# NCERT solution class 10 chapter 8 Introduction to Trigonometry exercise 8.1 mathematics

## 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

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 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.

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

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.

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.

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)

∠ACD = ∠BCD [Using equation (6)]

∠CDA = ∠CDB [Both 90°]

Therefore, the remaining angles should be equal.

⇒ ∠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

⇒ (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

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 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  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.

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:

(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

(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

7< 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

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