then ∠AOC = = 54°

In right angled triangle OCA,

⇒ sin 54° =

⇒ sin 54° =

⇒ AC = sin 54° × 5

= 0.8090 × 5

= 4.045 cm

∴ Length of the chord AB = AC × 2 = 4.045 × 2 = 8.90 cm

Let AB be a side of the regular polygon with 12 sides in the circle of radius 6 cm.

If O is a centre of the circle, then ∠AOB = = 30°

Draw OC perpendicular to AB.

Then ∠AOC = = 15°

⇒ sin 15° = =

⇒ 0.2588 =

⇒ AC = 0.2588 × 6 = 1.5528

∴ Length of the side AB = 2 × AC = 2 × 1.5528 = 3.1056 cm

Let AB be a side of the regular polygon with 12 sides in the circle of radius 6 cm.

If O is a centre of the circle, then ∠AOB = = 30°

Draw OC perpendicular to AB.

Then ∠AOC = = 15°

⇒ sin 15° = =

⇒ 0.2588 =

⇒ AC = 0.2588 × 6 = 1.5528

∴ Length of the side AB = 2 × AC = 2 × 1.5528 = 3.1056 cm

Let AB be the side of the regular hexagon and let O be the centre of the incircle.

Draw OC perpendicular to AB.

If r is the radius of the circle, then OC = r.

We know that total sum of angles of a regular polygon = 360°

So, ∠AOB = = 60°

∴ ∠AOC = = 30°

⇒ tan 30° =

⇒ =

⇒ r = 12 × 1.732 = 20.784 cm

Hence, radius of incircle is 20.784 cm.

Let AB be the side of the regular hexagon and let O be the centre of the incircle.

Draw OC perpendicular to AB.

If r is the radius of the circle, then OC = r.

We know that total sum of angles of a regular polygon = 360°

So, ∠AOB = = 60°

∴ ∠AOC = = 30°

⇒ tan 30° =

⇒ =

⇒ r = 12 × 1.732 = 20.784 cm

Hence, radius of incircle is 20.784 cm.

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D. is incorrect because,

We know,

∴

⇒

Hence option B is correct.

Note: Alternatively, the identity could also have been used.

We know,

∴

⇒

Hence option B is correct.

Note: Alternatively, the identity could also have been used.

We know,

∴

So, C is the correct option.

We know,

∴

So, C is the correct option.

We know,

∴

⇒

Using the above result,

So, B is the correct option.

Note: Alternatively, the identity could also have been used.

We know,

∴

⇒

Using the above result,

So, B is the correct option.

Note: Alternatively, the identity could also have been used.

It is given that,

Using

∴

⇒

And so, can be rewritten as

i.e

∴

So, C is the correct option.

Note: Alternatively, the identity could also have been used.

It is given that,

Using

∴

⇒

And so, can be rewritten as

i.e

∴

So, C is the correct option.

Note: Alternatively, the identity could also have been used.

We know, and

∴

⇒ .

Comparing the above result with the given options, we find that

A. is incorrect because,

B. is incorrect because,

C. is incorrect because,

D. is the correct because,

We know, and

∴

⇒ .

Comparing the above result with the given options, we find that

A. is incorrect because,

B. is incorrect because,

C. is incorrect because,

D. is the correct because,

We know, and

Given

So,

⇒

⇒

So, B is the correct option.

We know, and

Given

So,

⇒

⇒

So, B is the correct option.

We know,

Using the above result,

So, D is the correct option.

Note: Alternatively, the identity could also have been used.

We know,

Using the above result,

So, D is the correct option.

Note: Alternatively, the identity could also have been used.

We know,

So,

So, A is the correct option.

We know,

So,

So, A is the correct option.

We know,

So,

⇒

Comparing the above result with the given options, we find that

A. is the correct option because,

B. is incorrect because,

C. is incorrect because,

D. is incorrect because,

We know,

So,

⇒

Comparing the above result with the given options, we find that

A. is the correct option because,

B. is incorrect because,

C. is incorrect because,

D. is incorrect because,