## EXERCISE 8.1

#### Page No 181:

#### 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

AC^{2} = AB^{2} + BC^{2}

= (24 cm)^{2} + (7 cm)^{2}

= (576 + 49) cm^{2}

= 625 cm^{2}

∴ 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

PR^{2} = PQ^{2} + QR^{2}

(13 cm)^{2} = (12 cm)^{2} + QR^{2}

169 cm^{2} = 144 cm^{2} + QR^{2}

25 cm^{2} = QR^{2}

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

Applying Pythagoras theorem in ΔABC, we obtain

AC^{2} = AB^{2} + BC^{2}

(4*k*)^{2} = AB^{2} + (3*k*)^{2}

16*k* ^{2} − 9*k* ^{2} = AB^{2}

7*k* ^{2} = AB^{2}

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

Applying Pythagoras theorem in ΔABC, we obtain

AC^{2} = AB^{2} + BC^{2}

= (8*k*)^{2} + (15*k*)^{2}

= 64*k*^{2} + 225*k*^{2}

= 289*k*^{2}

AC = 17*k*

#### 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 13*k*, AB will be 12*k*, where *k* is a positive integer.

Applying Pythagoras theorem in ΔABC, we obtain

(AC)^{2} = (AB)^{2} + (BC)^{2}

(13*k*)^{2} = (12*k*)^{2} + (BC)^{2}

169*k*^{2} = 144*k*^{2} + BC^{2}

25*k*^{2} = BC^{2}

BC = 5*k*

#### 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

CD^{2} = AC^{2} − AD^{2} … (3)

And, CD^{2} = BC^{2} − BD^{2} … (4)

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

AC^{2} − AD^{2} = BC^{2} − BD^{2}

⇒ (*k *BC)^{2} − (*k* BD)^{2} = BC^{2} − BD^{2}

⇒ *k*^{2} (BC^{2} − BD^{2}) = BC^{2} − BD^{2}

⇒ *k*^{2} = 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) cot^{2} θ

#### Answer:

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

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

Applying Pythagoras theorem in ΔABC, we obtain

AC^{2} = AB^{2} + BC^{2}

= (8*k*)^{2} + (7*k*)^{2}

= 64*k*^{2} + 49*k*^{2}

= 113*k*^{2}

AC =

(i)

(ii) cot^{2} 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 4*k*, then BC will be 3*k*, where *k* is a positive integer.

In ΔABC,

(AC)^{2} = (AB)^{2} + (BC)^{2}

= (4*k*)^{2} + (3*k*)^{2}

= 16*k*^{2} + 9*k*^{2}

= 25*k*^{2}

AC = 5*k*

cos^{2} A − sin^{2} 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,

AC^{2} = AB^{2} + BC^{2}

=

= 3*k*^{2} + *k*^{2} = 4*k*^{2}

∴ AC = 2*k*

(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

PR^{2} = PQ^{2} + QR^{2}

*x*^{2} = (5)^{2} + (25 − *x*)^{2}

*x*^{2} = 25 + 625 + *x*^{2} − 50*x*

50*x* = 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 12*k*, AB will be 5*k*, where *k* is a positive integer.

Applying Pythagoras theorem in ΔABC, we obtain

AC^{2} = AB^{2} + BC^{2}

(12*k*)^{2} = (5*k*)^{2} + BC^{2}

144*k*^{2} = 25*k*^{2} + BC^{2}

BC^{2} = 119*k*^{2}

BC = 10.9*k*

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

BC should be such that,

AC − AB < BC < AC + AB

12*k* − 5*k* < BC < 12*k* + 5*k*

7*k *< BC < 17 *k*

However, BC = 10.9*k*. 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