OP Malhotra Class 9 Maths Solutions Chapter 19 Trigonometrical Ratios Ex 19(B)

Continuous practice using ICSE S Chand Maths Class 9 Solutions Chapter 19 Trigonometrical Ratios Ex 19(B) can lead to a stronger grasp of mathematical concepts.

S Chand Class 9 ICSE Maths Solutions Chapter 19 Trigonometrical Ratios Ex 19(B)

Question 1.
Given that tan θ = \(\frac { 5 }{ 12 }\) and angle θ is an acute angle, find sin θ and cos θ. (ICSE)
Solution:
tan θ = \(\frac { 5 }{ 12 }\) = \(\frac{\text { Perpendicular }}{\text { Base }}\)
In the △ABC, ∠B = 90°,

AB = 12, BC = 5, then
AC² = AB² + BC² (Pythagoras Theorem)
= (12)² + (5)² = 144 + 25 = 169
= (13)²
∴ AC = 13
Now sin θ = \(\frac{\text { Perpendicular }}{\text { Hypotenuse }}\) = \(\frac{\text { BC }}{\text { AC }}\) = \(\frac { 5 }{ 13 }\)

and cos θ = \(\frac{\text { Base }}{\text { Hypotenuse }}\) = \(\frac{\text { AB }}{\text { AC }}\) = c

Question 2.
If sin θ = \(\frac { 3 }{ 5 }\) and θ is an acute angle, find
(i) cos θ,
(ii) tan θ.
Solution:
sin θ = \(\frac { 3 }{ 5 }\) = \(\frac{\text { Perpendicular }}{\text { Hypotenuse }}\) = \(\frac{\text { AC }}{\text { AB }}\)
∴ In △ABC,
AC = 3, AB = 5

Now, AB² = BC² + AC² ⇒ (5)² = BC² + (3)²
⇒ BC² = (5)² – (3)² = 25 – 9 = 16 = (4)²
∴BC = 4
Now,
(i) cos θ = \(\frac{\text { Base }}{\text { Hypotenuse }}\) = \(\frac{\text { BC }}{\text { AB }}\) = \(\frac { 4 }{ 5 }\)
(ii) and tan θ = \(\frac{\text { Perpendicular }}{\text { Base }}\) = \(\frac{\text { AC }}{\text { BC }}\) = \(\frac { 3 }{ 4 }\)

Question 3.
△ABC is right-angled at ∠A. Find tan, sine and cos of angles B and C in each of the following cases.
(i) AB = 7 cm, AC = 24 cm
(ii) AB = 12 cm, AC = 9 cm
Solution:
(i) In right △ABC, ∠A = 90°
AB = 7 cm, AC = 24 cm

∴ BC² = AB² + AC²
= (7)² + (24)²
= 49 + 576 = 625
= (25)²
∴BC = 25 cm
Now, tan B = \(\frac{\text { AC }}{\text { AB }}\) = \(\frac { 24 }{ 7 }\)
tan C = \(\frac{\text { AB }}{\text { AC }}\) = \(\frac { 7 }{ 24 }\)
sin B = \(\frac{\text { AC }}{\text { BC }}\) = \(\frac { 24 }{ 25 }\)
sin C = \(\frac{\text { AB }}{\text { BC }}\) = \(\frac { 7 }{ 25 }\)
cos B = \(\frac{\text { AB }}{\text { BC }}\) = \(\frac { 7 }{ 25 }\)
cos C = \(\frac{\text { AC }}{\text { BC }}\) = \(\frac { 24 }{ 25 }\)

(ii) In ∠ABC, ∠A = 90
AB = 12 cm, AC = 9 cm

∴ BC² = AB² + AC²
= (12)² + (9)²
= 144 + 81 = 225
= (15)²
∴ BC = 15 cm
Now, tan B = \(\frac{\text { AC }}{\text { AB }}\) = \(\frac { 12 }{ 9 }\) or \(\frac { 4 }{ 3 }\)
tan C = \(\frac{\text { AB }}{\text { AC }}\) = \(\frac { 9 }{ 12 }\) or \(\frac { 3 }{ 4 }\)
sin B = \(\frac{\text { AC }}{\text { BC }}\) = \(\frac { 9 }{ 15 }\) or \(\frac { 3 }{ 5 }\)
sin C = \(\frac{\text { AB }}{\text { BC }}\) = \(\frac { 12 }{ 15 }\) or \(\frac { 4 }{ 5 }\)
cos B = \(\frac{\text { AB }}{\text { BC }}\) = \(\frac { 12 }{ 15 }\) or \(\frac { 4 }{ 5 }\)
cos C = \(\frac{\text { AC }}{\text { BC }}\) = \(\frac { 9 }{ 15 }\) or \(\frac { 3 }{ 5 }\)

Question 4.
△PQR is right-angled at R. Find tan, sine and cos of ∠P and ∠Q in each of the following cases.
(i) PQ = 10 cm, QR = 8 cm
(ii) PQ = 29 mm, PR = 21 mm
(iii) PQ = 3.7 cm, PR = 3.5 cm
Solution:
(i) In △PQR, ∠R = 90°
PQ = 10 cm, QR = 8 cm

PQ² = QR² + PR² (Pythagoras Theorem)
(10)² = (8)² + PR² ⇒ 100 = 64 + PR²
⇒ PR² = 100 – 64 = 36 = (6)²
∴ PR = 6 cm
tan P = \(\frac{\text { QR }}{\text { PR }}\) = \(\frac { 8 }{ 6 }\) or \(\frac { 4 }{ 3 }\),
tan Q = \(\frac{\text { PR }}{\text { QR }}\) = \(\frac { 6 }{ 8 }\) or \(\frac { 3 }{ 4 }\),
sin P = \(\frac{\text { QR }}{\text { PQ }}\) = \(\frac { 8 }{ 10 }\) or \(\frac { 4 }{ 5 }\),
sin Q = \(\frac{\text { PR }}{\text { PQ }}\) = \(\frac { 6 }{ 10 }\) or \(\frac { 3 }{ 5 }\),
cos P = \(\frac{\text { PR }}{\text { PQ }}\) = \(\frac { 6 }{ 10 }\) or \(\frac { 3 }{ 5 }\),
cos Q = \(\frac{\text { QR }}{\text { PQ }}\) = \(\frac { 8 }{ 10 }\) or \(\frac { 4 }{ 5 }\)

(ii) In △PQR, ∠R = 90°
PQ = 29 mm, PR = 21 mm

∴PQ² = PR² + QR² (Pythagoras Theorem)
(29)² = (21)² + QR²
⇒ 841 = 441 + QR²
⇒ QR² = 841 – 441 = 400 = (20)²
∴QR = 20 mm
Now,
tap P = \(\frac{\text { QR }}{\text { PR }}\) = \(\frac { 20 }{ 21 }\), tan Q = \(\frac{\text { PR }}{\text { QR }}\) = \(\frac { 21 }{ 20 }\)
sin P = \(\frac{\text { QR }}{\text { PQ }}\) = \(\frac { 20 }{ 29 }\), sin Q = \(\frac{\text { PR }}{\text { PQ }}\) = \(\frac { 21 }{ 29 }\)
cos P = \(\frac{\text { PR }}{\text { PQ }}\) = \(\frac { 21 }{ 29 }\), cos Q = \(\frac{\text { QR }}{\text { PQ }}\) = \(\frac { 20 }{ 29 }\)

(iii) In △PQR, ∠R = 90°
PQ = 37 cm, PR = 3.5 cm

But PQ² = PR² + QR²
⇒ (3.7)² = (3.5)² + QR²
⇒ 13.69 = 12.25 + QR²
QR² = 13.69 – 12.25
=1.44 = (1.2)²
∴ QR = 1.2 cm
Now,
tan P = \(\frac{\text { QR }}{\text { PR }}\) = \(\frac { 1.2 }{ 3.5 }\) or \(\frac { 12 }{ 35 }\),
tan Q = \(\frac{\text { PR }}{\text { QR }}\) = \(\frac { 3.5 }{ 1.2 }\) or \(\frac { 35 }{ 12 }\)
sin P = \(\frac{\text { QR }}{\text { PQ }}\) = \(\frac { 1.2 }{ 3.7 }\) or \(\frac { 12 }{ 37 }\),
sin Q = \(\frac{\text { PR }}{\text { PQ }}\) = \(\frac { 3.5 }{ 3.7 }\) or \(\frac { 35 }{ 37 }\)
cos P = \(\frac{\text { PR }}{\text { PQ }}\) = \(\frac { 3.5 }{ 3.7 }\) or \(\frac { 35 }{ 37 }\),
cos Q = \(\frac{\text { QR }}{\text { PQ }}\) = \(\frac { 1.2 }{ 3.7 }\) or \(\frac { 12 }{ 37 }\)

Question 5.
For any angle θ, state the value of sin²θ + cos²θ.
Solution:
In △ABC, ∠A = θ and ∠B = 90°
∴ sin θ = \(\frac{\text { BC }}{\text { CA }}\) and cos θ = \(\frac{\text { AB }}{\text { CA }}\)
∵ABC is a right angled triangle
∴ CA² = AB² + BC²

Now, sin²θ + cos²θ = \(\left(\frac{\mathrm{BC}}{\mathrm{CA}}\right)^2+\left(\frac{\mathrm{AB}}{\mathrm{CA}}\right)^2\)
= \(\frac{\mathrm{BC}^2}{\mathrm{CA}^2}+\frac{\mathrm{AB}^2}{\mathrm{CA}^2}\)
= \(\frac{\mathrm{BC}^2+\mathrm{AB}^2}{\mathrm{CA}^2}\) = \(\frac{\mathrm{CA}^2}{\mathrm{CA}^2}\) [From (i)]
= 1
Hence sin ²θ + cos²θ = 1

Question 6.
The diagonals AC, BD of a rhombus ABCD meet at O. If AC = 6, BD = 8, find sin OCD.
Solution:
In rhombus ABCD, diagonals AC and BD bisect each other at O at right angles

∴ AO = OC and BO = OD
and ∠AOB = ∠BOC = ∠COD = ∠DOA = 90°
∵ AC = 6 units and BD = 8 units
∴ AO = OC = \(\frac { 6 }{ 2 }\) = 3 units and BO = OD = \(\frac { 8 }{ 2 }\) =4 units
Now in right △DOC
DC² = DO² + OC²
= (4)² + (3)² =16 + 9 = 25 = (5)²
∴ DC = 5 units
Now sin ∠OCD = \(\frac{\text { OD }}{\text { DC }}\) = \(\frac { 4 }{ 5 }\)

Question 7.
If cos θ = 0.6, find the value of 5 sin θ – 3 tan θ.
Solution:
In △ABC, ∠B = 90° and ∠A = θ
cos θ = 0.6 = \(\frac { 6 }{ 10 }\) = \(\frac { 3 }{ 5 }\) = \(\frac{\text { AB }}{\text { CA }}\)
∴ AB = 3, CA = 5
Now,
CA² = AB² + BC² (Pythagoras Theorem)
⇒ (5)² = (3)² + BC² ⇒ 25 = 9 + BC²
⇒ BC² = 25 – 9 = 16 = (4)²

∴ BC = 4
∴ sin θ = \(\frac{\text { BC }}{\text { CA }}\) = \(\frac { 4 }{ 5 }\) and tan θ = \(\frac{\text { BC }}{\text { AB }}\) = \(\frac { 4 }{ 3 }\)
Now 5 sin θ – 3 tan θ = 5 × \(\frac { 4 }{ 5 }\) – 3 × \(\frac { 4 }{ 3 }\) = 4 – 4 = 0

Question 8.
In a right-angled triangle, it is given that A is an acute angle and that tan A = \(\frac { 3 }{ 4 }\). Without using tables, find the value of cos A.
Solution:
Let ABC is a right triangle in which ∠A is an acute angle, ∠B = 90°

tan A = \(\frac{\text { BC }}{\text { AB }}\) = \(\frac { 3 }{ 4 }\)
∴ BC = 3, AB = 4
But AC² = AB² + BC²
= (4)² + (3)² = 16 + 9 = 25 = (5)²
∴ AC = 5
Now cos A = \(\frac{\text { AB }}{\text { AC }}\) = \(\frac { 4 }{ 5 }\) or 0.8

Question 9.
If sin θ = \(\frac { 6 }{ 10 }\), find, without using table, the value of (cos θ + tan θ).
Solution:
In △ABC, ∠B = 90° and ∠A = θ
sin θ = \(\frac { 6 }{ 10 }\) = \(\frac{\text { BC }}{\text { AC }}\)
∴ BC = 6, AC = 10

But AC² = AB² + BC² (Pythagoras Theorem)

(10)² = AB² + (6)²
⇒ 100 = AB² + 36
⇒ AB² = 100 – 36 = 64
⇒ AB² = 64 = (8)²
∴ AB = 8
Now cos θ = \(\frac{\text { AB }}{\text { AC }}\) = \(\frac { 8 }{ 10 }\) = \(\frac { 4 }{ 5 }\)
and tan θ = \(\frac{\text { BC }}{\text { AB }}\) = \(\frac { 6 }{ 8 }\) = \(\frac { 3 }{ 4 }\)
∴ cos θ + tan θ = \(\frac { 4 }{ 5 }\) + \(\frac { 3 }{ 4 }\)
= \(\frac{16+15}{20}\) = \(\frac{31}{20}\) = 1\(\frac{11}{20}\)

Question 10.
Answer true or false.
(a) In the diagram
(i) tan θ = \(\frac{\text { BC }}{\text { AB }}\);
(ii) sec θ = \(\frac{\text { BC }}{\text { AC }}\);

(b) In a triangle right-angled at B, the ratio AB : BC is the same as the ratio of sin A : cos A.
Solution:
(a) In the figure,
(i) tan θ = \(\frac{\text { Perpendicular }}{\text { Base }}\) = \(\frac{\text { AB }}{\text { BC }}\)
tan θ = \(\frac{\text { BC }}{\text { AB }}\) is false

(ii) sec θ = \(\frac{\text { Hypotenuse }}{\text { Base }}\) = \(\frac{\text { AC }}{\text { BC }}\)
∴ sec θ = \(\frac{\text { BC }}{\text { AC }}\) is false

(b) AB : BC = sin A : cos A
Now, sin A : cos A = \(\frac{\sin A}{\cos A}\) = tan A
= \(\frac{\text { Perpendicular }}{\text { Base }}\) = \(\frac{\text { BC }}{\text { AB }}\)
= BC : AB
∴ AB : BC = sin A : cos A is false

Question 11.
Using the measurements given in figure,
(a) Find the value of (i) sin Φ, (ii) tan θ
(b) Write an expression for AD in terms of θ.

Solution:
(a) (i) sin Φ = \(\frac{\text { CD }}{\text { BC }}\) = \(\frac { 5 }{ 13 }\)
Draw DE || BC so that

(ii) ED = BC = 12 and
EB = DC = 5
∴ AE = AB – EB = 14 – 5 = 9
Now tan θ = \(\frac{\text { DE }}{\text { AE }}\) = \(\frac { 12 }{ 9 }\) = \(\frac { 4 }{ 3 }\)

(b) In right △AED,
AD² = AE² + ED²
= (9)² + (12)² = 81 + 144
= 225 = (15)²
∴ AD = 15
sin θ = \(\frac{\text { DE }}{\text { AD }}\) ⇒ AD = \(\frac{D E}{\sin \theta}\) = \(\frac{12}{\sin \theta}\)
and cos θ = \(\frac{\text { AE }}{\text { AD }}\) ⇒ AD = \(\frac{A E}{\cos \theta}\) = \(\frac{9}{\cos \theta}\)
∴ AD = \(\frac{12}{\sin \theta}\) or \(\frac{9}{\cos \theta}\)

Question 12.
ABC is a right-angled triangle, right-angled at B. Give that ∠ACB = θ, side AB = 2 units and side BC = 1 unit, find the vlaue of sin² θ + tan ² θ.

Solution:
In △ABC, ∠B = 90°, ∠C = θ
AB = 2, BC = 1 units
But AC² = BC² + AB² (Pythagoras Theorem)
= (1)² +(2)² = 1 + 4 = 5
∴ AC = √5
Now sin θ = \(\frac{\mathrm{AB}}{\mathrm{AC}}\) = \(\frac{2}{\sqrt{5}}\) and tan θ = \(\frac{\mathrm{AB}}{\mathrm{BC}}\) = \(\frac { 2 }{ 1 }\)
∴ sin²θ + tan²θ = \(\left(\frac{2}{\sqrt{5}}\right)^2\) + \(\left(\frac{2}{1}\right)^2\) = \(\frac { 4 }{ 5 }\) + \(\frac { 4 }{ 1 }\)
= \(\frac{4+20}{5}\) = \(\frac{24}{5}\) = \(4 \frac{4}{5}\)

Question 13.
(i) If sin θ = \(4 \frac{12}{13}\) and θ is less than 90°, find the value of (cos θ + tan θ).
(ii) If cos θ = \(4 \frac{12}{13}\), find the value of sin θ and tanθ. Also, find the value of 2 sin θ – 4 tan θ, where θ is acute.
Solution:
(i) In △ABC, ∠A = θ and ∠B = 90°

sin θ = \(\frac{12}{13}\) = \(\frac{\mathrm{BC}}{\mathrm{AC}}\)
∴BC = 12 and AC = 13
But AC² = AB² + BC² (Pythagoras Theorem)
⇒ (13)² = AB² + (12)² ⇒ 169 = AB² + 144
⇒ AB² = 169 – 144 = 25 = (5)²
∴ AC = 5
Now cos θ = \(\frac{\mathrm{AB}}{\mathrm{AC}}\) = \(\frac{5}{13}\)
and tan θ = \(\frac{\mathrm{BC}}{\mathrm{AB}}\) = \(\frac{12}{5}\)
∴ cos θ + tan θ = \(\frac{5}{13}\) + \(\frac{12}{5}\)
= \(\frac{25+156}{65}\) = \(\frac{181}{65}\) = \(2 \frac{51}{65}\)

(ii) In △ABC, ∠A = θ, ∠B = 90°

cos θ = \(\frac{12}{13}\) = \(\frac{\mathrm{AB}}{\mathrm{AC}}\)
∴ AB = 12, AC = 13
But AC² = AB² + BC²
⇒ (13)² = (12)² + BC² ⇒ 169 = 144 + BC²
⇒ BC² = 169 – 144 = 25 = (5)²
∴ BC = 5
Now sin θ = \(\frac{\mathrm{BC}}{\mathrm{AC}}\) = \(\frac{5}{13}\)
and tan θ = \(\frac{\mathrm{BC}}{\mathrm{AB}}\) = \(\frac{5}{12}\)
and 2 sin θ – 4 tan θ = 2 × \(\frac{5}{13}\) – 4 × \(\frac{5}{12}\)
= \(\frac{10}{13}\) – \(\frac{5}{3}\)
= \(\frac{30-65}{39}\) = \(\frac{-35}{39}\)

Question 14.
If tan θ = \(\frac{5}{12}\), find the value of
\(\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}\)
Solution:
tan θ = \(\frac{5}{12}\)
Now \(\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}\) = \(\frac{\frac{\cos \theta}{\cos \theta}-\frac{\sin \theta}{\cos \theta}}{\frac{\cos \theta}{\cos \theta}+\frac{\sin \theta}{\cos \theta}}\)
(Dividing each term by cos θ)
= \(\frac{1-\tan \theta}{1+\tan \theta}\) = \(\frac{1-\frac{5}{12}}{1+\frac{5}{12}}\)
= \(\frac{\frac{12-5}{12}}{\frac{12+5}{12}}\) = \(\frac{\frac{7}{12}}{\frac{17}{12}}\)
= \(\frac{7}{12}\) × \(\frac{12}{17}\) = \(\frac{7}{17}\)

Question 15.
If 5 sin θ = 4, find the value of \(\frac{1+\sin \theta}{1-\sin \theta}\).
Solution:
5 sin θ = 4 ⇒ sin θ = \(\frac{4}{5}\)
Now, \(\frac{1+\sin \theta}{1-\sin \theta}\) = \(\frac{1+\frac{4}{5}}{1-\frac{4}{5}}\) = \(\frac{\frac{5+4}{5}}{\frac{5-4}{5}}\)
= \(\frac{\frac{9}{5}}{\frac{1}{5}}\) = \(\frac{9}{5}\) × \(\frac{5}{1}\) = 9

Question 16.
If b tan θ = a, find the value of \(\frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}\).
Solution:
b tan θ = a ⇒ tan θ = \(\frac{a}{b}\)
Now, \(\frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}\) = \(\frac{\frac{\cos \theta}{\cos \theta}+\frac{\sin \theta}{\cos \theta}}{\frac{\cos \theta}{\cos \theta}-\frac{\sin \theta}{\cos \theta}}\) (Dividing each term by cos θ)
= \(\frac{1+\tan \theta}{1-\tan \theta}\) = \(\frac{1+\frac{a}{b}}{1-\frac{a}{b}}\)
= \(\frac{\frac{b+a}{b}}{\frac{b-a}{b}}\) = \(\frac{b+a}{b}\) × \(\frac{b}{b-a}\)
= \(\frac{b+a}{b-a}\)

Question 17.
If 5 tan θ = 4, find the value of \(\frac{5 \sin \theta-3 \cos \theta}{5 \sin \theta+2 \cos \theta}\).
Solution:
5 tan θ = 4 ⇒ tan θ = \(\frac{4}{5}\)
Now, \(\frac{5 \sin \theta-3 \cos \theta}{5 \sin \theta+2 \cos \theta}\) = \(\frac{5 \frac{\sin \theta}{\cos \theta}-3 \frac{\cos \theta}{\cos \theta}}{5 \frac{\sin \theta}{\cos \theta}+2 \frac{\cos \theta}{\cos \theta}}\)
= \(\frac{5 \tan \theta-3}{5 \tan \theta+2}\) {Dividing each term by cos θ}
= \(\frac{5 \times \frac{4}{5}-3}{5 \times \frac{4}{5}+2}\) = \(\frac{4-3}{4+2}\) = \(\frac{1}{6}\)

Question 18.
If 13 sin A = 5 and ∠A is acute, find the value of \(\frac{5 \sin A-2 \cos A}{\tan A}\).
Solution:
13 sin A = 5 ⇒ sin A = \(\frac{5}{13}\)
In right △ABC, ∠B = 90°

sin A = \(\frac{5}{13}\) = \(\frac{\mathrm{BC}}{\mathrm{AC}}\)
∴ BC = 5, AC 13
But AC² = AB² + BC² ⇒ (13)² = AB² + (5)²
⇒ 169 = AB² + 25
⇒ AB = 169 – 25 = 144 = (12)²
∴AB = 12
Now cos A = \(\frac{\mathrm{AB}}{\mathrm{AC}}\) = \(\frac{12}{13}\) and tan A = \(\frac{\mathrm{BC}}{\mathrm{AB}}\) = \(\frac{5}{12}\)
Now \(\frac{5 \sin A-2 \cos A}{\tan A}\) = \(\frac{5 \times \frac{5}{13}-2 \times \frac{12}{13}}{\frac{5}{12}}\)
= \(\frac{\frac{25}{13}-\frac{24}{13}}{\frac{5}{12}}\) = \(\frac{\frac{1}{13}}{\frac{5}{12}}\) = \(\frac{1}{13}\) x \(\frac{12}{5}\) = \(\frac{12}{65}\)

Question 19.
Given 5 cos A – 12 sin A = 0, find the value of \(\frac{\sin A+\cos A}{2 \cos A-\sin A}\).
Solution:
5 cos A – 12 sin A = 0
⇒ 5 cos A = 12 sin A
⇒ \(\frac{\sin A}{\cos A}\) = \(\frac{5}{12}\)
Now, \(\frac{\sin A+\cos A}{2 \cos A-\sin A}\) = \(\frac{\frac{\sin A}{\cos A}+\frac{\cos A}{\cos A}}{\frac{2 \cos A}{\cos A}-\frac{\sin A}{\cos A}}\) (Dividing each term by cos A)
= \(\frac{\tan \mathrm{A}+1}{2-\tan \mathrm{A}}\) = \(\frac{\frac{5}{12}+1}{2-\frac{5}{12}}\)
= \(\frac{\frac{5+12}{12}}{\frac{24-5}{12}}\) = \(\frac{\frac{17}{12}}{\frac{19}{12}}\)
= \(\frac{17}{12}\) × \(\frac{12}{19}\) = \(\frac{17}{19}\)