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.
(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,
= (9)2 + (12)2 = 81 + 144
= 225 = (15)2
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)
= $$\frac{\tan \mathrm{A}+1}{2-\tan \mathrm{A}}$$ = $$\frac{\frac{5}{12}+1}{2-\frac{5}{12}}$$
= $$\frac{\frac{5+12}{12}}{\frac{24-5}{12}}$$ = $$\frac{\frac{17}{12}}{\frac{19}{12}}$$
= $$\frac{17}{12}$$ × $$\frac{12}{19}$$ = $$\frac{17}{19}$$