Chapter 8 – Introduction to Trigonometry

🍀Select EXERCISE🍀 Exercise 8.1 Exercise 8.2 Exercise 8.3 Exercise 8.4

Exercise 8.1

Question 1:

In ΔABC right angled at B, AB = 24 cm, BC = 7 cm. Determine

(i) sin A, cos A

(ii) sin C, cos C

Answer:

Applying Pythagoras theorem for ΔABC, we obtain

AC² = AB² + BC²

= (24 cm)² + (7 cm)²

= (576 + 49) cm²

= 625 cm²

∴ AC = cm = 25 cm

(i) sin A =

cos A = 

(ii)

sin C = 

cos C =

Question 2:

In the given figure find tan P − cot R

Answer:

Applying Pythagoras theorem for ΔPQR, we obtain

PR² = PQ² + QR²

(13 cm)² = (12 cm)² + QR²

169 cm² = 144 cm² + QR²

25 cm² = QR²

QR = 5 cm

tan P − cot R =

Question 3:

If sin A =, calculate cos A and tan A.

Answer:

Let ΔABC be a right-angled triangle, right-angled at point B.

Given that,

Let BC be 3k. Therefore, AC will be 4k, where k is a positive integer.

Applying Pythagoras theorem in ΔABC, we obtain

AC² = AB² + BC²

(4k)² = AB² + (3k)²

16k ² − 9k ² = AB²

7k ² = AB²

AB = 

Question 4:

Given 15 cot A = 8. Find sin A and sec A

Answer:

Consider a right-angled triangle, right-angled at B.

It is given that,

cot A =

Let AB be 8k.Therefore, BC will be 15k, where k is a positive integer.

Applying Pythagoras theorem in ΔABC, we obtain

AC² = AB² + BC²

= (8k)² + (15k)²

= 64k² + 225k²

= 289k²

AC = 17k

Question 5:

Given sec θ =, calculate all other trigonometric ratios.

Answer:

Consider a right-angle triangle ΔABC, right-angled at point B.

If AC is 13k, AB will be 12k, where k is a positive integer.

Applying Pythagoras theorem in ΔABC, we obtain

(AC)² = (AB)² + (BC)²

(13k)² = (12k)² + (BC)²

169k² = 144k² + BC²

25k² = BC²

BC = 5k

Question 6:

If ∠A and ∠B are acute angles such that cos A = cos B, then show that

∠A = ∠B.

Answer:

Let us consider a triangle ABC in which CD ⊥ AB.

It is given that

cos A = cos B

 … (1)

We have to prove ∠A = ∠B. To prove this, let us extend AC to P such that BC = CP.

From equation (1), we obtain

By using the converse of B.P.T,

CD||BP

⇒∠ACD = ∠CPB (Corresponding angles) … (3)

And, ∠BCD = ∠CBP (Alternate interior angles) … (4)

By construction, we have BC = CP.

∴ ∠CBP = ∠CPB (Angle opposite to equal sides of a triangle) … (5)

From equations (3), (4), and (5), we obtain

∠ACD = ∠BCD … (6)

In ΔCAD and ΔCBD,

∠ACD = ∠BCD [Using equation (6)]

∠CDA = ∠CDB [Both 90°]

Therefore, the remaining angles should be equal.

∴∠CAD = ∠CBD

⇒ ∠A = ∠B

Alternatively,

Let us consider a triangle ABC in which CD ⊥ AB.

It is given that,

cos A = cos B

Let 

⇒ AD = k BD … (1)

And, AC = k BC … (2)

Using Pythagoras theorem for triangles CAD and CBD, we obtain

CD² = AC² − AD² … (3)

And, CD² = BC² − BD² … (4)

From equations (3) and (4), we obtain

AC² − AD² = BC² − BD²

⇒ (k BC)² − (k BD)² = BC² − BD²

⇒ k² (BC² − BD²) = BC² − BD²

⇒ k² = 1

⇒ k = 1

Putting this value in equation (2), we obtain

AC = BC

⇒ ∠A = ∠B(Angles opposite to equal sides of a triangle)

Question 7:

If cot θ =, evaluate

(i)  (ii) cot² θ

Answer:

Let us consider a right triangle ABC, right-angled at point B.

If BC is 7k, then AB will be 8k, where k is a positive integer.

Applying Pythagoras theorem in ΔABC, we obtain

AC² = AB² + BC²

= (8k)² + (7k)²

= 64k² + 49k²

= 113k²

AC = 

(i) 

(ii) cot² θ = (cot θ)² = = 

Question 8:

If 3 cot A = 4, Check whether 

Answer:

It is given that 3cot A = 4

Or, cot A =

Consider a right triangle ABC, right-angled at point B.

If AB is 4k, then BC will be 3k, where k is a positive integer.

In ΔABC,

(AC)² = (AB)² + (BC)²

= (4k)² + (3k)²

= 16k² + 9k²

= 25k²

AC = 5k

cos² A − sin² A =

∴ 

Question 9:

In ΔABC, right angled at B. If, find the value of

(i) sin A cos C + cos A sin C

(ii) cos A cos C − sin A sin C

Answer:

If BC is k, then AB will be, where k is a positive integer.

In ΔABC,

AC² = AB² + BC²

= 

= 3k² + k² = 4k²

∴ AC = 2k

(i) sin A cos C + cos A sin C

(ii) cos A cos C − sin A sin C

Question 10:

In ΔPQR, right angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.

Answer:

Given that, PR + QR = 25

PQ = 5

Let PR be x.

Therefore, QR = 25 − x

Applying Pythagoras theorem in ΔPQR, we obtain

PR² = PQ² + QR²

x² = (5)² + (25 − x)²

x² = 25 + 625 + x² − 50x

50x = 650

x = 13

Therefore, PR = 13 cm

QR = (25 − 13) cm = 12 cm

Question 11:

State whether the following are true or false. Justify your answer.

(i) The value of tan A is always less than 1.

(ii) sec A =for some value of angle A.

(iii) cos A is the abbreviation used for the cosecant of angle A.

(iv) cot A is the product of cot and A

(v) sin θ =, for some angle θ

Answer:

(i) Consider a ΔABC, right-angled at B.

But > 1

∴tan A > 1

So, tan A < 1 is not always true.

Hence, the given statement is false.

(ii) 

Let AC be 12k, AB will be 5k, where k is a positive integer.

Applying Pythagoras theorem in ΔABC, we obtain

AC² = AB² + BC²

(12k)² = (5k)² + BC²

144k² = 25k² + BC²

BC² = 119k²

BC = 10.9k

It can be observed that for given two sides AC = 12k and AB = 5k,

BC should be such that,

AC − AB < BC < AC + AB

12k − 5k < BC < 12k + 5k

7k < BC < 17 k

However, BC = 10.9k. Clearly, such a triangle is possible and hence, such value of sec A is possible.

Hence, the given statement is true.

(iii) Abbreviation used for cosecant of angle A is cosec A. And cos A is the abbreviation used for cosine of angle A.

Hence, the given statement is false.

(iv) cot A is not the product of cot and A. It is the cotangent of ∠A.

Hence, the given statement is false.

(v) sin θ =

We know that in a right-angled triangle,

In a right-angled triangle, hypotenuse is always greater than the remaining two sides. Therefore, such value of sin θ is not possible.

Hence, the given statement is false

Exercise 8.2 

Question 1:

Evaluate the following

(i) sin60° cos30° + sin30° cos 60°

(ii) 2tan²45° + cos²30° − sin²60°

(iii) 

(iv) 

(v)

Answer:

(i) sin60° cos30° + sin30° cos 60°

(ii) 2tan²45° + cos²30° − sin²60°

(iii) 

(iv) 

(v) 

Question 2:

Choose the correct option and justify your choice.

(i) 

(A). sin60°

(B). cos60°

(C). tan60°

(D). sin30°

(ii) 

(A). tan90°

(B). 1

(C). sin45°

(D). 0

(iii) sin2A = 2sinA is true when A =

(A). 0°

(B). 30°

(C). 45°

(D). 60°

(iv) 

(A). cos60°

(B). sin60°

(C). tan60°

(D). sin30°

Answer:

(i) 

Out of the given alternatives, only 

Hence, (A) is correct.

(ii)

Hence, (D) is correct.

(iii)Out of the given alternatives, only A = 0° is correct.

As sin 2A = sin 0° = 0

2 sinA = 2sin 0° = 2(0) = 0

Hence, (A) is correct.

(iv) 

Out of the given alternatives, only tan 60° 

Hence, (C) is correct.

Question 3:

If  and;

0° < A + B ≤ 90°, A > B find A and B.

Answer:

⇒ 

⇒ A + B = 60 … (1)

⇒ tan (A − B) = tan30

⇒ A − B = 30 … (2)

On adding both equations, we obtain

2A = 90

⇒ A = 45

From equation (1), we obtain

45 + B = 60

B = 15

Therefore, ∠A = 45° and ∠B = 15°

Question 4:

State whether the following are true or false. Justify your answer.

(i) sin (A + B) = sin A + sin B

(ii) The value of sinθ increases as θ increases

(iii) The value of cos θ increases as θ increases

(iv) sinθ = cos θ for all values of θ

(v) cot A is not defined for A = 0°

Answer:

(i) sin (A + B) = sin A + sin B

Let A = 30° and B = 60°

sin (A + B) = sin (30° + 60°)

= sin 90°

= 1

sin A + sin B = sin 30° + sin 60°

Clearly, sin (A + B) ≠ sin A + sin B

Hence, the given statement is false.

(ii) The value of sin θ increases as θ increases in the interval of 0° < θ < 90° as

sin 0° = 0

sin 90° = 1

Hence, the given statement is true.

(iii) cos 0° = 1

cos90° = 0

It can be observed that the value of cos θ does not increase in the interval of 0° < θ < 90°.

Hence, the given statement is false.

(iv) sin θ = cos θ for all values of θ.

This is true when θ = 45°

As 

It is not true for all other values of θ.

As  and ,

Hence, the given statement is false.

(v) cot A is not defined for A = 0°

As ,

 = undefined

Hence, the given statement is true.

Exercise 8.3

Question 1:

Evaluate

(I) 

(II) 

(III) cos 48° − sin 42°

(IV)cosec 31° − sec 59°

Answer:

(I)

(II)

(III)cos 48° − sin 42° = cos (90°− 42°) − sin 42°

= sin 42° − sin 42°

= 0

(IV) cosec 31° − sec 59° = cosec (90° − 59°) − sec 59°

= sec 59° − sec 59°

= 0

Question 2:

Show that

(I) tan 48° tan 23° tan 42° tan 67° = 1

(II)cos 38° cos 52° − sin 38° sin 52° = 0

Answer:

(I) tan 48° tan 23° tan 42° tan 67°

= tan (90° − 42°) tan (90° − 67°) tan 42° tan 67°

= cot 42° cot 67° tan 42° tan 67°

= (cot 42° tan 42°) (cot 67° tan 67°)

= (1) (1)

= 1

(II) cos 38° cos 52° − sin 38° sin 52°

= cos (90° − 52°) cos (90°−38°) − sin 38° sin 52°

= sin 52° sin 38° − sin 38° sin 52°

= 0

Question 3:

If tan 2A = cot (A− 18°), where 2A is an acute angle, find the value of A.

Answer:

Given that,

tan 2A = cot (A− 18°)

cot (90° − 2A) = cot (A −18°)

90° − 2A = A− 18°

108° = 3A

A = 36°

Question 4:

If tan A = cot B, prove that A + B = 90°

Answer:

Given that,

tan A = cot B

tan A = tan (90° − B)

A = 90° − B

A + B = 90°

Question 5:

If sec 4A = cosec (A− 20°), where 4A is an acute angle, find the value of A.

Answer:

Given that,

sec 4A = cosec (A − 20°)

cosec (90° − 4A) = cosec (A − 20°)

90° − 4A= A− 20°

110° = 5A

A = 22°

Question 6:

If A, Band C are interior angles of a triangle ABC then show that

Answer:

We know that for a triangle ABC,

∠ A + ∠B + ∠C = 180°

∠B + ∠C= 180° − ∠A

Question 7:

Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.

Answer:

sin 67° + cos 75°

= sin (90° − 23°) + cos (90° − 15°)

= cos 23° + sin 15°

Exercise 8.4

Question 1:

Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

Answer:

We know that,

 will always be positive as we are adding two positive quantities.

Therefore, 

We know that, 

However, 

Therefore, 

Also,

Question 2:

Write all the other trigonometric ratios of ∠A in terms of sec A.

Answer:

We know that,

Also, sin² A + cos² A = 1

sin² A = 1 − cos² A

tan²A + 1 = sec²A

tan²A = sec²A − 1

Question 3:

Evaluate

(i) 

(ii) sin25° cos65° + cos25° sin65°

Answer:

(i) 

 (As sin²A + cos²A = 1)

= 1

(ii) sin25° cos65° + cos25° sin65°

= sin²25° + cos²25°

= 1 (As sin²A + cos²A = 1)

Question 4:

Choose the correct option. Justify your choice.

(i) 9 sec² A − 9 tan² A =

(A) 1

(B) 9

(C) 8

(D) 0

(ii) (1 + tan θ + sec θ) (1 + cot θ − cosec θ)

(A) 0

(B) 1

(C) 2

(D) −1

(iii) (secA + tanA) (1 − sinA) =

(A) secA

(B) sinA

(C) cosecA

(D) cosA

(iv) 

(A) sec² A

(B) −1

(C) cot² A

(D) tan² A

Answer:

(i) 9 sec²A − 9 tan²A

= 9 (sec²A − tan²A)

= 9 (1) [As sec² A − tan² A = 1]

= 9

Hence, alternative (B) is correct.

(ii)

(1 + tan θ + sec θ) (1 + cot θ − cosec θ)

Hence, alternative (C) is correct.

(iii) (secA + tanA) (1 − sinA)

= cosA

Hence, alternative (D) is correct.

(iv) 

Hence, alternative (D) is correct.

Question 5:

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

Answer:

(i) 

(ii) 

(iii) 

= secθ cosec θ +

= R.H.S.

(iv) 

= R.H.S

(v) 

Using the identity cosec² = 1 + cot²,

L.H.S = 

= cosec A + cot A

= R.H.S

(vi) 

(vii) 

(viii) 

(ix) 

Hence, L.H.S = R.H.S

(x) 

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