# RD Sharma Solutions Class 10 Trigonometric Ratios Exercise 5.2

### Exercise 5.2

Evaluate each of the following:

Q 1 . sin $45^{\circ}$ sin $30^{\circ}$ + cos $45^{\circ}$ cos $30^{\circ}$

Solution:

Sin $45^{\circ}$sin $30^{\circ}$ + cos $45^{\circ}$ cos $30^{\circ}$

Value of trigonometric ratios are:

$sin 45^{\circ}= \frac{1}{\sqrt{2}}$ $sin 30^{\circ}= \frac{1}{2}$

$cos 45^{\circ}= \frac{1}{\sqrt{2}}$ $cos 30^{\circ}= \frac{\sqrt{3}}{2}$

Substituting in the given equation, we get

$\frac{1}{\sqrt{2}}\cdot \frac{1}{2}+\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2}$

= $\frac{1}{\sqrt{2}}\cdot \frac{\sqrt{3}}{2\sqrt{2}}$

= $\frac{\sqrt{3}+1}{2\sqrt{2}}$

Q 2 . sin $60^{\circ}$ cos $30^{\circ}$ + cos $60^{\circ}$ sin $30^{\circ}$

Solution:

sin $60^{\circ}$ cos $30^{\circ}$ + cos $60^{\circ}$ sin $30^{\circ}$

By trigonometric ratios we have ,

$sin 60^{\circ}= \frac{\sqrt{3}}{2}$ $sin 30^{\circ}= \frac{1}{2}$

$cos 30^{\circ}= \frac{\sqrt{3}}{2}$ $cos 60^{\circ}= \frac{1}{2}$

Substituting the values in given equation

= $\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{3}}{2}+\frac{1}{2}\cdot \frac{1}{2}$

= $\frac{3}{4}+\frac{1}{4}$ = $\frac{4}{4}$ = 1

Q 3 . cos $60^{\circ}$ cos $45^{\circ}$ – sin $60^{\circ}$ sin $45^{\circ}$

Solution:

cos $60^{\circ}$ cos $45^{\circ}$ – sin $60^{\circ}$ sin $45^{\circ}$

We know that by trigonometric ratios

$cos 60^{\circ}= \frac{1}{2}$ $cos 45^{\circ}= \frac{1}{\sqrt{2}}$

$sin 60^{\circ}= \frac{\sqrt{3}}{2}$ $sin 45^{\circ}= \frac{1}{\sqrt{2}}$

Substituting the values in given equation

$\frac{1}{2}\cdot \frac{1}{\sqrt{2}}-\frac{\sqrt{3}}{2}\cdot \frac{1}{\sqrt{2}}$

= $\frac{1-\sqrt{3}}{2\sqrt{2}}$

Q.4: $sin^{2}30^{\circ}+sin^{2}45^{\circ}+sin^{2}60^{\circ}+sin^{2}90^{\circ}$

Solution:

$sin^{2}30^{\circ}+sin^{2}45^{\circ}+sin^{2}60^{\circ}+sin^{2}90^{\circ}$

We know that by trigonometric ratios

$sin 30^{\circ}= \frac{1}{2}$ $sin 45^{\circ}= \frac{1}{\sqrt{2}}$

$sin 60^{\circ}= \frac{\sqrt{3}}{2}$ $sin90^{\circ}$ = 1

Substituting the values in given equation, we get

= $\left [ \frac{1}{2} \right ]^{2}+\left [ \frac{1}{\sqrt{2}} \right ]^{2}+\left [ \frac{\sqrt{3}}{2} \right ]^{2}+1$

= $\frac{1}{4}+\frac{1}{2}+\frac{3}{4}+1$

= $\frac{5}{2}$

Q 5. $cos^{2}30^{\circ}+cos^{2}45^{\circ}+cos^{2}60^{\circ}+cos^{2}90^{\circ}$

Solution:

$cos^{2}30^{\circ}+cos^{2}45^{\circ}+cos^{2}60^{\circ}+cos^{2}90^{\circ}$

We know that by trigonometric ratios

$cos 30^{\circ}= \frac{\sqrt{3}}{2}$ $cos 45^{\circ}= \frac{1}{\sqrt{2}}$

$cos 60^{\circ}= \frac{1}{2}$ $cos90^{\circ}$ = 0

Substituting the values in given equation

$\left [ \frac{\sqrt{3}}{2} \right ]^{2}+\left [ \frac{1}{\sqrt{2}} \right ]^{2}+\left [ \frac{1}{2} \right ]^{2}+0$

= $\frac{3}{4}+\frac{1}{2}+\frac{1}{4}$

= $\frac{3}{2}$

Q 6 . $tan^{2}30^{\circ}+tan^{2}45^{\circ}+tan^{2}60^{\circ}$

Solution:

$tan^{2}30^{\circ}+tan^{2}45^{\circ}+tan^{2}60^{\circ}$

We know that by trigonometric ratios

$tan30^{\circ}=\frac{1}{\sqrt{3}}$ $tan60^{\circ}=\sqrt{3}$

$tan45^{\circ}=1$

Substituting the values in given equation

$\left [ \frac{1}{\sqrt{3}} \right ]^{2}+\left [ \sqrt{3} \right ]^{2}+1$

= $\frac{1}{3} +3+1$

= $\frac{13}{3}$

Q 7 . $2sin^{2}30^{\circ}-3cos^{2}45^{\circ}+tan^{2}60^{\circ}$

Solution:

$2sin^{2}30^{\circ}-3cos^{2}45^{\circ}+tan^{2}60^{\circ}$

We know that by trigonometric ratios

$sin 30^{\circ}= \frac{1}{2}$ $cos 45^{\circ}= \frac{1}{\sqrt{2}}$

$tan60^{\circ}=\sqrt{3}$

Substituting the values in given equation

= $2\left ( \frac{1}{2} \right )^{2}-3\left ( \frac{1}{\sqrt{2}} \right )^{2}+\left ( \sqrt{3} \right )^{2}$

= $2\left ( \frac{1}{4} \right )-3\left ( \frac{1}{2} \right )+3$

= $\frac{1-3+6}{2}$

= 2

Q8:$sin^{2}30^{\circ}cos^{2}45^{\circ}+4tan^{2}30^{\circ}+\frac{1}{2}sin^{2}90^{\circ}-2cos^{2}90^{\circ}+\frac{1}{24}cos^{2}0^{\circ}$

Solution:

$sin^{2}30^{\circ}cos^{2}45^{\circ}+4tan^{2}30^{\circ}+\frac{1}{2}sin^{2}90^{\circ}-2cos^{2}90^{\circ}+\frac{1}{24}cos^{2}0^{\circ}$ [1]

We know that by trigonometric ratios

$sin 30^{\circ}= \frac{1}{2}$

$cos 45^{\circ}= \frac{1}{\sqrt{2}}$

$tan30^{\circ}=\frac{1}{\sqrt{3}}$

$sin90^{\circ}$ = 1

$cos90^{\circ}$ = 0

$cos0^{\circ}$ = 1

Substituting the values in given equation

$\left [ \frac{1}{2} \right ]^{2}\cdot \left [ \frac{1}{\sqrt{2}} \right ]^{2}+4\left [ \frac{1}{\sqrt{3}} \right ]^{2}+\frac{1}{2}\left [ 1 \right ]^{2}-2\left [ 0 \right ]^{2}+\frac{1}{24}\left [ 1 \right ]^{2}$

= $\frac{1}{8}+\frac{4}{3}+\frac{1}{2}+\frac{1}{24}$

= $\frac{48}{24}$

= 2

Q 9 . $4\left ( sin^{4} 60^{\circ}+cos^{4} 30^{\circ}\right )-3\left ( tan^{2}60^{\circ}-tan^{2}45^{\circ} \right )+5cos^{2}45^{\circ}$

Solution:

$4\left ( sin^{4} 60^{\circ}+cos^{4} 30^{\circ}\right )-3\left ( tan^{2}60^{\circ}-tan^{2}45^{\circ} \right )+5cos^{2}45^{\circ}$

We know that by trigonometric ratios we have ,

$sin 60^{\circ}= \frac{\sqrt{3}}{2}$ $cos 45^{\circ}= \frac{1}{\sqrt{2}}$

$tan60^{\circ}=\sqrt{3}$ $cos 30^{\circ}= \frac{\sqrt{3}}{2}$

Substituting the values in given equation

= $4\cdot \frac{18}{16}-6+\frac{5}{2}$

= $\frac{1}{4}-6+\frac{5}{2}$

= $\frac{14}{2}-6$ = 7 – 6 = 1

Q 10 . $\left (cosec^{2}45^{\circ}sec^{2}30^{\circ} \right )\left ( sin^{2}30^{\circ} +4cot^{2}45^{\circ}-sec^{2}60^{\circ}\right )$

Solution:

$\left (cosec^{2}45^{\circ}sec^{2}30^{\circ} \right )\left ( sin^{2}30^{\circ} +4cot^{2}45^{\circ}-sec^{2}60^{\circ}\right )$

We know that by trigonometric ratios,

$cosec45^\circ=\sqrt{2}$ $sec30^\circ=\frac{2}{\sqrt{3}}$

$sin 30^{\circ}= \frac{1}{2}$ $cot45^{\circ}$ = 1

$sec60^{\circ}$ = 2

Substituting the values in given equation

$\left ( \left [ \sqrt{2} \right ]^{2}.\left [ \frac{2}{\sqrt{3}} \right ]^{2} \right )\left ( \left [ \frac{1}{2} \right ]^{2} +4\left ( 1 \right )\left ( 2 \right )^{2}\right )$

= $3\cdot \frac{4}{3}\cdot \frac{1}{4}$

= $\frac{2}{3}$

Q11. $cosec^{3}30^{\circ} cos 60^{\circ} tan^{3}45^{\circ} sin^{2}90^{\circ} sec^{2}45^{\circ} cot 30^{\circ}$

Solution:

$cosec^{3}30^{\circ} cos 60^{\circ} tan^{3}45^{\circ} sin^{2}90^{\circ} sec^{2}45^{\circ} cot 30^{\circ}$

Using trigonometric values, we have

= $(2)^{3}\times (\frac{1}{2})\times (1^{3})\times (1^{2})\times (\sqrt{2}^{2}) \times (\sqrt{3})$

= $8\times (\frac{1}{2})\times (1)\times (1)\times (2) \times (\sqrt{3})$

= $8\sqrt{3}$

Q12. $cot^{2}30^{\circ}- 2cos^{2}60^{\circ}-\frac{3}{4}sec^{2}45^{\circ} – 4sec^{2}30^{\circ}$

Solution:

$cot^{2}30^{\circ}- 2cos^{2}60^{\circ}-\frac{3}{4}sec^{2}45^{\circ} – 4sec^{2}30^{\circ}$

= $(\sqrt{3}^{2})\times 2(\frac{1}{2})^{2}\times (\frac{3}{4}\times \sqrt{2}^{2})\times (4\times (\frac{2}{\sqrt{3}})^{2})$

= $3-\frac{1}{2}-\frac{3}{2}-\frac{16}{3}$

= $\frac{-13}{3}$

Q 13. $(cos 0^{\circ}+sin 45^{\circ}+sin 30^{\circ})(sin 90^{\circ}-cos 45^{\circ}+cos 60^{\circ})$

Solution:

$(cos 0^{\circ}+sin 45^{\circ}+sin 30^{\circ})(sin 90^{\circ}-cos 45^{\circ}+cos 60^{\circ})\\ (1 + \frac{1}{\sqrt{2}} + \frac{1}{2})(1 – \frac{1}{\sqrt{2}} + \frac{1}{2})\\ (\frac{3}{2}+\frac{1}{\sqrt{2}})(\frac{3}{2}-\frac{1}{\sqrt{2}})\\ ((\frac{3}{2})^{2}-(\frac{1}{\sqrt{2}})^{2}) = \frac{9}{4}-\frac{1}{2} = \frac{7}{4}$

Q14. $\frac{sin 30^{\circ}-sin 90^{\circ} + 2 cos 0^{\circ}}{tan 30^{\circ} tan 60^{\circ}}$

Solution:

$\frac{sin 30^{\circ}-sin 90^{\circ} + 2 cos 0^{\circ}}{tan 30^{\circ} tan 60^{\circ}}\\ \frac{\frac{1}{2}-1+2}{\frac{1}{\sqrt{3}}\times\sqrt{3}}\\ = \frac{3}{2}$

Q15. $\frac{4}{cot^{2}30^{\circ}}+\frac{1}{sin^{2}60^{\circ}}-cos^{2}45^{\circ}$

Solution:

$\frac{4}{cot^{2}30^{\circ}}+\frac{1}{sin^{2}60^{\circ}}-cos^{2}45^{\circ}\\ = \frac{4}{(\sqrt{3})^{2}} + \frac{1}{(\frac{\sqrt{3}}{2})^{2}} – (\frac{1}{\sqrt{2}})^{2}\\ = \frac{4}{3} + \frac{4}{3} – \frac{1}{2}\\ = \frac{16 – 3}{6}\\ = \frac{13}{6}$

Q16. $4(sin^{4}30^{\circ}+cos^{2}60^{\circ})-3(cos^{2}45^{\circ}-sin^{2}90^{\circ})-sin^{2}60^{\circ}$

Solution:

$4(sin^{4}30^{\circ}+cos^{2}60^{\circ})-3(cos^{2}45^{\circ}-sin^{2}90^{\circ})-sin^{2}60^{\circ}\\ = 4((\frac{1}{2})^{4}+(\frac{1}{2})^{2})-3((\frac{1}{\sqrt{2}})^{2}-1)-(\frac{\sqrt{3}}{2})^{2}\\ = 4(\frac{1}{16}+\frac{1}{4})+\frac{3}{2}-\frac{3}{4}\\ = \frac{8}{4} = 2$

Q17. $\frac{tan^{2}60^{\circ} + 4cos^{2}45^{\circ} + 3sec^{2}30^{\circ} + 5cos^{2}90^{\circ}}{cosec 30^{\circ}+ sec 60^{\circ}-cot^{2}30^{\circ}}$

Solution:

$\frac{tan^{2}60^{\circ} + 4cos^{2}45^{\circ} + 3sec^{2}30^{\circ} + 5cos^{2}90^{\circ}}{cosec 30^{\circ}+ sec 60^{\circ}-cot^{2}30^{\circ}}\\ = \frac{(\sqrt{3})^{2}+4(\frac{1}{\sqrt{2}})^{2}+3(\frac{2}{\sqrt{3}})^{2}+5(0)}{2 + 2 – (\sqrt{3})^{2}}\\ = 3 + 2 + 4\\ = 9$

Q18. $\frac{sin 30^{\circ}}{sin 45^{\circ}} + \frac{tan 45^{\circ}}{sec 60^{\circ}} – \frac{sin 60^{\circ}}{cot 45^{\circ}} – \frac{cos 30^{\circ}}{sin 90^{\circ}}$

Solution:

Given,

$\frac{sin 30^{\circ}}{sin 45^{\circ}} + \frac{tan 45^{\circ}}{sec 60^{\circ}} – \frac{sin 60^{\circ}}{cot 45^{\circ}} – \frac{cos 30^{\circ}}{sin 90^{\circ}}\\ = \frac{\frac{1}{2}}{\frac{1}{\sqrt{2}}} + \frac{1}{2} – \frac{\frac{\sqrt{3}}{2}}{1} – \frac{\frac{\sqrt{3}}{2}}{1}\\ = \frac{\sqrt{2}}{2} + \frac{1}{2} – \frac{\sqrt{3}}{2} – \frac{\sqrt{3}}{2}\\ = \frac{\sqrt{2} + 1 – 2\sqrt{3}}{2}$

Q19. $\frac{tan 45^{\circ}}{cosec 30^{\circ}} + \frac{sec 60^{\circ}}{cot 45^{\circ}} + \frac{s sin 90^{\circ}}{2 cos 0^{\circ}}$

Solution:

Given,

$\frac{tan 45^{\circ}}{cosec 30^{\circ}} + \frac{sec 60^{\circ}}{cot 45^{\circ}} + \frac{s sin 90^{\circ}}{2 cos 0^{\circ}}\\ = \frac{1}{2} + \frac{2}{1} – \frac{5(1)}{2(1)}\\ = \frac{5}{2} – \frac{5}{2}\\ = 0$

Q20. $2 sin 3x = \sqrt{3}$

Solution:

Given,

$2 sin 3x = \sqrt{3}\\ => sin 3x = \frac{\sqrt{3}}{2}\\ => sin 3x = sin 60^{\circ}\\ => 3x = 60^{\circ}\\ => x = 20^{\circ}$

Q21) $2sin \frac{x}{2}=1,\;x=?$

Solution:

$sin \frac{x}{2}=\frac{1}{2}$

$sin \frac{x}{2}=sin30^{0}$

$\frac{x}{2}=30^{0}$

x = 600

Q22) $\sqrt{3}sin x=cos x$

Solution:

$\sqrt{3}\;tan x=1$

$tan x=\frac{1}{\sqrt{3}}$

$∴ tanx=tan45^{0}$

x = 450

Q23) Tan x = sin 450 cos 450 + sin 300

Solution:

$Tan\ x=\frac{1}{\sqrt{2}}.\frac{1}{\sqrt{2}}+\frac{1}{2}\;\;\;\;\;[∵ sin 45^{0}=\frac{1}{\sqrt{2}}\;cos45^{0}=\frac{1}{\sqrt{2}}\;sin30^{0}=\frac{1}{2}]$

$Tan\ x=\frac{1}{2}+\frac{1}{2}$

Tan x = 1

Tan x = 450

x = 450

Q24) $\sqrt{3}\;Tan2x=cos60^{0}+sin45^{0}cos45^{0}$

Solution:

$\sqrt{3}\;Tan2x=\frac{1}{2}+\frac{1}{\sqrt{2}}.\frac{1}{\sqrt{2}}\;\;\;\;\;\;[∵ cos60^{0}=\frac{1}{2}\;and\ sin45^{0}=cos45^{0}=\frac{1}{\sqrt{2}}]$

$\sqrt{3}\;Tan2x=\frac{1}{\sqrt{3}}\Rightarrow tan2x=tan30^{0}$

2x = 300

x = 150

Q25) $cos2x=cos60^{0}\;cos30^{0}+sin60^{0}sin30^{0}$

Solution:

$cos2x=\frac{1}{2}.\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}.\frac{1}{2}\;\;\;\;\;\;[∵ cos60^{0}=sin30^{0}=\frac{1}{2}\ and\ sin60^{0}=cos30^{0}=\frac{\sqrt{3}}{2}]$

$cos2x=2.\frac{\sqrt{3}}{4}$

$cos2x=\frac{\sqrt{3}}{2}$

$cos2x=cos30^{0}$

$2x=30^{0}$

$x=15^{0}$

Q26) $If\;\theta=30^{0},\;verify\\\\ (i)Tan2\theta=\frac{2Tan\theta}{1-tan^{2}\theta}$

Solution:

$Tan2\theta=\frac{2Tan\theta}{1-tan^{2}\theta}…..(i)$

Substitute $\theta=30^{0}$ in equation (i)

LHS = Tan 600 = $\sqrt{3}$

RHS = $\frac{2Tan 30^{0}}{1+(Tan 30^{0})^{2}}=\frac{2-\frac{1}{\sqrt{2}}}{1-(\frac{1}{\sqrt{2}})^{2}}=\sqrt{3}$

Therefore, LHS = RHS

(ii) $sin\theta=\frac{2tan\theta}{1-tan^{2}\theta}$

Substitute $\theta=30^{0}$

$sin60^{0}=\frac{2tan30^{0}}{(1-tan30^{0})^{2}}$

=>$\frac{\sqrt{3}}{2}=\frac{2.\frac{1}{\sqrt{2}}}{1+(\frac{1}{\sqrt{2}})^{2}}$

=> $\frac{\sqrt{3}}{2}=\frac{2}{\sqrt{3}}.\frac{3}{4}\\ \Rightarrow \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}$

Therefore, LHS = RHS.

(iii) $cos2\theta=\frac{1-tan^{2}\theta}{1+tan^{2}\theta}$

Substitute $\theta=30^{0}$

LHS = cos 2(300)

cos 600 = $\frac{1}{2}$

RHS = $\frac{1-tan^{2}\theta}{1+tan^{2}\theta}$

$=\frac{1-tan^{2}30^{0}}{1+tan^{2}30^{0}}$

$= \frac{1-(\frac{1}{\sqrt{2}})^{2}}{1+(\frac{1}{\sqrt{2}})^{2}}=\frac{\frac{2}{2}}{\frac{1}{2}}=\frac{1}{2}$

Therefore, LHS = RHS

(iv) $cos3\theta=4cos^{3}\theta-3cos\theta$

Solution:

LHS = $cos3\theta$

Substitute $\theta=30^{0}$

= cos 3 (300) = cos 900

= 0

RHS = $4cos^{3}\theta-3cos\theta$

= $4cos^{3}30^{0}-3cos30^{0}$

= $4(\frac{\sqrt{3}}{2})^{3}-3.\frac{\sqrt{3}}{2}$

= $3.\frac{\sqrt{3}}{2}-3.\frac{\sqrt{3}}{2}$

= 0

Therefore, LHS = RHS.

Q27) If A = B = 600. Verify (i) Cos (A – B) = Cos A Cos B + Sin A Sin B

Solution:

Cos (A – B) = Cos A Cos B + Sin A Sin B…….(i)

Substitute A and B in (i)

=>cos (600 – 600) = cos 600 cos 600 + sin 600 sin 600

=>cos 00 = $(\frac{1}{2})^{2}+(\frac{\sqrt{3}}{2})^{2}$

=>1 = $\frac{1}{4}+\frac{3}{4}$

=>1 = 1

Therefore, LHS = RHS

(ii) Substitute A and B in (i)

=>sin (600 – 600) = sin 600 cos 600 – cos 600 sin 600

=> sin 00 = 0

=>0 = 0

Therefore, LHS = RHS

(iii) $Tan (A-B)=\frac{TanA-TanB}{1+TanA\;TanB}$

A = 600, B = 600 we get,

$Tan (60^{0}-60^{0})=\frac{Tan60^{0}-Tan60^{0}}{1+Tan60^{0}\;Tan60^{0}}$

Tan 00 = 0

0 = 0

Therefore, LHS = RHS

Q28 ) If A = 300, B = 600 verify:

(i) Sin (A + B) = Sin A Cos B + Cos A Sin B

Solution:

A = 300, B = 600 we get

Sin (300 + 600) = Sin 300 Cos 600 + Cos 300 Sin 600

Sin (900) = $\frac{1}{2}.\frac{1}{2}+\frac{\sqrt{3}}{2}.\frac{\sqrt{3}}{2}$

Sin (900) = 1 => 1 = 1

Therefore, LHS = RHS

(ii) Cos (A + B) = Cos A Cos B – Sin A Sin B

A = 300, B = 600 we get

Cos (300 + 600) = Cos 300 Cos 600 – Sin 300 Sin 600

Cos (900) = $\frac{1}{2}.\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}.\frac{1}{2}$

0 = 0

Therefore, LHS = RHS

Q29. If sin(A+B) = 1 and cos(A-B) = 1, $0^{\circ}< A+B\leq 90^{\circ}$, A≥B find A and B.

Sol:

sin(A+B) = 1 or sin(A+B) = $sin(90^{\circ})$ [As sin 90^o = 1]

cos(A-B) = 1 or cos(A-B) = $cos(0^{\circ})$ [As cos 0^o = 1]

=> A + B = $90^{\circ}$

A – B = $0^{\circ}$

2A = $90^{\circ}$

A = $\frac{90^{\circ}}{2}$

A = $45^{\circ}$

Substitute A value in A – B = $0^{\circ}$

$45^{\circ}$ – B = $0^{\circ}$

B =$45^{\circ}$

Hence, the value of A = $45^{\circ}$ and B =$45^{\circ}$

Q30. If tan(A-B) = $\frac{1}{\sqrt{3}}$ and tan(A+B) = $\sqrt{3}$, $0^{\circ}< A+B\leq 90^{\circ}$, A>B find A and B

Solution:

tan(A-B) = $\frac{1}{\sqrt{3}}$

A – B = $tan^{-1}(\frac{1}{\sqrt{3}})$

A – B = $30^{\circ}$ ——- 1

tan(A+B) = $\sqrt{3}$

A + B = $tan^{-1}{\sqrt{3}}$

A + B = $60^{\circ}$ ——- 2

Solve equations 1 and 2

A + B = $30^{\circ}$

A – B = $60^{\circ}$

2A = $90^{\circ}$

A = $\frac{90^{\circ}}{2}$

A = $45^{\circ}$

Substitute the value of A in equation 1

$45^{\circ}$ + B = $30^{\circ}$

B = $30^{\circ}$ – $45^{\circ}$

B = $15^{\circ}$

The value of A = $45^{\circ}$ and B = $15^{\circ}$

Q31. If sin(A-B) = $\frac{1}{2}$ and cos(A+B) = $\frac{1}{2}$, $0^{\circ}< A+B\leq 90^{\circ}$, A<B find A and B.

Solution:

sin(A-B) = $\frac{1}{2}$

A – B = $sin^{-1}(\frac{1}{2})$

A – B = $30^{\circ}$ ——- 1

cos(A+B) = $\frac{1}{2}$

A + B = $cos^{-1}(\frac{1}{2})$

A + B = $60^{\circ}$ ——- 2

Solve equations 1 and 2

A + B = $60^{\circ}$

A – B = $30^{\circ}$

2A = $90^{\circ}$

A = $\frac{90^{\circ}}{2}$

A = $45^{\circ}$

Substitute the value of A in equation 2

$45^{\circ}$ + B = $60^{\circ}$

B = $60^{\circ}$ – $45^{\circ}$

B = $15^{\circ}$

The value of A = $45^{\circ}$ and B = $15^{\circ}$

Q32. In a $\Delta$ ABC right angled triangle at B, $\angle A\;=\;\angle C$.Find the values of:

1. sinAcosC + cosAsinC

Solution:

since, it is given as $\angle A\;=\;\angle C$

the value of A and C is $45^{\circ}$, the value of angle B is $90^{\circ}$

because the sum of angles of triangle is $180^{\circ}$

=> sin($45^{\circ}$)cos($45^{\circ}$) + cos($45^{\circ}$)sin($45^{\circ}$)

=> $(\frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}})$ + $(\frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}})$

=> $\frac{1}{2}$ + $\frac{1}{2}$

=> 1

The value of sinAcosC + cosAsinC is 1

2. sinAsinB + cosAcosB

Solution:

since, it is given as $\angle A\;=\;\angle C$

the value of A and C is $45^{\circ}$, the value of angle B is $90^{\circ}$

because the sum of angles of triangle is $180^{\circ}$

=> sin($45^{\circ}$)sin($90^{\circ}$) + cos($45^{\circ}$)sin($90^{\circ}$)

=> $\frac{1}{\sqrt{2}}$(1) + $\frac{1}{\sqrt{2}}$(0)

=> $\frac{1}{\sqrt{2}}$ + 0

=> $\frac{1}{\sqrt{2}}$

The value of sinAsinB + cosAcosB is $\frac{1}{\sqrt{2}}$

Q33. Find the acute angle A and B, if sin(A+2B) = $\frac{\sqrt{3}}{2}$ and cos(A+4B) = 0, A>B.

Solution:

sin(A+2B) = $\frac{\sqrt{3}}{2}$

A + 2B = $sin^{-1}\frac{\sqrt{3}}{2}$

A + 2B = $60^{\circ}$ ………(1)

Again, Cos(A + 4B) = 0

A + 4B = $sin^{-1}(90)$

A + 4B = $90^{\circ}$ ………(2)

Solve equations (1) and (2)

A + 2B = $60^{\circ}$

A + 4B = $90^{\circ}$

(-)(-) (-)

-2B = –$30^{\circ}$

2B = $30^{\circ}$

B = $\frac{30^{\circ}}{2}$ = $15^{\circ}$

Substitute B value in equation (2)

A + 4B = $90^{\circ}$

A + 4($15^{\circ}$) = $90^{\circ}$

A + $60^{\circ}$ = $90^{\circ}$

A = $90^{\circ}$ – $60^{\circ}$

A = $30^{\circ}$

The value of A = $30^{\circ}$ and B = $15^{\circ}$

Q 34. In $\Delta PQR$, right angled at Q, PQ = 3 cm and PR = 6 cm. Determine $\angle$ P and $\angle$ R.

Solution:

In $\Delta PQR$, right angled at Q, PQ = 3 cm and PR = 6 cm

By Pythagoras theorem,

$PR^{2} = PQ^{2} + QR^{2}\\ => 6^{2} = 3^{2} + QR^{2}\\ => QR^{2} = 36 – 9\\ => QR = \sqrt{27}\\ => QR = 3\sqrt{3}$

sin R = $\frac{3}{6} = \frac{1}{2} = sin 30^{\circ}$

$\angle R = 30^{\circ}$

Sum of angles in a triangle = 180 degrees

$\angle P + \angle Q + \angle R = 180^{\circ} \\ => \angle P + 90^{\circ} + 30^{\circ} = 180^{\circ}\\ => \angle P = 180^{\circ} – 120^{\circ}\\ => \angle P = 60^{\circ}$

Therefore, $\angle R = 30^{\circ}$ and, $\angle P = 60^{\circ}$

Q35. If sin(A – B) = sin A cos B – cos A sin B and cos (A – B) = cos A cos B + sin A sin B, find the values of sin 15 and cos 15.

Solution:

sin(A – B) = sin A cos B – cos A sin B

and, cos (A – B) = cos A cos B + sin A sin B

To find: sin 15 and cos 15

Let A = 45 and B = 30

sin 15 can be written as sin (45 – 30)

So, sin 15 = sin (45 – 30) = sin 45 cos 30 – cos 45 sin 30

= $= (\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}) – (\frac{1}{\sqrt{2}}\times \frac{1}{2})\\ = \frac{\sqrt{3}-1}{2\sqrt{2}}$

cos 15 = cos (45- 30) = cos 45 cos 30 – sin 45 sin 30

= $= (\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}) + (\frac{1}{\sqrt{2}}\times \frac{1}{2})\\ = \frac{\sqrt{3}+1}{2\sqrt{2}}$

Q36. In a right triangle ABC, right angled at C, if $\angle B=60^{\circ}$ and AB=15 units. Find the remaining angles and sides.

Solution:

$sin 60^{\circ}=\frac{x}{15}\\ \frac{\sqrt{3}}{2}=\frac{x}{15}\\ x=\frac{15\sqrt{3}}{2}units\\ \\ \\ cos 60^{\circ}=\frac{x}{15}\\ \frac{1}{2}=\frac{x}{15}\\ x=\frac{15}{2}\\ x=7.5 units$

Q37. In $\Delta$ABC is a right triangle such that $\angle C = 90^{\circ}$, $\angle A= 45^{\circ}$ and BC = 7 units. Find the remaining angles and sides.

Solution:

Here, $\angle C = 90^{\circ}$ and $\angle A= 45^{\circ}$

We know that,

$\angle A +\angle B +\angle C$ = $180^{\circ}$

=> $45^{\circ}$ + $90^{\circ}$ + $\angle C$ = $180^{\circ}$

=> $135^{\circ}$ + $\angle C$ = $180^{\circ}$

=> $\angle C$ = $180^{\circ}$ – $135^{\circ}$

=> $\angle C$ = $45^{\circ}$

The value of the remaining angle C is $45^{\circ}$

Now, we need to find the sides x and y

here,

cos(45) = $\frac{BC}{AB}$

$\frac{1}{\sqrt{2}}$ = $\frac{7}{y}$

y = $7\sqrt{2}$ units

sin(45) = $\frac{AC}{AB}$

$\frac{1}{\sqrt{2}}$ = $\frac{x}{y}$

$\frac{1}{\sqrt{2}}$ = $\frac{x}{7\sqrt{2}}$

x = $\frac{7\sqrt{2}}{\sqrt{2}}$

x = 7 units

the value of x = 7 units and y = $7\sqrt{2}$ units

Q 38 . In a rectangle ABCD , AB = 20 cm , $\angle$BAC = $60^{\circ}$ , calculate side BC and diagonals AC and BD .

Solution:

Let AC = x cm and CB = y cm

Since , $cos\theta$ = $\frac{base}{hypotenuse}$

Therefore , $cos 60^{\circ}=\frac{20}{x}$

$\Rightarrow \frac{1}{2}=\frac{20}{x}$ [since,$cos60^{\circ}=\frac{1}{2}$]

$\Rightarrow$ x= 40 cm = AC

Similarly BD = 40 cm

Now ,

Since , $sin\theta$ = $\frac{perpendicular}{hypotenuse}$

Therefore , $sin 60^{\circ}=\frac{BC}{AC}$

$\Rightarrow \frac{\sqrt{3}}2{}=\frac{y}{40}$

$\Rightarrow y=\frac{40\sqrt{3}}{2}$

$\Rightarrow y=20\sqrt{3}$ cm .

Q39:If A & B are acute angles such that tanA=1/2 tanB=1/3 and tan(A+B)= $\frac{tanA+tanB}{1-tanA\;tanB}$, find A+B.

Solution:

$Tan(A+B)=\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}.\frac{1}{3}}$

$=\frac{\frac{3+2}{6}}{\frac{5}{6}}$

$=\frac{\frac{5}{6}}{\frac{5}{6}}$

(Tan (A+B)= 5 /6 x 6/5 = 1

$(A+B)=Tan^{-1}(1)$

(A+B) = 450

Q 40: Prove that : $(\sqrt{3}-1)(3-cot30{^{\circ}})=tan^{3}60-2sin60^{\circ}$

Ans:

L.H.S => $(\sqrt{3}+1)(3-cot30^{\circ})$

=$(\sqrt{3}+1)(3-\sqrt{3})\;\;∵ cot 30^{\circ}=\sqrt{3}$

=$(\sqrt{3}+1)(\sqrt{3}-1)\sqrt{3}$

= $((\sqrt{3})^{2}-(1)^{2})\sqrt{3}$

= $2\sqrt{3}$

R.H.S => $tan^{3}60-2sin60^{\circ}$

= $(\sqrt{3})^{3}-2\times\frac{\sqrt{3}}{2}$

= $3\sqrt{3}-\sqrt{3}$

= $2\sqrt{3}$

L.H.S = R.H.S

Hence proved.