SELINA Solution Class 9 Chapter 22 Trigonometrical Ratios [Sine, Consine, Tangent of an Angle and their Reciprocals] Exercise 22 B

Question 1

From the following figure, find:
(i) y
(ii) sin x°
(iii) (sec x° - tan x°) (sec x° + tan x°)

SOl:

Consider the given figure :

(i) Since the triangle is a right-angled triangle, so using Pythagorean Theorem
2² = y² + 1²
y² = 4 – 1 = 3
y = ${\displaystyle \sqrt{3}}$ 

(ii) sin x° = ${\displaystyle \frac{\text{perpendicular}}{\text{base}}=\frac{\sqrt{3}}{2}}$

(iii) tan x° = ${\displaystyle \frac{\text{perpendicular}}{\text{base}}=\sqrt{3}}$

     sec x° = ${\displaystyle \frac{\text{hypotenuse}}{\text{base}}=2}$

Therefore
( sec x° – tan x°) ( sec x° + tan x°)

= (2–${\displaystyle \sqrt{3}}$) (2+${\displaystyle \sqrt{3}}$)

= 4 – 3

= 1

Question 2

Use the given figure to find :
(i) sin x^o 
(ii) cos y^o
(iii) 3 tan x^o - 2 sin y^o + 4 cos y^o.

Sol:

Consider the given figure : 

Since the triangle is a right-angled triangle, so using Pythagorean Theorem 
AD² = 8² + 6²

AD² = 64 + 36 = 100

AD = 10

Also
BC² = AC² – AB²

BC² = 17² – 8² = 225

BC = 15

(i) sin x° = ${\displaystyle \frac{\text{perpendicular}}{\text{hypotenuse}}=\frac{8}{17}}$

(ii) cos y° = ${\displaystyle \frac{\text{base}}{\text{hypotenuse}}=\frac{6}{10}=\frac{3}{5}}$

(iii) sin y° =  ${\displaystyle \frac{\text{perpendicular}}{\text{hypotenuse}}=\frac{\text{AB}}{\text{AD}}=\frac{8}{10}=\frac{4}{5}}$

    cos y° = ${\displaystyle \frac{\text{base}}{\text{hypotenuse}}=\frac{6}{10}=\frac{3}{5}}$

    tan x° = ${\displaystyle \frac{\text{perpendicular}}{\text{base}}=\frac{\text{AB}}{\text{BC}}=\frac{8}{15}}$ 

Therefore
3tan x° – 2sin y° + 4 cos y°

= ${\displaystyle 3\left(\frac{8}{15}\right)–2\left(\frac{4}{5}\right)+4\left(\frac{3}{5}\right)}$

= ${\displaystyle \frac{8}{5} –\frac{8}{5}+\frac{12}{5}}$

= ${\displaystyle 2\frac{2}{5}}$

Question 3

In the diagram, given below, triangle ABC is right-angled at B and BD is perpendicular to AC.
Find:
(i) cos ∠DBC
(ii) cot ∠DBA

Sol:

Consider the given figure :

Since the triangle is a right-angled triangle, so using Pythagorean Theorem

AC² = 5² + 12²

AC² = 25 + 144 + 169

AC = 13

In ΔCBD and ΔCBA, the ∠C is common to both the triangles, ∠CDB = ∠CBA = 90° so therefore ∠CBD = ∠CAB.

Therefore ΔCBD and ΔCBA are similar triangles according to AAA Rule 
So
${\displaystyle \frac{\text{AC}}{\text{BC}}=\frac{\text{AB}}{\text{BD}}}$

${\displaystyle \frac{13}{5}=\frac{12}{\text{BD}}}$

${\displaystyle BD=\frac{60}{13}}$

(i) cos ∠DBC = ${\displaystyle \frac{\text{base}}{\text{hypotenuse}}=\frac{\text{BD}}{\text{BC}}=\frac{\frac{60}{13}}{5}=\frac{12}{13}}$

(ii) cot ∠DBA =${\displaystyle \frac{\text{base}}{\text{perpendicular}}=\frac{\text{BD}}{\text{AB}}=\frac{\frac{60}{13}}{12}=\frac{5}{13}}$

Question 4

In the given figure, triangle ABC is right-angled at B. D is the foot of the perpendicular from B to AC. Given that BC = 3 cm and AB = 4 cm.
find :
(i) tan ∠DBC
(ii) sin ∠DBA

SOl:

Consider the given figure : 

Since the triangle is a right-angled triangle, so using Pythagorean Theorem
AC² = 4² + 3²

AC² = 16 + 19 = 25

AC = 5

In ΔCBD and ΔCBA, the ∠C is common to both the triangles, ∠CDB = ∠CBA = 90° so therefore ∠CBD = ∠CAB.

Therefore ΔCBD and ΔCBA are similar triangles according to AAA Rule

So
${\displaystyle \frac{\text{AC}}{\text{BC}}=\frac{\text{AB}}{\text{BD}}}$

${\displaystyle \frac{5}{3}=\frac{4}{\text{BD}}}$

${\displaystyle \text{BD}=\frac{12}{5}}$

Now using Pythagorean Theorem

DC² = ${\displaystyle 32–{\left(\frac{12}{5}\right)}^2}$

DC² = 9 – ${\displaystyle \frac{144}{25}=\frac{81}{25}}$

DC = ${\displaystyle \frac{9}{5}}$

Therefore
AD = AC – DC

= ${\displaystyle 5 –\frac{9}{5}}$

= ${\displaystyle \frac{16}{5}}$

(i) tan ∠DBC = ${\displaystyle \frac{\text{perpendicular}}{\text{base}}=\frac{\text{DC}}{\text{BD}}=\frac{\frac{9}{5}}{\frac{12}{5}}=\frac{3}{4}}$

(ii) sin ∠DBA = ${\displaystyle \frac{\text{AD}}{\text{AB}}=\frac{\frac{16}{5}}{4}=\frac{4}{5}}$

Question 5

In triangle ABC, AB = AC = 15 cm and BC = 18 cm, find cos ∠ABC.

Sol:

Consider the figure below :

In the isosceles ΔABC, AB = AC = 15cm  and BC =18cm  the perpendicular drawn from angle A to the side BC divides the side BC into two equal parts BD = DC = 9 cm

cos ∠ABC = ${\displaystyle \frac{\text{base}}{\text{hypotenuse}}=\frac{\text{BD}}{\text{AB}}=\frac{9}{15}=\frac{3}{5}}$ 

Question 6

In the figure given below, ABC is an isosceles triangle with BC = 8 cm and AB = AC = 5 cm. Find:
(i) sin B 
(ii) tan C
(iii) sin² B + cos²B 
(iv) tan C - cot B

Sol:

Consider the figure below : 

In the isosceles ΔABC, AB =AC = 5cm and BC = 8cm the perpendicular drawn from angle A to the side BC divides the side BC into two equal parts BD = DC = 4 cm

Since ∠ADB = 90°
⇒ AB² + AD² + BD²   ...( AB is hypotenuse in ΔABD)

⇒ AD² = 5² – 4²

∴ AD² = 9 and AD = 3

(i) sin B = ${\displaystyle \frac{\text{AD}}{\text{AB}}=\frac{3}{5}}$

(ii) tan C = ${\displaystyle \frac{\text{AD}}{\text{DC}}=\frac{3}{4}}$

(iii) sin B = ${\displaystyle \frac{\text{AD}}{\text{AB}}=\frac{3}{5}}$

 cos B = ${\displaystyle \frac{\text{BD}}{\text{AB}}=\frac{4}{5}}$

Therefore
sin² B + cos² B

= ${\displaystyle {\left(\frac{3}{5}\right)}^2+{\left(\frac{4}{5}\right)}^2}$

= ${\displaystyle \frac{25}{25}}$

= 1

(iv) tan C = ${\displaystyle \frac{\text{AD}}{\text{DC}}=\frac{3}{4}}$

cot B = ${\displaystyle \frac{\text{BD}}{\text{AD}}=\frac{4}{3}}$

Therefore
tan C – cot B

= ${\displaystyle \frac{3}{4}–\frac{4}{3}}$

= ${\displaystyle –\frac{7}{12}}$   

Question 7

In triangle ABC; ∠ABC = 90°, ∠CAB = x°, tan x° = ${\displaystyle \frac{3}{4}}$ and BC = 15 cm. Find the measures of AB and AC.

Sol:

Consider the figure : 

tan x° = ${\displaystyle \frac{3}{4}}$

i.e. ${\displaystyle \frac{\text{perpendicular}}{\text{base}}=\frac{\text{BC}}{\text{AB}}=\frac{3}{4}}$

Therefore if length of base = 4x, length of perpendicular = 3x

Since
BC² + AB² = AC²  ...[ Using Pythagoras Theorem ]

(3x)² + (4x)² = AC²

AC² = 9x² + 16x² = 25x²

∴ AC = 5x 

Now 
BC = 15

3x = 15

x = 15

Therefore
AB = 4x

= 4 x 5

= 20 cm

And
AC = 5x

= 5 x 5

= 25 cm

Question 8

Using the measurements given in the following figure:
(i) Find the value of sin θ and tan θ.
(ii) Write an expression for AD in terms of θ

Sol:

Consider the figure :

A perpendicular is drawn from D to the side AB at point E which makes BCDE is a rectangle.

Now in right-angled triangle BCD using Pythagorean Theorem
⇒ BD² = BC² + CD² ...( AB is hypotenuse in ΔABD)

⇒ CD₂ = 13₂ – 12₂ = 25

∴ CD = 5

Since BCDE is rectangle so ED 12 cm, EB = 5 and AE = 14 - 5 = 9 

(i) sin Ø = ${\displaystyle \frac{\text{CD}}{\text{BD}}=\frac{5}{13}}$

tan θ = ${\displaystyle \frac{\text{ED}}{\text{AE}}=\frac{12}{9}=\frac{4}{3}}$

(ii) sec θ = ${\displaystyle \frac{\text{AD}}{\text{AE}}}$

sec θ = ${\displaystyle \frac{\text{AD}}{9}}$

 AD   = 9 secθ

Or

cosec θ = ${\displaystyle \frac{\text{AD}}{\text{ED}}}$

cosec θ = ${\displaystyle \frac{\text{AD}}{12}}$

 AD = 12cosec θ

Question 9

Consider the figure :

A perpendicular is drawn from D to the side AB at point E which makes BCDE is a rectangle.

Now in right-angled triangle BCD using Pythagorean Theorem
⇒ BD² = BC² + CD² ...( AB is hypotenuse in ΔABD)

⇒ CD₂ = 13₂ – 12₂ = 25

∴ CD = 5

Since BCDE is rectangle so ED 12 cm, EB = 5 and AE = 14 - 5 = 9 

(i) sin Ø = ${\displaystyle \frac{\text{CD}}{\text{BD}}=\frac{5}{13}}$

tan θ = ${\displaystyle \frac{\text{ED}}{\text{AE}}=\frac{12}{9}=\frac{4}{3}}$

(ii) sec θ = ${\displaystyle \frac{\text{AD}}{\text{AE}}}$

sec θ = ${\displaystyle \frac{\text{AD}}{9}}$

 AD   = 9 secθ

Or

cosec θ = ${\displaystyle \frac{\text{AD}}{\text{ED}}}$

cosec θ = ${\displaystyle \frac{\text{AD}}{12}}$

 AD = 12cosec θ

Sol:

Given 
sin B = ${\displaystyle \frac{4}{5}}$

i.e.${\displaystyle \frac{\text{perpendicular}}{\text{base}}=\frac{\text{AC}}{\text{AB}}=\frac{4}{5}}$

Therefore if length of perpendicular = 4x, length of hypotenuse = 5x

Since
BC² + AC² = AB²  ...[ Using Pythagoras Theorem]

(5x)² – (4x)² = BC

BC² = 9x²

∴ BC = 3x

Now
BC = 15

3x = 15

x = 5

(i) AC = 4x
  = 4 x 5
  = 20 cm

And
 AB = 5x 
= 5 x 5
= 25 cm

(ii) Given 
tan ∠ADC = ${\displaystyle \frac{1}{1}}$

i.e. ${\displaystyle \frac{\text{perpendicular}}{\text{base}}=\frac{\text{AC}}{\text{CD}}=\frac{1}{1}}$

Therefore if length of perpendicular = x, length of hypotenuse = x

Since
AC² + CD² = AD²  ...[Using Pythagoras Theorem]

(x)2 – (x)² = BC

AD² = 2x²

∴ AD = ${\displaystyle \sqrt{2}x}$

Now 

AC = 20
x = 20

So
AD = ${\displaystyle \sqrt{2}x}$
= ${\displaystyle \sqrt{2}}$ x 20

= 20${\displaystyle \sqrt{2}\text{cm}}$

And
CD = 20 cm

Now 
tan B = ${\displaystyle \frac{\text{AC}}{\text{BC}}=\frac{20}{15}=\frac{4}{3}}$

cos B = ${\displaystyle \frac{\text{BC}}{\text{AB}}=\frac{15}{25}=\frac{3}{5}}$

So
tan2 B – ${\displaystyle \frac{1}{{\cos }^{2}B}}$

=${\displaystyle {\left(\frac{4}{3}\right)}^2– \frac{1}{{\left(\frac{3}{5}\right)}^2}}$

= ${\displaystyle \frac{16}{9}– \frac{25}{9}}$

= ${\displaystyle –\frac{9}{9}}$

= –  1

Question 10

If sin A + cosec A = 2;
Find the value of sin² A + cosec² A.

Sol:

sin A+cosecA = 2
Squaring both sides
                     (sin A+cosecA)² = 2²
sin² A +cosec² A+2sin A . cosecA = 4
sin² A + cosec²A+2sin A. ${\displaystyle \frac{1}{\sin \text{A}}}$ = 4
                           sin² A +cosec² A =2

Question 11

If tan A + cot A = 5;
Find the value of tan² A + cot² A.

Sol:

tan A+cotA = 5
Squaring both sides

                        (tan A +cotA)² = 5²
tan² A+ cot² A+2 tan A. cotA = 25
tan² A+cot² A+2 tan A. ${\displaystyle \frac{1}{\tan \text{A}}}$ = 25
                        tan² A+cot² A = 23

Question 12

Given: 4 sin θ = 3 cos θ ; find the value of:
(i) sin θ (ii) cos θ
(iii) cot² θ - cosec² θ .
(iv) 4 cos²θ- 3 sin²θ+2

Sol:

Consider the diagram below :

4 sin θ = 3cos θ
tan θ =${\displaystyle \frac{3}{4}}$

i.e.${\displaystyle \frac{\text{perpendicular}}{\text{base}}=\frac{3}{4}\Rightarrow \frac{\text{BC}}{\text{AB}}=\frac{3}{4}}$

Therefore if length of BC = 3x, length of AB = 4x

Since
AB² + BC² = AC²  ... [ Using Pythagoras Theorem]

(4x)² + (3x)² = AC²

AC² = 25x²

∴ AC = 5x          ...( hypotenuse)

(i) sin θ = ${\displaystyle \frac{\text{BC}}{\text{AC}}=\frac{3}{5}}$

(ii) cos θ = ${\displaystyle \frac{\text{AB}}{\text{AC}}=\frac{4}{5}}$

(iii) cot θ = ${\displaystyle \frac{\text{AB}}{\text{BC}}=\frac{4}{3}}$

cosec θ = ${\displaystyle \frac{\text{AC}}{\text{BC}}=\frac{5}{3}}$

Therefore

cot²θ – cosec²θ

= ${\displaystyle {\left(\frac{4}{3}\right)}^2–{\left(\frac{5}{3}\right)}^2}$

= ${\displaystyle \frac{16–25}{9}}$

= ${\displaystyle –\frac{9}{9}}$

= – 1

(iv) 4 cos² θ – 3sin² θ + 2

= ${\displaystyle 4{\left(\frac{4}{5}\right)}^2–3{\left(\frac{3}{5}\right)}^2+2}$

= ${\displaystyle \frac{64}{25}–\frac{27}{25}+2}$

= ${\displaystyle \frac{64–27+50}{25}}$

= ${\displaystyle \frac{87}{25}}$

= ${\displaystyle 3\frac{12}{25}}$

Question 13

Given : 17 cos θ = 15;
Find the value of: tan θ + 2 secθ .

SOl:

Consider the diagram below :

17 cos θ = 15

cos θ = ${\displaystyle \frac{15}{17}}$

i.e ${\displaystyle \frac{\text{base}}{\text{hypotenuse}}=\frac{15}{17}\Rightarrow \frac{\text{AB}}{\text{AC}}=\frac{15}{17}}$

Therefore if length of AB = 15x, length of AC = 17x

Since
AB² + BC² = AC²  ...[ Using Pythagoras Theorem ]

(17x)² – (15x)² = BC²

BC² = 64x²

∴ BC = 8x     ...( perpendicular)

Now

sec θ  = ${\displaystyle \frac{\text{AC}}{\text{AB}}=\frac{17}{15}}$

tan θ  = ${\displaystyle \frac{\text{BC}}{\text{AB}}=\frac{8}{15}}$

Therefore
tan θ +2 sec θ 

= ${\displaystyle \frac{8}{15}+2.\frac{17}{15}}$

= ${\displaystyle \frac{42}{15}}$

= ${\displaystyle \frac{14}{5}}$

= ${\displaystyle 2\frac{4}{5}}$

Question 14

Given : 5 cos A - 12 sin A = 0; evaluate :

${\displaystyle \frac{\sin \text{A}+\cos \text{A}}{2\cos \text{A}–\sin \text{A}}}$

Sol:

5 cos A – 12 sin A = 0
5 cos A = 12 sin A

${\displaystyle \frac{\sin \text{A}}{\cos \text{A}}=\frac{5}{12}}$

tan A = ${\displaystyle \frac{15}{12}}$

Now

${\displaystyle \frac{\sin \text{A}+\cos \text{A}}{2\cos \text{A}–\sin \text{A}}= \frac{\frac{\sin \text{A}}{\cos \text{A}}+\frac{\cos \text{A}}{\cos \text{A}}}{2\frac{\cos \text{A}}{\cos \text{A}}–\frac{\sin \text{A}}{\cos \text{A}}}}$

= ${\displaystyle \frac{\tan \text{A}+1}{2–\tan \text{A}}}$ 

= ${\displaystyle \frac{\frac{5}{12}+1}{2–\frac{5}{12}}}$

= ${\displaystyle \frac{\frac{17}{12}}{\frac{19}{12}}}$

= ${\displaystyle \frac{17}{19}}$

Question 15

In the given figure; ∠C = 90^o and D is mid-point of AC.
Find : 
(i) ${\displaystyle \frac{\tan \angle CAB}{\tan \angle CDB}}$ (ii) ${\displaystyle \frac{\tan \angle ABC}{\tan \angle DBC}}$

Sol:

Since D is mid-point of AC so AC = 2DC

(i) ${\displaystyle \frac{\tan \angle CAB}{\tan \angle CDB}}$

= ${\displaystyle \frac{\frac{\text{BC}}{\text{AC}}}{\frac{\text{BC}}{\text{DC}}}}$

= ${\displaystyle \frac{\text{BC}}{\text{2 DC}}.\frac{\text{DC}}{\text{BC}}}$

= ${\displaystyle \frac{1}{2}}$

(ii) ${\displaystyle \frac{\tan \angle ABC}{\tan \angle DBC}}$

= ${\displaystyle \frac{\frac{\text{AC}}{\text{BC}}}{\frac{\text{DC}}{\text{BC}}}}$

= ${\displaystyle \frac{\text{2 DC}}{\text{BC}}.\frac{\text{BC}}{\text{DC}}}$

= 2

Question 16

If 3 cos A = 4 sin A, find the value of :
(i) cos A(ii) 3 - cot² A + cosec²A.

SOl:

Consider the diagram below :

3cos A = 4 sin A

cot A = ${\displaystyle \frac{4}{3}}$

i.e. ${\displaystyle \frac{\text{base}}{\text{perpendicular}}=\frac{4}{3}\Rightarrow \frac{\text{AB}}{\text{BC}}=\frac{4}{3}}$

Therefore if length of AB = 4x, length of BC = 3x

Since
AB² + BC = AC²  ...[ Using Pythagoras Theorem]

(4x)² + (3x)² = AC²

AC² = 25x²

∴ AC = 5x           ...( hypotenuse)

(i) cos A = ${\displaystyle \frac{\text{AB}}{\text{AC}}=\frac{4}{5}}$

(ii) cosec A = ${\displaystyle \frac{\text{AC}}{\text{BC}}=\frac{5}{3}}$

Therefore
3–cot² A + cosec² A

= ${\displaystyle 3 –{\left(\frac{4}{3}\right)}^2+{\left(\frac{5}{3}\right)}^2}$

= ${\displaystyle \frac{27– 16+25}{9}}$

=${\displaystyle \frac{36}{9}}$

= 4

 

Question 17

In triangle ABC, ∠B = 90° and tan A = 0.75. If AC = 30 cm, find the lengths of AB and BC.

Sol:

Consider the figure :

tan A = ${\displaystyle \frac{75}{100}=\frac{3}{4}}$

i.e.${\displaystyle \frac{\text{perpendicular}}{\text{base}}=\frac{\text{BC}}{\text{AB}}=\frac{3}{4}}$

Therefore if length of base = 4x, length of perpendicular = 3x

Since
BC² + AB² = AC²     ...[ Using Pythagoras Theorem ]

(3x)² + (4x)² = AC²

AC² = 9x² + 16x² = 25x²

∴ Ac = 5x

Now
Ac = 30

5x = 30

x = 6

Therefore
AB = 4x

= 4 x 6

= 24 cm

And
Bc = 3x

= 3 x 6

= 18 cm

Question 18

In rhombus ABCD, diagonals AC and BD intersect each other at point O.
If cosine of angle CAB is 0.6 and OB = 8 cm, find the lengths of the side and the diagonals of the rhombus.

Sol:

Consider the figure :

The diagonals of a rhombus bisect each other perpendicularly

cos ∠CAB = ${\displaystyle \frac{6}{10}=\frac{3}{5}}$

i.e.${\displaystyle \frac{\text{base}}{\text{hypotenuse}}=\frac{\text{OA}}{\text{AB}}=\frac{3}{5}}$

Therefore if length of base = 3x, length of hypotenuse = 5x

Since
OB² + OA² = AB²  ...[ Using Pythagoras Theorem ]

(5x)² – (3x)² = OB²

OB² = 16x²

∴ OB = 4x

Now
OB = 8

4x = 8

x = 2

Therefore
AB = 5x

= 5 x 2

= 10 cm

And
OA = 3x

= 3 x 2

= 6 cm

Since the sides of a rhombus are equal so the length of the side of the rhombus 

The diagonals are
BD = 8 x 2

= 16 cm

AC = 6 x 2

= 12 cm

Question 19

In triangle ABC, AB = AC = 15 cm and BC = 18 cm. Find:
(i) cos B
(ii) sin C
(iii) tan² B - sec² B + 2

SOl:

Consider the figure below : 

In the isosceles ΔABC, the perpendicular drawn from angle A to the side BC divides the side BC into two equal parts BD = DC = 9 cm

Since ∠ADB = 90°
⇒ AB² = AD² + BD²  ...(AB is hypotenuse in ΔABD)

⇒ AD² = 15² – 9²

∴ AD² = 144 and AD = 12

(i) cos B = ${\displaystyle \frac{\text{base}}{\text{hypotenue}}=\frac{\text{BD}}{\text{AB}}=\frac{9}{15}=\frac{3}{5}}$

(ii) sin C = ${\displaystyle \frac{\text{perpendicular}}{\text{hypotenuse}}=\frac{\text{AD}}{\text{AB}}=\frac{12}{15}=\frac{4}{5}}$

(iii) tan B = ${\displaystyle \frac{\text{perpendicular}}{\text{base}}=\frac{\text{AD}}{\text{BD}}=\frac{12}{9}=\frac{4}{3}}$

sec B = ${\displaystyle \frac{\text{hypotenuse}}{\text{base}}=\frac{\text{AB}}{\text{BD}}=\frac{15}{9}=\frac{5}{3}}$

Therefore
tan² B – sec² B + 2

= ${\displaystyle {\left(\frac{4}{3}\right)}^2– {\left(\frac{5}{3}\right)}^2+2}$

= ${\displaystyle \frac{16 –25+28}{9}}$

= ${\displaystyle \frac{9}{9}}$

=1

Question 20

In triangle ABC, AD is perpendicular to BC. sin B = 0.8, BD = 9 cm and tan C = 1.
Find the length of AB, AD, AC, and DC.

SOl:

Consider the figure below :

sin B = ${\displaystyle \frac{8}{10}=\frac{4}{5}}$

i.e.${\displaystyle \frac{\text{perpendicular}}{\text{hypotenuse}}=\frac{\text{AD}}{\text{AB}}=\frac{4}{5}}$

Therefore if length of perpendicular = 4x, length of hypotenuse = 5x

Since
AD² + BD² = AB²  ...[ Using Pythagoras Theorem ]

(5x)² – (4x)² = BD²

BD² = 9x²

∴ BD = 3x 

Now
BD = 9

3x = 9

x = 3

Therefore
AB = 5x

= 5 x 3

= 15 cm

And
AD= 4x

= 4 x 3

= 12 cm

Again
tan C = ${\displaystyle \frac{1}{1}}$

i.e.${\displaystyle \frac{\text{perpendicular}}{\text{base}}=\frac{\text{AD}}{\text{DC}}=\frac{1}{1}}$

Therefore if length of perpendicular = x, length of base = x

Since
AD² + DC² = AC² ...[ Using Pythagoras Theorem ]

(x)² + (x)² = AC²

AC² = 2x² 

∴ AC = ${\displaystyle \sqrt{2}x}$

Now
AD = 12

x = 12

Therefore
DC = x

= 12 cm

And
AC = ${\displaystyle \sqrt{2}}$

= ${\displaystyle \sqrt{2}}$ x 12

= 12${\displaystyle \sqrt{2}\text{cm}}$

Question 21

Given q tan A = p, find the value of :

${\displaystyle \frac{\text{p}\sin \text{A}–\text{q}\cos \text{A}}{\text{p}\sin \text{A}+\text{q}\cos \text{A}}}$.

Sol:

q tan A = p

tan A = ${\displaystyle \frac{P}{q}}$

Now

${\displaystyle \frac{\text{p}\sin \text{A}–\text{q}\cos \text{A}}{\text{p}\sin \text{A}+\text{q}\cos \text{A}}=\frac{\frac{\text{p sinA}}{\cos \text{A}}–\frac{\text{q cos A}}{\cos \text{A}}}{\frac{\text{p sin A}}{\cos \text{A}}+\frac{\text{q cos A}}{\cos \text{A}}}}$

= ${\displaystyle \frac{\text{p}\tan \text{A}–\text{q}}{\text{p}\tan \text{A}+\text{q}}}$

= ${\displaystyle \frac{\text{p}\left(\frac{p}{q}\right)–\text{q}}{\text{p}\left(\frac{p}{q}\right)+\text{q}}}$

= ${\displaystyle \frac{\frac{{\text{p}}^2–{\text{q}}^2}{\text{q}}}{\frac{{\text{p}}^2+{\text{q}}^2}{\text{q}}}}$

= ${\displaystyle \frac{{\text{p}}^2–{\text{q}}^2}{{\text{p}}^2+{\text{q}}^2}}$

Question 22

If sin A = cos A, find the value of 2 tan²A - 2 sec² A + 5.

SOl:

Consider the figure : 

sin A = cos A

tan A = ${\displaystyle \frac{1}{1}}$

i.e.${\displaystyle \frac{\text{perpendicular}}{\text{base}}=\frac{\text{BC}}{\text{AB}}=\frac{1}{1}}$

Therefore if length of perpendicular = x, length of base = x

Since
AB² + BC² = AC²  ...[ Using Pythagoras Theorem]

(x)² + (x)² = AC²

AC² = 2x²

∴ AC = ${\displaystyle \sqrt{2}x}$

Now

sec A = ${\displaystyle \frac{\text{AC}}{\text{AB}}=\sqrt{2}}$

Therefore
2 tan² A – 2sec² A + 5

= 2(1)² –2 (${\displaystyle \sqrt{2}}$)² + 5

= 2 – 4 + 5

= 3 

Question 23

Consider the figure : 

sin A = cos A

tan A = ${\displaystyle \frac{1}{1}}$

i.e.${\displaystyle \frac{\text{perpendicular}}{\text{base}}=\frac{\text{BC}}{\text{AB}}=\frac{1}{1}}$

Therefore if length of perpendicular = x, length of base = x

Since
AB² + BC² = AC²  ...[ Using Pythagoras Theorem]

(x)² + (x)² = AC²

AC² = 2x²

∴ AC = ${\displaystyle \sqrt{2}x}$

Now

sec A = ${\displaystyle \frac{\text{AC}}{\text{AB}}=\sqrt{2}}$

Therefore
2 tan² A – 2sec² A + 5

= 2(1)² –2 (${\displaystyle \sqrt{2}}$)² + 5

= 2 – 4 + 5

= 3 

Sol:

Consider the diagram : 

cot ∠ABD = ${\displaystyle \frac{15}{10}=\frac{3}{2}}$

i.e.${\displaystyle \frac{\text{base}}{\text{perpendicular}}=\frac{\text{AB}}{\text{BD}}=\frac{3}{2}}$

Therefore if length of base = 3x, length of perpendicular = 2x

Since
AB² + AD² = BD²  ...[ Using Pythagoras Theorem ]

(3x)² + (2x)² = BD²

BD² = 13x²

∴ BD = ${\displaystyle \sqrt{13}x}$ 

Now
BD = 26

${\displaystyle \sqrt{13}x}$ = 26

x = ${\displaystyle \frac{26}{\sqrt{13}}}$

Therefore
AD = 2x

= 2 x ${\displaystyle \frac{26}{\sqrt{13}}}$

= ${\displaystyle \frac{52}{\sqrt{13}}}$ cm

AB = 3x

= 3 x ${\displaystyle \frac{26}{\sqrt{13}}}$

=${\displaystyle \frac{78}{\sqrt{13}}}$ cm

Now
Area of rectangle ABCD = AB x AD

= ${\displaystyle \frac{78}{\sqrt{13}}x\frac{52}{\sqrt{13}}}$

= 312 cm² 

Perim of rectangle ABCD = 2 ( AB + AD )

= 2 ${\displaystyle (\frac{78}{\sqrt{13}}+\frac{52}{\sqrt{13}})}$

= ${\displaystyle \frac{260}{\sqrt{13}}}$

= 20${\displaystyle \sqrt{13}}$ cm

Question 24

If 2 sin x = ${\displaystyle \sqrt{3}}$ , evaluate.
(i) 4 sin³ x - 3 sin x.
(ii) 3 cos x - 4 cos³ x.

SOl:

Consider the figure :

2sin x  = ${\displaystyle \sqrt{3}}$

sin x = ${\displaystyle \frac{\sqrt{3}}{2}}$

i.e.${\displaystyle \frac{\text{perpendicular}}{\text{base}}=\frac{\text{BC}}{\text{AC}}=\frac{\sqrt{3}}{2}}$

Therefore if length of perpendicular = ${\displaystyle \sqrt{3}x}$ , length of = 2x

Since
AB² + BC² = AC²  ...[ Using Pythagoras Theorem]

(2x)² – (`sqrt3x)² = AB²

AB² = x²

∴ AB = x

Now
cos x = ${\displaystyle \frac{\text{AB}}{\text{AC}}=\frac{1}{2}}$

(i) 4 sin³ x – 3sin x = ${\displaystyle 4{\left(\frac{\sqrt{3}}{2}\right)}^2–3\left(\frac{\sqrt{3}}{2}\right)}$

 = ${\displaystyle 3\frac{\sqrt{3}}{2}–3\frac{\sqrt{3}}{2}}$

= 0

(ii) 3cos x – 4 cos³  x = ${\displaystyle 3.\frac{1}{2}–4.{\left(\frac{1}{2}\right)}^3}$

= ${\displaystyle \frac{3}{2}–\frac{1}{2}}$

= 1

Question 25

If sin A = ${\displaystyle \frac{\sqrt{3}}{2}}$ and cos B = ${\displaystyle \frac{\sqrt{3}}{2}}$ , find the value of : ${\displaystyle \frac{\tan \text{A}–\tan \text{B}}{1+\tan \text{A}\tan \text{B}}}$

Sol:

Consider the diagram below :
'in A = ${\displaystyle \frac{\sqrt{3}}{2}}$

i.e.${\displaystyle \frac{\text{perpendicular}}{\text{hypotenuse}}=\frac{\sqrt{3}}{2}\Rightarrow \frac{\text{BC}}{\text{AC}}=\frac{\sqrt{3}}{2}}$

Therefore if length of BC = ${\displaystyle \sqrt{3}x}$, length of AC = 2x

Since
AB² + BC² = AC²    ...[ Using Pythagoras Therorm]

${\displaystyle {\left(\sqrt{3}x\right)}^2+A{B}^2={\left(2x\right)}^2}$

AB² = x²

∴ AB = x (base)

Consider the diagram below :

cos B = ${\displaystyle \frac{\sqrt{3}}{2}}$

i.e.${\displaystyle \frac{\text{base}}{\text{perpendicular}}=\frac{\sqrt{3}}{2}\Rightarrow \frac{\text{AB}}{\text{BC}}=\frac{\sqrt{3}}{2}}$

Therefore if length of AB = ${\displaystyle \sqrt{3}x}$ , length of  BC = 2x

Since
AB² + AC² = BC²  ...[ Using Pythagoras Theorem ]

AC² + ${\displaystyle {\left(\sqrt{3}x\right)}^2={\left(2x\right)}^2}$

AC² = x²

∴ AC = x(perpendicular)

Now

tan A = ${\displaystyle \frac{\text{perpendicular}}{\text{base}}=\frac{\sqrt{3}x}{x}=\sqrt{3}}$

tan B = ${\displaystyle \frac{\text{perpendicular}}{\text{base}}=\frac{x}{\sqrt{3}x} =\frac{1}{\sqrt{3}}}$

Therefore

${\displaystyle \frac{\tan A–\tan B}{1+\tan A\tan B}=\frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{1+\sqrt{3}\frac{1}{\sqrt{3}}}}$

= ${\displaystyle \frac{\frac{2}{\sqrt{3}}}{2}}$

= ${\displaystyle \frac{1}{\sqrt{3}}}$

Question 26

Use the information given in the following figure to evaluate: ${\displaystyle \frac{10}{\sin x}+\frac{6}{\sin y}–6\cot y}$.

Sol:

Consider the given diagram as 

Using Pythagorean Theorem
AD² + DC² = AC²

DC² = 20² – 12² = 256

DC = 16

Now
BC = BD + DC

21 = BD + 16

BD = 5

Again using Pythagorean Theorem
AD² + BD² = AB²

12² + 5² = AB²

AB² = 169

AB = 13

Now

sin x = ${\displaystyle \frac{\text{BD}}{\text{AB}}=\frac{5}{13}}$

sin y = ${\displaystyle \frac{\text{AD}}{\text{AC}}=\frac{12}{20}=\frac{3}{5}}$

cot y = ${\displaystyle \frac{\text{DC}}{\text{AD}}=\frac{16}{12}=\frac{4}{3}}$

Therefore

${\displaystyle \frac{10}{\sin x}+\frac{6}{\sin y}–6\cot y=\frac{10}{\frac{5}{13}}+\frac{6}{\frac{3}{5}}–6\left(\frac{4}{3}\right)}$ 

= ${\displaystyle \frac{130}{5}+\frac{130}{3}–\frac{24}{3}}$

= 26+ 10 – 8

= 28

Question 27

If sec A = ${\displaystyle \sqrt{2}}$ , find : ${\displaystyle \frac{3{\cot }^2\text{A}+2{\sin }^2\text{A}}{{\tan }^2\text{A}–{\cos }^2\text{A}}}$.

Sol:

Consider the figure :

sec A = ${\displaystyle \sqrt{2}}$

i.e.${\displaystyle \frac{\text{hypotenuse}}{\text{base}}=\frac{\text{AC}}{\text{AB}}=\sqrt{2}}$

Therefore if length of base = x , length of hypotenuse = ${\displaystyle \sqrt{2}x}$

Since
AB² + BC² = AC²  ...[ Using Pythagoras Theorem ]

${\displaystyle {\left(\sqrt{2}x\right)}^2–{\left(x\right)}^2=B{C}^2}$

BC² = x²

∴ BC = x

Now

cos A = ${\displaystyle \frac{1}{\sec \text{A}}=\frac{1}{\sqrt{2}}}$

sin A = ${\displaystyle \frac{\text{BC}}{\text{AC}}=\frac{1}{\sqrt{2}}}$

tan A = "BC"/"AB" = 1

cot A = ${\displaystyle \frac{1}{\tan \text{A}}}$ = 1

Therefore

${\displaystyle \frac{3{\cot }^2\text{A}+2{\sin }^2\text{A}}{{\tan }^2\text{A}–{\cos }^2\text{A}}=\frac{3{\left(1\right)}^2+2{\left(\frac{1}{\sqrt{2}}\right)}^2}{1^2–{\left(\frac{1}{\sqrt{2}}\right)}^2}}$

= ${\displaystyle \frac{3+1}{1–\frac{1}{2}}}$

= ${\displaystyle \frac{4}{1}}$

= 8

Question 28

If 5 cos θ = 3, evaluate : ${\displaystyle \frac{co\sec \theta –\cot \theta }{co\sec \theta +\cot \theta }}$

SOl:

cos θ = ${\displaystyle \frac{3}{5}}$

Now

${\displaystyle \frac{co\sec \theta –\cot \theta }{co\sec \theta +\cot \theta }=\frac{\frac{1}{\sin \theta }–\frac{\cos \theta }{\sin \theta }}{\frac{1}{\sin \theta }+\frac{\cos \theta }{\sin \theta }}}$

= ${\displaystyle \frac{\frac{1–\cos \theta }{\sin \theta }}{\frac{1+\cos \theta }{\sin \theta }}}$

= ${\displaystyle \frac{1–\cos \theta }{1+\cos \theta }}$

= ${\displaystyle \frac{1–\frac{3}{5}}{1+\frac{3}{5}}}$

= ${\displaystyle \frac{\frac{2}{5}}{\frac{8}{5}}}$

= ${\displaystyle \frac{2}{8}}$

= ${\displaystyle \frac{1}{4}}$

Question 29

If cosec A + sin A = 5${\displaystyle \frac{1}{5}}$, find the value of cosec²A + sin²A.

Sol:

co secA + sin A = 5${\displaystyle \frac{1}{5}}$

Squaring both sides

${\displaystyle {(\cos ec\text{A}+\sin \text{A})}^2={\left(5\frac{1}{5}\right)}^2}$

${\displaystyle \cos e{c}^2\text{A}+{\sin }^2\text{A}+2\cos ecA.\frac{1}{\cos ec\text{A}}=\frac{26}{25}}$

${\displaystyle \cos e{c}^2\text{A}+{\sin }^2\text{A}=\frac{626}{25}}$

${\displaystyle \cos e{c}^2\text{A}+{\sin }^2\text{A}=25\frac{1}{25}}$

Question 30

If 5 cos  = 6 sin ; evaluate:
(i) tan θ
(ii) ${\displaystyle \frac{12\sin \theta –3\cos \theta }{12\sin \theta +3\cos \theta }}$

Sol:

5 cos θ = 6 sin θ

tan θ = ${\displaystyle \frac{5}{6}}$

Now
(i) tan θ = ${\displaystyle \frac{5}{6}}$

(ii) ${\displaystyle \frac{12\sin \theta –3\cos \theta }{12\sin \theta +3\cos \theta }=\frac{\frac{12\sin \theta }{\cos \theta }–\frac{3\cos \theta }{\cos \theta }}{\frac{12\sin \theta }{\cos \theta }+\frac{3\cos \theta }{\cos \theta }}}$ 

= ${\displaystyle \frac{12\tan \theta –3}{12\tan \theta +3}}$

= ${\displaystyle \frac{12\left(\frac{5}{6}\right)–3}{12\left(\frac{5}{6}\right)+3}}$

= ${\displaystyle \frac{\frac{42}{6}}{\frac{78}{6}}}$

= ${\displaystyle \frac{42}{78}}$

= ${\displaystyle \frac{7}{13}}$