Rs Aggarwal 2019 for Class 10 Math Chapter 11 - T Ratios Of Some Particular Angles

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Page No 552:

Answer:

On substituting the values of various T-ratios, we get:
sin 60^o cos 30^o + cos 60^o sin 30^o
$= (\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} + \frac{1}{2} \times \frac{1}{2}) = (\frac{3}{4} + \frac{1}{4}) = \frac{4}{4} = 1$

Page No 552:

Answer:

On substituting the values of various T-ratios, we get:
cos 60^o cos 30^o − sin 60^o sin 30^o
$= (\frac{1}{2} \times \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} \times \frac{1}{2}) = (\frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4}) = 0$

Page No 552:

Answer:

On substituting the values of various T-ratios, we get:
cos 45^o cos 30^o + sin 45^o sin 30^o
$= (\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \times \frac{1}{2}) = (\frac{\sqrt{3}}{2 \sqrt{2}} + \frac{1}{2 \sqrt{2}}) = (\frac{\sqrt{3} + 1}{2 \sqrt{2}})$

Page No 552:

Answer:

$\frac{\sin 30 °}{\cos 45 °} + \frac{\cot 45 °}{\sec 60 °} - \frac{\sin 60 °}{\tan 45 °} + \frac{\cos 30 °}{\sin 90 °} = \frac{(\frac{1}{2})}{(\frac{1}{\sqrt{2}})} + \frac{1}{2} - \frac{(\frac{\sqrt{3}}{2})}{1} + \frac{(\frac{\sqrt{3}}{2})}{1} = \frac{\sqrt{2}}{2} + \frac{1}{2} - \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} = \frac{\sqrt{2} + 1}{2}$

Page No 553:

Answer:

$\frac{5 \cos^{2} 60 ° + 4 \sec^{2} 30 ° - \tan^{2} 45 °}{\sin^{2} 30 ° + \cos^{2} 30 °} = \frac{5 {(\frac{1}{2})}^{2} + 4 {(\frac{2}{\sqrt{3}})}^{2} - {(1)}^{2}}{{(\frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} = \frac{(\frac{5}{4} + \frac{4 \times 4}{3} - 1)}{(\frac{1}{4} + \frac{3}{4})} = \frac{(\frac{5}{4} + \frac{16}{3} - \frac{1}{1})}{(\frac{4}{4})} = \frac{(\frac{15 + 64 - 12}{12})}{(\frac{4}{4})} = \frac{(\frac{67}{12})}{(1)} = \frac{67}{12}$

Page No 553:

Answer:

On substituting the values of various T-ratios, we get:
2 cos² 60^o + 3 sin² 45^o − 3 sin² 30^o + 2 cos² 90^o
$= 2 \times {(\frac{1}{2})}^{2} + 3 \times {(\frac{1}{\sqrt{2}})}^{2} - 3 \times {(\frac{1}{2})}^{2} + 2 \times {(0)}^{2} = 2 \times \frac{1}{4} + 3 \times \frac{1}{2} - 3 \times \frac{1}{4} + 0 = (\frac{1}{2} + \frac{3}{2} - \frac{3}{4}) = (\frac{2 + 6 - 3}{4}) = \frac{5}{4}$

Page No 553:

Answer:

On substituting the values of various T-ratios, we get:
cot² 30^o − 2 cos² 30^o − $\frac{3}{4}$ sec² 45^o + $\frac{1}{4}$ cosec² 30^o
$= {(\sqrt{3})}^{2} - 2 \times {(\frac{\sqrt{3}}{2})}^{2} - \frac{3}{4} \times {(\sqrt{2})}^{2} + \frac{1}{4} \times {(2)}^{2} = 3 - 2 \times \frac{3}{4} - \frac{3}{4} \times 2 + \frac{1}{4} \times 4 = 3 - \frac{3}{2} - \frac{3}{2} + 1 = 4 - (\frac{3}{2} + \frac{3}{2}) = 4 - 3 = 1$

Page No 553:

Answer:

On substituting the values of various T-ratios, we get:
(sin² 30^o + 4 cot² 45^o − sec² 60^o )(cosec² 45^osec² 30^o)
$= [{(\frac{1}{2})}^{2} + 4 \times {(1)}^{2} - {(2)}^{2}] [{(\sqrt{2})}^{2} {(\frac{2}{\sqrt{3}})}^{2}] = (\frac{1}{4} + 4 - 4) (2 \times \frac{4}{3}) = \frac{1}{4} \times \frac{8}{3} = \frac{2}{3}$

Page No 553:

Answer:

On substituting the values of various T-ratios, we get:

$\frac{4}{\cot^{2} 30^{o}} + \frac{1}{\sin^{2} 30^{o}} - 2 \cos^{2} 45^{o} - \sin^{2} 0^{o} = \frac{4}{{(\sqrt{3})}^{2}} + \frac{1}{{(\frac{1}{2})}^{2}} - 2 \times {(\frac{1}{\sqrt{2}})}^{2} - {(0)}^{2} = \frac{4}{3} + \frac{1}{\frac{1}{4}} - 2 \times \frac{1}{2} - 0 = \frac{4}{3} + 4 - 1 = \frac{4}{3} + 3 = \frac{4 + 9}{3} = \frac{13}{3}$

Page No 553:

Answer:

(i)
$LHS = \frac{1 - \sin 60^{o}}{\cos 60^{o}} = \frac{1 - \frac{\sqrt{3}}{2}}{\frac{1}{2}} = \frac{(\frac{2 - \sqrt{3}}{2})}{\frac{1}{2}} = (\frac{2 - \sqrt{3}}{2}) \times 2 = 2 - \sqrt{3} RHS = \frac{\tan 60^{o} - 1}{\tan 60^{o} + 1} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1} = \frac{{(\sqrt{3} - 1)}^{2}}{{(\sqrt{3})}^{2} - 1^{2}} = \frac{3 + 1 - 2 \sqrt{3}}{3 - 1} = \frac{4 - 2 \sqrt{3}}{2} = 2 - \sqrt{3}$

Hence, LHS = RHS

∴ $\frac{1 - \sin 60^{o}}{\cos 60^{o}} = \frac{\tan 60^{o} - 1}{\tan 60^{o} + 1}$

(ii)
$LHS = \frac{\cos 30^{o} + \sin 60^{o}}{1 + \sin 30^{o} + \cos 60^{o}} = \frac{(\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2})}{(1 + \frac{1}{2} + \frac{1}{2})} = \frac{\frac{\sqrt{3} + \sqrt{3}}{2}}{\frac{2 + 1 + 1}{2}} = \frac{\sqrt{3}}{2} Also, RHS = \cos 30^{o} = \frac{\sqrt{3}}{2}$

Hence, LHS = RHS

∴ $\frac{\cos 30^{o} + \sin 60^{o}}{1 + \sin 30^{o} + \cos 60^{o}} = \cos 30^{o}$ 1−sin60°cos60°

Page No 553:

Answer:

(i) sin 60^o cos 30^o − cos 60^o sin 30^o
$= (\frac{\sqrt{3}}{2}) \times (\frac{\sqrt{3}}{2}) - (\frac{1}{2}) \times (\frac{1}{2}) = \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2} Also, \sin 30^{o} = \frac{1}{2}$
∴ sin 60^o cos 30^o − cos 60^o sin 30^o = sin 30^o

(ii) cos 60^o cos 30^o + sin 60^o sin 30^o

$= (\frac{1}{2}) \times (\frac{\sqrt{3}}{2}) + (\frac{\sqrt{3}}{2}) \times (\frac{1}{2}) = \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{2} Also, \cos 30^{o} = \frac{\sqrt{3}}{2}$
∴ cos 60^o cos 30^o + sin 60^o sin 30^o = cos 30^o

(iii) 2 sin 30^o cos 30^o
$= 2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} Also, \sin 60^{o} = \frac{\sqrt{3}}{2}$
∴ 2 sin 30^o cos 30^o = sin 60^o

(iv) 2 sin 45^o cos 45^o
$= 2 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = 1$
Also, sin 90^o = 1
∴ 2 sin 45^o cos 45^o = sin 90^o

Page No 553:

Answer:

A = 45^o
⇒ 2A = 2 $\times$ 45^o = 90^o

(i) sin 2A = sin 90^o = 1
2 sin A cos A = 2 sin 45^o cos 45^o = $2 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = 1$
∴ sin 2A = 2 sin A cos A

(ii) cos 2A = cos 90^o = 0
2 cos² A − 1 = 2 cos² 45^o − 1 = $2 \times {(\frac{1}{\sqrt{2}})}^{2} - 1 = 2 \times \frac{1}{2} - 1 = 1 - 1 = 0$
Now, 1 − 2 sin² A = $1 - 2 \times {(\frac{1}{\sqrt{2}})}^{2} = 1 - 2 \times \frac{1}{2} = 1 - 1 = 0$
∴ cos 2A = 2 cos² A − 1 = 1 − 2 sin² A

Page No 553:

Answer:

A = 30^o
⇒ 2A = 2 $\times$ 30^o = 60^o

(i) sin 2A = sin 60^o = $\frac{\sqrt{3}}{2}$
$\frac{2 \tan A}{1 + \tan^{2} A} = \frac{2 \tan 30^{o}}{1 + \tan^{2} 30^{o}} = \frac{2 \times (\frac{1}{\sqrt{3}})}{1 + {(\frac{1}{\sqrt{3}})}^{2}} = \frac{(\frac{2}{\sqrt{3}})}{1 + \frac{1}{3}} = \frac{(\frac{2}{\sqrt{3}})}{\frac{4}{3}} = (\frac{2}{\sqrt{3}}) \times \frac{3}{4} = \frac{\sqrt{3}}{2}$
∴ $\sin 2 A = \frac{2 \tan A}{1 + \tan^{2} A}$

(ii) cos 2A = cos 60^o = $\frac{1}{2}$

\frac{1 - \tan^{2} A}{1 + \tan^{2} A} = \frac{1 - \tan^{2} 30^{o}}{1 + \tan^{2} 30^{o}} = \frac{1 - {(\frac{1}{\sqrt{3}})}^{2}}{1 + {(\frac{1}{\sqrt{3}})}^{2}} = \frac{(1 - \frac{1}{3})}{(1 + \frac{1}{3})} = \frac{(\frac{2}{3})}{\frac{4}{3}} = (\frac{2}{3}) \times \frac{3}{4} = \frac{1}{2}

∴

\cos 2 A = \frac{1 - \tan^{2} A}{1 + \tan^{2} A}

(iii) tan 2A = tan 60^o =

\sqrt{3}

\frac{2 \tan A}{1 - \tan^{2} A} = \frac{2 \tan 30^{o}}{1 - \tan^{2} 30^{o}} = \frac{2 \times (\frac{1}{\sqrt{3}})}{1 - {(\frac{1}{\sqrt{3}})}^{2}} = \frac{(\frac{2}{\sqrt{3}})}{1 - \frac{1}{3}} = \frac{(\frac{2}{\sqrt{3}})}{\frac{2}{3}} = (\frac{2}{\sqrt{3}}) \times \frac{3}{2} = \sqrt{3}

∴

\tan 2 A = \frac{2 \tan A}{1 - \tan^{2} A}

=2tanA1+tan2A

Page No 553:

Answer:

A = 60^o and B = 30^o
Now, A + B = 60^o + 30^o = 90^o
Also, A − B = 60^o − 30^o = 30^o

(i) sin (A + B) = sin 90^o = 1
sin A cos B + cos A sin B = sin 60^o cos 30^o + cos 60^o sin 30^o
= $(\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} + \frac{1}{2} \times \frac{1}{2}) = (\frac{3}{4} + \frac{1}{4}) = 1$
∴ sin (A + B) = sin A cos B + cos A sin B

(ii) cos (A + B) = cos 90^o = 0
cos A cos B − sin A sin B =cos 60^o cos 30^o − sin 60^o sin 30^o $= (\frac{1}{2} \times \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} \times \frac{1}{2}) = (\frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4}) = 0$
∴ cos (A + B) = cos A cos B − sin A sin B

Page No 553:

Answer:

(i) sin (A − B) = sin 30^o = $\frac{1}{2}$
sin A cos B − cos A sin B = sin 60^o cos 30^o − cos 60^o sin 30^o
= $(\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} - \frac{1}{2} \times \frac{1}{2}) = (\frac{3}{4} - \frac{1}{4}) = \frac{2}{4} = \frac{1}{2}$
∴ sin (A − B) = sin A cos B − cos A sin B

(ii) cos (A − B) = cos 30^o = $\frac{\sqrt{3}}{2}$
cos A cos B + sin A sin B = cos 60^o cos 30^o + sin 60^o sin 30^o
= $(\frac{1}{2} \times \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \times \frac{1}{2}) = (\frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4}) = 2 \times \frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{2}$
∴ cos (A − B) = cos A cos B + sin A sin B

(iii) tan (A − B) = tan 30^o = $\frac{1}{\sqrt{3}}$
$\frac{\tan A - \tan B}{1 + \tan A \tan B} = \frac{\tan 60^{o} - \tan 30^{o}}{1 + \tan 60^{o} \tan 30^{o}} = \frac{\sqrt{3} - \frac{1}{\sqrt{3}}}{1 + (\sqrt{3} \times \frac{1}{\sqrt{3}})} = \frac{1}{2} \times (\frac{3 - 1}{\sqrt{3}}) = \frac{1}{\sqrt{3}}$
∴ tan (A − B) = $\frac{\tan A - \tan B}{1 + \tan A \tan B}$

Page No 553:

Answer:

Given:
$\tan A = \frac{1}{3} and \tan B = \frac{1}{2} \tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} On substituting these values in RHS of the expression, we get : \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{(\frac{1}{3} + \frac{1}{2})}{(1 - \frac{1}{3} \times \frac{1}{2})} = \frac{(\frac{5}{6})}{(1 - \frac{1}{6})} = \frac{(\frac{5}{6})}{(\frac{5}{6})} = 1 \Rightarrow \tan (A + B) = 1 = \tan 45^{o} [∵ \tan 45^{o} = 1] ∴ A + B = 45^{o}$

\tan A = \frac{1}{3} a n d \tan B = \frac{1}{2} By substituting the values, we get; \tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{(\frac{1}{3} + \frac{1}{2})}{(1 - \frac{1}{3} \times \frac{1}{2})} = \frac{(\frac{5}{6})}{(1 - \frac{1}{6})} = \frac{(\frac{5}{6})}{(\frac{5}{6})} = 1 \tan (A + B) = 1 = \tan 45^{o} Hence, A + B = 45

Page No 553:

Answer:

A = 30^o
⇒ 2A = 2 $\times$ 30^o = 60^o

By substituting the value of the given T-ratio, we get:
$\tan 2 A = \frac{2 \tan A}{1 - \tan^{2} A} \Rightarrow \tan 60^{o} = \frac{2 \tan 30^{o}}{1 - \tan^{2} 30^{o}} = \frac{2 \times \frac{1}{\sqrt{3}}}{1 - {(\frac{1}{\sqrt{3}})}^{2}} = \frac{\frac{2}{\sqrt{3}}}{1 - \frac{1}{3}} = \frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}} = \frac{2}{\sqrt{3}} \times \frac{3}{2} = \sqrt{3}$
∴ tan 60^o = $\sqrt{3}$

Page No 553:

Answer:

A = 30^o
⇒ 2A = 2 $\times$ 30^o = 60^o

By substituting the value of the given T-ratio, we get:
$\cos A = \sqrt{\frac{1 + \cos 2 A}{2}} \Rightarrow \cos 30^{o} = \sqrt{\frac{1 + \cos 60^{o}}{2}} = \sqrt{\frac{1 + \frac{1}{2}}{2}} = \sqrt{\frac{\frac{3}{2}}{2}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$
∴ cos 30^o = $\frac{\sqrt{3}}{2}$

Page No 554:

Answer:

A = 30^o
⇒ 2A = 2 $\times$ 30^o = 60^o

By substituting the value of the given T-ratio, we get:

$\sin A = \sqrt{\frac{1 - \cos 2 A}{2}} \Rightarrow \sin 30^{o} = \sqrt{\frac{1 - \cos 60^{o}}{2}} = \sqrt{\frac{1 - \frac{1}{2}}{2}} = \sqrt{\frac{\frac{1}{2}}{2}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$
∴ sin 30^o = $\frac{1}{2}$

Page No 554:

Answer:

From the given right-angled triangle, we have:
$\frac{B C}{A C} = \sin 30^{o} \Rightarrow \frac{B C}{20} = \frac{1}{2} \Rightarrow B C = \frac{20}{2} = 10 cm Also, \frac{A B}{A C} = \cos 30^{o} \Rightarrow \frac{A B}{20} = \frac{\sqrt{3}}{2} \Rightarrow A B = (20 \times \frac{\sqrt{3}}{2}) = 10 \sqrt{3} cm ∴ BC = 10 cm and AB = 10 \sqrt{3} cm$

Page No 554:

Answer:

From the given right-angled triangle, we have:

$\frac{B C}{A B} = \tan 30^{o} \Rightarrow \frac{6}{A B} = \frac{1}{\sqrt{3}} \Rightarrow A B = 6 \sqrt{3} cm Also, \frac{B C}{A C} = \sin 30^{o} \Rightarrow \frac{6}{A C} = \frac{1}{2} \Rightarrow A C = (2 \times 6) = 12 cm ∴ A B = 6 \sqrt{3} cm and A C = 12 cm$

Page No 554:

Answer:

From right-angled ∆ABC, we have:

$\frac{B C}{A C} = \sin 45^{o} \Rightarrow \frac{B C}{3 \sqrt{2}} = \frac{1}{\sqrt{2}} \Rightarrow B C = 3 cm Also, \frac{A B}{A C} = \cos 45^{o} \Rightarrow \frac{A B}{3 \sqrt{2}} = \frac{1}{\sqrt{2}} \Rightarrow A B = 3 cm ∴ B C = 3 cm and A B = 3 cm$

Page No 554:

Answer:

Here, sin (A + B) = 1
⇒ sin (A+ B) = sin 90^o [∵ sin 90^o = 1]
⇒ A + B = 90^o                       ...(i)

Also, cos (A − B) = 1
⇒ cos (A − B) = cos 0^o    [∵ cos 0^o = 1]
⇒ A − B = 0^o                      ...(ii)

Solving (i) and (ii), we get:
A = 45^o and B = 45^o

Page No 554:

Answer:

Here, sin (A − B) = $\frac{1}{2}$
⇒ sin (A − B) = sin 30^o      [∵ sin 30^o = $\frac{1}{2}$ ]
⇒ A − B = 30^o                                  ...(i)

Also, cos (A + B) = $\frac{1}{2}$
⇒ cos (A + B) = cos 60^o        [∵ cos 60^o = $\frac{1}{2}$ ]
⇒ A + B = 60^o                                 ...(ii)

Solving (i) and (ii), we get:
A = 45^o and B = 15^o

Page No 554:

Answer:

Here, tan (A − B) = $\frac{1}{\sqrt{3}}$
⇒ tan (A − B) = tan 30^o   [∵ tan 30^o = $\frac{1}{\sqrt{3}}$ ]
⇒ A − B = 30^o                     ...(i)

Also, tan (A + B) = $\sqrt{3}$
⇒ tan (A + B) = tan 60^o      [∵ tan 60^o = $\sqrt{3}$ ]
⇒ A + B = 60^o...(ii)

Solving (i) and (ii), we get:
A = 45^o and B = 15^o

Page No 554:

Answer:

$3 (x^{2} - \frac{1}{x^{2}}) = \frac{9}{3} (x^{2} - \frac{1}{x^{2}}) = \frac{1}{3} (9 x^{2} - \frac{9}{x^{2}}) = \frac{1}{3} [{(3 x)}^{2} - {(\frac{3}{x})}^{2}]$
$= \frac{1}{3} [{(cosec θ)}^{2} - {(\cot θ)}^{2}] = \frac{1}{3} ({cosec}^{2} θ - \cot^{2} θ) = \frac{1}{3} (1) = \frac{1}{3}$

Page No 554:

Answer:

$Let A = 45 ° and B = 30 °$

$(i) As, \sin (A + B) = sinA cosB + cosA sinB \Rightarrow \sin (45 ° + 30 °) = \sin 45 ° \cos 30 ° + \cos 45 ° \sin 30 ° \Rightarrow \sin (75 °) = \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \times \frac{1}{2} \Rightarrow \sin 75 ° = \frac{\sqrt{3}}{2 \sqrt{2}} + \frac{1}{2 \sqrt{2}} ∴ \sin 75 ° = \frac{\sqrt{3} + 1}{2 \sqrt{2}}$

$(ii) As, \cos (A - B) = cosA cosB + sinA sinB \Rightarrow \cos (45 ° - 30 °) = \cos 45 ° \cos 30 ° + \sin 45 ° \sin 30 ° \Rightarrow \cos (15 °) = \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \times \frac{1}{2} \Rightarrow \cos 15 ° = \frac{\sqrt{3}}{2 \sqrt{2}} + \frac{1}{2 \sqrt{2}} ∴ \cos 15 ° = \frac{\sqrt{3} + 1}{2 \sqrt{2}}$

Disclaimer: $\cos 15 °$ can also be calculated by taking $A = 60 ° and B = 45 °$ .

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