KC Sinha Mathematics Solution Class 10 Chapter 4 Trigonometric Ratios and Identities Exercise 4.2

Q1 Q2 Q3 A Q3 B Q3 C Q3 D Q3 E Q3 F Q3 G Q3 H Q3 I Q3 J Q3 K Q3 L Q3 M Q3 N Q4 A Q4 B Q4 C Q4 D Q4 E Q4 F Q4 G Q4 H Q4 I Q5 A Q5 B Q5 C Q5 D Q5 E Q5 F Q5 G Q5 H Q5 I Q5 J Q5 K Q5 L Q5 M Q5 N Q6 A Q6 B Q6 C Q7 A Q7 B Q8 A Q8 B Q9 Q10 Q11 Q12 Q13 A Q13 B Q13 C Q13 D Q14 Q15

Exercise 4.1 Exercise 4.2 Exercise 4.3 Exercise 4.4

Exercise 4.2

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Question 1 

Find the value of the following :

(i) sin 30^o + cos 60^o
(ii) sin² 45^o+cos²45^o
(iii) sin 30^o + cos 60^o – tan45^o
(iv) $\sqrt{1+\tan ^{2} 60^{\circ}}$
(v) tan 60^o x cos30^o
Sol :
(i) sin 30^o + cos 60^o
We know that,

$\sin \left(30^{\circ}\right)=\frac{1}{2}>\cos \left(60^{\circ}\right)=\frac{1}{2}$

So,
sin(30^o) + cos(60^o)
$=\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)$
=1

(ii) sin² 45^o+cos²45^o
We know that,
$\sin \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$
$\cos \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$

So, sin² 45^o+cos²45^o
$=\left(\frac{1}{\sqrt{2}}\right)^{2}+\left(\frac{1}{\sqrt{2}}\right)^{2}$
$=\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)$
=1

(iii) sin 30^o + cos 60^o – tan45^o
$\sin \left(30^{\circ}\right)=\frac{1}{2}$
$\cos \left(60^{\circ}\right)=\frac{1}{2}$
tan(45^o)=1

So,
sin 30^o + cos 60^o – tan 45^o
$=\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)-1$
$=\frac{1+1-2}{2}$
=0

(iv) $\sqrt{1+\tan ^{2} 60^{\circ}}$
We know that
tan(60^o) = √3
So,
$=\sqrt{1+\tan ^{2} 60^{\circ}}$
$=\sqrt{1+(\sqrt{3})^{2}}$
$=\sqrt{1+3}$
=√4
= 2
(v) tan 60^o × cos30^o
tan(60^o) = √3
$\cos \left(30^{\circ}\right)=\frac{\sqrt{3}}{2}$
So,
tan 60^o × cos30^o
$=\sqrt{3} \times \frac{\sqrt{3}}{2}$
$=\frac{3}{2}$

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Question 2 

If θ = 45°, find the value of

(i) $\tan ^{2} \theta+\frac{1}{\sin ^{2} \theta}$
(ii) cos² θ - sin² θ
Sol :
(i) $\tan ^{2} \theta+\frac{1}{\sin ^{2} \theta}$
Given θ =45°
We know that,
tan(45^o) = 1
$\sin \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$
$=(1)^{2}+\frac{1}{\left(\frac{1}{\sqrt{2}}\right)^{2}}$
= 1+ 2
= 3
(ii) cos² θ – sin² θ
Given θ = 45°
$\sin \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$
$\cos \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$
So, cos² 45° – sin² 45°
$=\left(\frac{1}{\sqrt{2}}\right)^{2}-\left(\frac{1}{\sqrt{2}}\right)^{2}$
= 0

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Question 3 A 

Find the numerical value of the following :

sin45°.cos45° – sin²30°.
Sol :
We know that,
$\sin \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$
$\cos \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$
$\sin \left(30^{\circ}\right)=\frac{1}{2}$

Now, putting the values
$=\left(\frac{1}{\sqrt{2}}\right) \times\left(\frac{1}{\sqrt{2}}\right)-\left(\frac{1}{2}\right)^{2}$
$=\left(\frac{1}{2}\right)-\left(\frac{1}{4}\right)$
$=\left(\frac{1}{4}\right)$

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Question 3 B 

Find the numerical value of the following :

$\frac{\tan 60^{\circ}}{\sin 60^{\circ}+\cos 60^{\circ}}$
Sol :
We know that,
$\sin \left(60^{\circ}\right)=\frac{\sqrt{3}}{2}$
$\cos \left(60^{\circ}\right)=\frac{1}{2}$
tan (60^°) = √3

Now putting the values;

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

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

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

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

Multiplying and dividing by the conjugate of (1+√3)
$=\frac{2 \sqrt{3}}{1+\sqrt{3}} \times \frac{1-\sqrt{3}}{1-\sqrt{3}}$
$=\frac{2 \sqrt{3}-6}{(1)^{2}-(\sqrt{3})^{2}}$ [∵(a)² – (b)² = (a+b)(a-b)]
$=\frac{2 \sqrt{3}-6}{-2}$

Multiplying and dividing by (-2)
= 3 - √3

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Question 3 C 

Find the numerical value of the following :

$\frac{\tan 60^{\circ}}{\sin 60^{\circ}+\cos 30^{\circ}}$
Sol :
We know that,
$\sin \left(60^{\circ}\right)=\frac{\sqrt{3}}{2}$
$\cos \left(30^{\circ}\right)=\frac{\sqrt{3}}{2}$
tan (60^o) = √3

Now putting the values;

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

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

= 1

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Question 3 D 

Find the numerical value of the following :

$\frac{4}{\sin ^{2} 60^{\circ}}+\frac{3}{\cos ^{2} 60^{\circ}}$
Sol :
We know that

$\sin \left(60^{\circ}\right)=\frac{\sqrt{3}}{2}$

$\cos \left(60^{\circ}\right)=\frac{1}{2}$

Now putting the value, we get
$=\frac{4}{\left(\frac{\sqrt{3}}{2}\right)^{2}}+\frac{3}{\left(\frac{1}{2}\right)^{2}}$
$=4 \times\left(\frac{2}{\sqrt{3}}\right)^{2}+3 \times(2)^{2}$
$=4\left(\frac{4}{3}\right)+3 \times 4$
$=\frac{16}{3}+12$
$=\frac{16+36}{3}$
$=\frac{52}{3}$

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Question 3 E 

Find the numerical value of the following :

sin² 60° – cos² 60°
Sol :
We know that,

$\sin \left(60^{\circ}\right)=\frac{\sqrt{3}}{2}$

$\cos \left(60^{\circ}\right)=\frac{1}{2}$

Now putting the value;
$=\left(\frac{\sqrt{3}}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}$
$=\left(\frac{3}{4}\right)-\left(\frac{1}{4}\right)$
$=\frac{1}{2}$

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Question 3 F 

Find the numerical value of the following :

4sin² 30° + 3 tan 30° – 8 sin 45° cos 45°
Sol :
We know that,

$\sin \left(30^{\circ}\right)=\frac{1}{2}$

$\tan \left(30^{\circ}\right)=\frac{1}{\sqrt{3}}$

$\sin \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$

$\cos \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$

Now putting the value, we get
$=4 \times\left(\frac{1}{2}\right)^{2}+3 \times\left(\frac{1}{\sqrt{3}}\right)^{2}-8 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}$
$=4 \times \frac{1}{4}+3 \times \frac{1}{3}-8 \times \frac{1}{2}$
= 1 + 1 – 4
= -2

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Question 3 G 

Find the numerical value of the following :

2sin²30° – 3cos² 45° + tan² 60ׄ°
Sol :
We know that,

$\sin \left(30^{\circ}\right)=\frac{1}{2}$

$\cos \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$

Tan (60^o) = √3
Now putting the value;
$=2 \times\left(\frac{1}{2}\right)^{2}-3 \times\left(\frac{1}{\sqrt{2}}\right)^{2}+(\sqrt{3})^{2}$
$=2 \times \frac{1}{4}-3 \times \frac{1}{2}+3$
$=\frac{1}{2}-\frac{3}{2}+3$

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

$=\frac{4}{2}$

=2

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Question 3 H 

Find the numerical value of the following :

sin 90° + cos 0° + sin 30° + cos 60°
Sol :
We know that,
Sin (90^o) = 1
Cos (0^o) = 1

$\sin \left(30^{\circ}\right)=\frac{1}{2}$

$\cos \left(60^{\circ}\right)=\frac{1}{2}$

Now putting the value;

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

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

$=\frac{6}{2}$

= 3

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Question 3 I 

Find the numerical value of the following :

sin 90° – cos 0° + tan 0° + tan 45°
Sol :
We know that
Sin (90^o) = 1
Cos (0^o) = 1
Tan(0^o) = 0
Tan(45^o) = 1
Now putting the value, we get
= 1 – 1 + 0 + 1
= 1

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Question 3 J 

Find the numerical value of the following :

$\cos ^{2} 0^{\circ}+\tan ^{2} \frac{\pi}{4}+\sin ^{2} \frac{\pi}{4}$, where π = 180°
Sol :
We know that
Cos (0^o) = 1
Tan (45^o) = 1 $\left[\frac{\pi}{4}=\frac{180^{\circ}}{4}=45^{\circ}\right]$
$\sin \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}\left[\frac{\pi}{4}=\frac{180^{\circ}}{4}=45^{\circ}\right]$

Now putting the values;
$=(1)^{2}+(1)^{2}+\left(\frac{1}{\sqrt{2}}\right)^{2}$
$=1+1+\frac{1}{2}$
$=\frac{2+2+1}{2}$
$=\frac{5}{2}$

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Question 3 K 

Find the numerical value of the following :

$\frac{\cos 60^{\circ}}{\sin ^{2} 45^{\circ}}-3 \cot 45^{\circ}+2 \sin 90^{\circ}$
Sol :
We know that,
$\cos \left(60^{\circ}\right)=\frac{1}{2}$
$\sin \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$
Cot (45^o) = 1
Sin (90^o) = 1
Now putting the values, we get
$=\frac{\frac{1}{2}}{\left(\frac{1}{\sqrt{2}}\right)^{2}}-3(1)+2(1)$
= 1-3+2
=0

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Question 3 L 

Find the numerical value of the following :

$\frac{4}{\tan ^{2} 60^{\circ}}+\frac{1}{\cos ^{2} 30^{\circ}}-\sin ^{2} 45^{\circ}$
Sol :
We can write the above equation as:
= 4 cot² 60^o + sec² 30^o – sin² 45^o …(a) $\left[\because \cos \theta=\frac{1}{\sec \theta}\right.$ and $\left.\tan \theta=\frac{1}{\cot \theta}\right]$

$\sin \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$

$\cot \left(60^{\circ}\right)=\frac{1}{\sqrt{3}}$

$\operatorname{Sec}\left(30^{\circ}\right)=\frac{2}{\sqrt{3}}$

Now putting the values in (a);
$=4\left(\frac{1}{\sqrt{3}}\right)^{2}+\left(\frac{2}{\sqrt{3}}\right)^{2}-\left(\frac{1}{\sqrt{2}}\right)^{2}$

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

$=\frac{8+8-3}{6}$

$=\frac{13}{6}$

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Question 3 M 

Find the numerical value of the following :

cos60° . cos 30° – sin 60° . sin 30°
Sol :
We know that,

$\cos \left(60^{\circ}\right)=\frac{1}{2}$

$\cos \left(30^{\circ}\right)=\frac{\sqrt{3}}{2}$

$\sin \left(60^{\circ}\right)=\frac{\sqrt{3}}{2}$

$\sin \left(30^{\circ}\right)=\frac{1}{2}$

Now putting the values, we get

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

$=\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}$

= 0

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Question 3 N 

Find the numerical value of the following :

$\frac{4\left(\sin ^{2} 60^{\circ}-\cos ^{2} 45^{\circ}\right)}{\tan ^{2} 30^{\circ}+\cos ^{2} 90^{\circ}}$
Sol :
We know that,

$\sin \left(60^{\circ}\right)=\frac{\sqrt{3}}{2}$

$\cos \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$

$\tan \left(30^{\circ}\right)=\frac{1}{\sqrt{3}}$

cos(90^o) = 0

Now putting the values;
$=4 \times \frac{\left(\frac{\sqrt{3}}{2}\right)^{2}-\left(\frac{1}{\sqrt{2}}\right)^{2}}{\left(\frac{1}{\sqrt{3}}\right)^{2}-(0)^{2}}$
$=4 \times \frac{\frac{3}{4}-\frac{1}{2}}{\frac{1}{3}}$
$=4 \times \frac{1}{4} \times 3$
= 3

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Question 4 A 

Evaluate the following :

sin30°.cos45° + cos30°.sin45°
Sol :

We know that,

$\sin \left(30^{\circ}\right)=\frac{1}{2}$

$\cos \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$

$\cos \left(30^{\circ}\right)=\frac{\sqrt{3}}{2}$

$\sin \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$

Now putting the values, we get

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

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

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

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Question 4 B 

Evaluate the following :

cosec²30°.tan²45° – sec²60°
Sol :
We know that
cosec (30^o) = 2
Tan(45^o) = 1
sec (60 ^o) = 2
Now putting the values;
= (2)² × (1)² - (2)²
= 4 – 4
= 0

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Question 4 C 

Evaluate the following :

2sin²30°.tan60° – 3cos²60°.sec²30°
Sol :
We know that
$\sin \left(30^{\circ}\right)=\frac{1}{2}$
tan (60^o) = √3

$\cos \left(60^{\circ}\right)=\frac{1}{2}$

$\sec \left(30^{\circ}\right)=\frac{2}{\sqrt{3}}$

Now putting the values;

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

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

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

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

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Question 4 D 

Evaluate the following :

tan60° . cosec²45° + sec²60°.tan45°
Sol :
We know that
tan (60^o) = √3
cosec (45^o) = √2
sec (60 ^o) = 2
tan(45^o) = 1
Now putting the values;
= (√3) × (√2)² + (2)² × (1)
= 2√3 +4
=2 (√3 + 2)

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Question 4 E 

Evaluate the following :

tan30°.sec45° + tan60°.sin30°
Sol :
We know that
$\tan \left(30^{\circ}\right)=\frac{1}{\sqrt{3}}$
sec (45^o) = √2
tan (60^o) = √3
$\sec \left(30^{\circ}\right)=\frac{2}{\sqrt{3}}$

Now putting the values, we get
$=\frac{1}{\sqrt{3}} \times \sqrt{2}+\sqrt{3} \times \frac{2}{\sqrt{3}}$
$=\frac{\sqrt{2}}{\sqrt{3}}+2$
$=2+\frac{\sqrt{2}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$
$=2+\frac{\sqrt{6}}{3}$

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Question 4 F 

Evaluate the following :

cos30°.cos45° – sin30°.sin45°
Sol :
We know that

$\cos \left(30^{\circ}\right)=\frac{\sqrt{3}}{2}$

$\cos \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$

$\sin \left(30^{\circ}\right)=\frac{1}{2}$

$\sin \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$

Now putting the values, we get
$=\frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{2}}-\frac{1}{2} \times \frac{1}{\sqrt{2}}$
$=\frac{\sqrt{3}-1}{2 \sqrt{2}}$

Multiplying and dividing by (√2), we get

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

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

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Question 4 G 

Evaluate the following :

$\frac{4}{3} \tan ^{2} 30^{\circ}+\sin ^{2} 60^{\circ}-3 \cos ^{2} 60^{\circ}+\frac{3}{4} \tan ^{2} 60^{\circ}-2 \tan ^{2} 45^{\circ}$
Sol :
We know that
$\begin{aligned} \tan \left(30^{\circ}\right) &=\frac{1}{\sqrt{3}} \\ \sin \left(60^{\circ}\right) &=\frac{\sqrt{3}}{2} \\ \cos \left(60^{\circ}\right) &=\frac{1}{2} \end{aligned}$
tan (60^o) = √3
tan(45^o) = 1

Now putting the values;
$=\left(\frac{4}{3} \times\left(\frac{1}{\sqrt{3}}\right)^{2}\right)+\left[\left(\frac{\sqrt{3}}{2}\right)^{2}\right]-\left[3 \times\left(\frac{1}{2}\right)^{2}\right]+\left[\frac{3}{4}(\sqrt{3})^{2}\right]-\left[2 \times(1)^{2}\right]$
$=\left[\frac{4}{3} \times \frac{1}{3}\right]+\left[\frac{3}{4}\right]-\left[3 \times \frac{1}{4}\right]+\left[\frac{3}{4} \times 3\right]-[2 \times(1)]$

$=\frac{4}{9}+\frac{3}{4}-\frac{3}{4}+\frac{9}{4}-2$

$=\frac{16+27-27+81-72}{36}$

$=\frac{25}{36}$

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Question 4 H 

Evaluate the following :

$\frac{\tan ^{2} 60^{\circ}+4 \cos ^{2} 45^{\circ}+3 \sec ^{2} 30^{\circ}+5 \cos ^{2} 90^{\circ}}{\operatorname{cosec} 30^{\circ}+\sec 60^{\circ}-\cot ^{2} 30^{\circ}}$
Sol :
We know that
tan (60^o) = √3

$\cos \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$

$\sec \left(30^{\circ}\right)=\frac{2}{\sqrt{3}}$

cos(90^o) = 0
cosec (30^o) = 2
sec (60 ^o) = 2
cot (30^o) = √3

Now putting the values, we get
$=\frac{(\sqrt{3})^{2}+\left[4 \times\left(\frac{1}{\sqrt{2}}\right)^{2}\right]+\left[3 \times\left(\frac{2}{\sqrt{3}}\right)^{2}\right]+\left[5 \times(0)^{2}\right]}{(2)+(2)-(\sqrt{3})^{2}}$
$=\frac{(3)+\left[4 \times \frac{1}{2}\right]+\left[3 \times \frac{4}{3}\right]+[5 \times 0]}{2+2-3}$
$=\frac{(3)+[2]+[4]+[0]}{2+2-3}$
= 9

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Question 4 I 

Evaluate the following :

$\frac{5 \sin ^{2} 30^{\circ}+\cos ^{2} 45^{\circ}-4 \tan ^{2} 30^{\circ}}{2 \sin 30^{\circ} \cdot \cos 30^{\circ}+\tan 45^{\circ}}$
Sol :
We know that

$\sin \left(30^{\circ}\right)=\frac{1}{2}$

$\cos \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$

$\tan \left(30^{\circ}\right)=\frac{1}{\sqrt{3}}$

$\cos \left(30^{\circ}\right)=\frac{\sqrt{3}}{2}$

tan (45^o) = 1

Now putting the values, we get
$=\frac{\left[5 \times\left(\frac{1}{2}\right)^{2}\right]+\left(\frac{1}{\sqrt{2}}\right)^{2}-\left[4 \times\left(\frac{1}{\sqrt{3}}\right)^{2}\right]}{2\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right)+(1)}$
$=\frac{\left(\frac{5}{4}\right)+\left(\frac{1}{2}\right)-\left(\frac{4}{3}\right)}{\left(\frac{\sqrt{3}}{2}\right)+1}$
$=\frac{\left(\frac{15+6-16}{12}\right)}{\left(\frac{\sqrt{3}+2}{2}\right)}$
$=\frac{5}{12} \times \frac{2}{\sqrt{3}+1}$
$=\frac{5}{6} \times \frac{1}{\sqrt{3}+2}$

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Question 5 A 

Prove the following :

$\frac{(1-\cos B)(1+\cos B)}{(1-\sin B)(1+\sin B)}=\frac{1}{3}$ When B = 30°
Sol :
Solving, L.H.S.
$=\frac{(1)^{2}-(\cos B)^{2}}{(1)^{2}-(\sin B)^{2}}$ [(a)² – (b)² = (a+b)(a-b)]
$=\frac{1-\cos ^{2} B}{1-\sin ^{2} B}$

We know that,
$\cos \left(30^{\circ}\right)=\frac{\sqrt{3}}{2}$
$\sin \left(30^{\circ}\right)=\frac{1}{2}$

Putting the values, we get
$=\frac{1-\left(\frac{\sqrt{3}}{2}\right)^{2}}{1-\left(\frac{1}{2}\right)^{2}}$
$=\frac{1-\frac{3}{4}}{1-\frac{1}{4}}$
$=\frac{4-3}{4-1}$
$=\frac{1}{3}$
=R.H.S.
Hence Proved

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Question 5 B 

Prove the following :

$\frac{(1-\cos \alpha)(1+\cos \alpha)}{(1-\sin \alpha)(1+\sin \alpha)}=3$ When α =60°
Sol :
Solving, L.H.S.
$=\frac{(1)^{2}-(\cos \alpha)^{2}}{(1)^{2}-(\sin \alpha)^{2}}$ [(a)² – (b)² = (a+b)(a-b)]
$=\frac{1-\cos ^{2} \alpha}{1-\sin ^{2} \alpha}$

We know that

$\cos \left(60^{\circ}\right)=\frac{1}{2}$

$\sin \left(60^{\circ}\right)=\frac{\sqrt{3}}{2}$

Putting the values, we get
$=\frac{1-\left(\frac{1}{2}\right)^{2}}{1-\left(\frac{\sqrt{3}}{2}\right)^{2}}$

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

$=\frac{4-1}{4-3}$
= 3 = R.H.S.

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Question 5 C 

Prove the following :

cos(A – B) = cos A. cos B + sinA . sin B if A=B=60^o
Sol :
Solving, L.H.S.
= cos (60^o – 60^o) [Putting the value A=B=60^o]
= cos (0^o)
= 1
Solving, R.H.S.
= cos (60^o) × cos (60^o) + sin (60^o) × sin (60^o) [Putting the value A=B=60^o]
= cos²(60^o) + sin²(60^o)

We know that,

$\cos \left(60^{\circ}\right)=\frac{1}{2}$

$\sin \left(60^{\circ}\right)=\frac{\sqrt{3}}{2}$

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

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

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

= 1
∴ LHS = RHS
Hence Proved

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Question 5 D 

Prove the following :

4(sin⁴30° + cos⁴ 60°) – 3(cos² 45° – sin²90°) = 2
Sol :
We know that,

$\sin \left(30^{\circ}\right)=\frac{1}{2}$

$\cos \left(60^{\circ}\right)=\frac{1}{2}$

$\cos \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$

Sin (90^o) = 1

Now solving, L.H.S.

= 4[{(sin 30^o)²}² + {(cos 60^o)²}²] – 3[(cos 45^o)² - (sin 90^o)²]

Putting the values
$=4 \times\left[\left\{\left(\frac{1}{2}\right)^{2}\right\}^{2}+\left\{\left(\frac{1}{2}\right)^{2}\right\}^{2}\right]-3\left[\left(\frac{1}{\sqrt{2}}\right)^{2}-1\right]$
$=4 \times\left[\left\{\frac{1}{4}\right\}^{2}+\left\{\frac{1}{4}\right\}^{2}\right]-3\left[\frac{1}{2}-1\right]$
$=4 \times\left[\frac{1}{16}+\frac{1}{16}\right]-3\left[-\frac{1}{2}\right]$
$=4 \times\left[\frac{1}{8}\right]-3\left[-\frac{1}{2}\right]$

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

$=\left[\frac{4}{2}\right]$

=2 = R.H.S.
Hence Proved

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Question 5 E 

Prove the following :

sin90° = 2sin45°.cos45°
Sol :
We know that,
sin (90^o) = 1

$\sin \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$

$\cos \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$

Taking LHS = sin 90° = 1

Now, taking RHS

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

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

= 1
= R.H.S.
Hence Proved

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Question 5 F 

Prove the following :

cos60° = 2cos²30° – 1 = 1 – 2 sin²30°
Sol :
We know that,

$\cos \left(60^{\circ}\right)=\frac{1}{2}$

$\cos \left(30^{\circ}\right)=\frac{\sqrt{3}}{2}$

$\sin \left(30^{\circ}\right)=\frac{1}{2}$

Taking LHS = cos 60° $=\frac{1}{2}$

Now, solving RHS = 2cos² 30° - 1 , we get

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

$=2 \times \frac{3}{4}-1$

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

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

$=\frac{1}{2}$

= RHS

Now taking RHS = 1- 2sin² 30°

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

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

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

$=\frac{1}{2}$

= RHS
Hence, proved.

---

Question 5 G 

Prove the following :

cos90° = 1 – 2 sin²45° = 2cos²45° – 1
Sol :
We know that
cos(90^o) = 0

$\sin \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$

$\cos \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$

taking LHS = cos 90° = 0

Now solving RHS 1- 2sin² 45°

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

$=1-2 \times \frac{1}{2}$

= 1- 1
= 0
= RHS

Now, solving RHS = 2cos² 45° - 1 , we get

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

$=1-2 \times \frac{1}{2}$

= 1- 1
= 0
Hence, proved.

---

Question 5 H 

Prove the following :

sin30°.cos60° + cos30°.sin60° = sin90°
Sol :
We know that

$\sin \left(30^{\circ}\right)=\frac{1}{2}$

$\cos \left(60^{\circ}\right)=\frac{1}{2}$

$\cos \left(30^{\circ}\right)=\frac{\sqrt{3}}{2}$

$\sin \left(60^{\circ}\right)=\frac{\sqrt{3}}{2}$

sin (90^o) = 1

Taking LHS =
$=\left[\left(\frac{1}{2}\right) \times\left(\frac{1}{2}\right)\right]+\left[\left(\frac{\sqrt{3}}{2}\right) \times\left(\frac{\sqrt{3}}{2}\right)\right]$

$=\left[\left(\frac{1}{4}\right)\right]+\left[\left(\frac{3}{4}\right)\right]$

$=\left[\frac{1+3}{4}\right]$

= 1

Now, RHS = sin 90° = 1
∴ LHS = RHS
Hence, proved.

---

Question 5 I 

Prove the following :

cos60°.cos30° – sin60°. sin30° = cos 90°
Sol :
We know that

$\cos \left(60^{\circ}\right)=\frac{1}{2}$

$\cos \left(30^{\circ}\right)=\frac{\sqrt{3}}{2}$

$\sin \left(60^{\circ}\right)=\frac{\sqrt{3}}{2}$

$\sin \left(30^{\circ}\right)=\frac{1}{2}$

cos(90^o) = 0

Taking LHS
$=\left[\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right)\right]-\left[\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right)\right]$
$=\left[\left(\frac{\sqrt{3}}{4}\right)\right]-\left[\left(\frac{\sqrt{3}}{4}\right)\right]$
= 0

Now, RHS = cos 90° = 0

∴ LHS =RHS
Hence, proved.

---

Question 5 J 

Prove the following :

$\cos 60^{\circ}=\frac{1-\tan ^{2} 30^{\circ}}{1+\tan ^{2} 30^{\circ}}$
Sol :
We know that,

$\cos \left(60^{\circ}\right)=\frac{1}{2}$

$\tan \left(30^{\circ}\right)=\frac{1}{\sqrt{3}}$

Taking LHS = $\cos 60^{\circ}=\frac{1}{2}$

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

$=\frac{2}{4}$

$=\frac{1}{2}$

∴ L.H.S. = R.H.S.
Hence, proved.

---

Question 5 K 

Prove the following :

$\frac{\tan 60^{\circ}-\tan 30^{\circ}}{1+\tan 60^{\circ} \cdot \tan 30^{\circ}}=\tan 30^{\circ}$
Sol :
We know that
tan(60^o) = √3
$\tan \left(30^{\circ}\right)=\frac{1}{\sqrt{3}}$

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

Now, RHS $=\tan 30^{\circ}=\frac{1}{\sqrt{3}}$

∴L.H.S. = R.H.S.
Hence, proved.

---

Question 5 L 

Prove the following :

$\frac{1-\tan 30^{\circ}}{1+\tan 30^{\circ}}=\frac{1-\sin 60^{\circ}}{\cos 60^{\circ}}$
Sol :

$\tan \left(30^{\circ}\right)=\frac{1}{\sqrt{3}}$

$\sin \left(60^{\circ}\right)=\frac{\sqrt{3}}{2}$

$\cos \left(60^{\circ}\right)=\frac{1}{2}$

Taking LHS
$=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}$
$=\frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}}$
$=\frac{\sqrt{3}-1}{\sqrt{3}+1}$

Multiplying and Dividing, LHS by (√3- 1)
$=\frac{\sqrt{3}-1}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1}$
$=\frac{(\sqrt{3}-1)^{2}}{(\sqrt{3})^{2}-(1)^{2}}$ [(a)² – (b)² = (a+b)(a-b)]
$=\frac{(\sqrt{3}-1)^{2}}{(\sqrt{3})^{2}-(1)^{2}}$
$=\frac{3+1-2 \sqrt{3}}{3-1}$
$=\frac{4-2 \sqrt{3}}{2}$

Multiplying and Dividing, LHS by 2
= 2- √3

Now, RHS
$=\frac{1-\frac{\sqrt{3}}{2}}{\frac{1}{2}}$
$=\frac{\frac{2-\sqrt{3}}{2}}{\frac{1}{2}}$
= 2- √3

∴ LHS = RHS
Hence, proved.

---

Question 5 M 

Prove the following :

$\frac{\sin 60^{\circ}+\cos 30^{\circ}}{\sin 30^{\circ}+\cos 60^{\circ}+1}=\cos 30^{\circ}$
Sol :
We know that

$\sin \left(30^{\circ}\right)=\frac{1}{2}$

$\sin \left(60^{\circ}\right)=\frac{\sqrt{3}}{2}$

$\cos \left(30^{\circ}\right)=\frac{\sqrt{3}}{2}$

$\cos \left(60^{\circ}\right)=\frac{1}{2}$

Taking LHS
$=\frac{\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}}{\frac{1}{2}+\frac{1}{2}+1}$
$=\frac{2 \times \frac{\sqrt{3}}{2}}{\frac{1+1+2}{2}}$
$=\frac{\sqrt{3}}{2}$

Now, RHS= $\cos \left(30^{\circ}\right)=\frac{\sqrt{3}}{2}$
∴ LHS =RHS

Hence Proved

---

Question 5 N 

Prove the following :

$\sin 60^{\circ}=2 \sin 30^{\circ} \cdot \cos 30^{\circ}=\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}$
Sol :
We know that

$\sin \left(30^{\circ}\right)=\frac{1}{2}$

$\sin \left(60^{\circ}\right)=\frac{\sqrt{3}}{2}$

$\cos \left(30^{\circ}\right)=\frac{\sqrt{3}}{2}$

$\tan \left(30^{\circ}\right)=\frac{1}{\sqrt{3}}$

Taking LHS $=\sin 60^{\circ}=\frac{\sqrt{3}}{2}$

Now, solving RHS = 2 sin 30° cos 30°

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

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

= LHS

Now, RHS$=\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}$
$=\frac{2\left(\frac{1}{\sqrt{3}}\right)}{1+\left(\frac{1}{\sqrt{3}}\right)^{2}}$
$=\frac{\frac{2}{\sqrt{3}}}{1+\frac{1}{3}}$
$=\frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}}$
$=\frac{2}{\sqrt{3}} \times \frac{3}{4}$
$=\frac{\sqrt{3}}{2}$
∴ LHS =RHS
Hence proved

---

Question 6A 

If A=60^o and B = 30^o, verify that :

cos (A+B) = cos A cos B – sin A sin B
Sol :
Given: A=60^o and B =30^o
Now, LHS = Cos (A+B)
⇒ Cos (60 ^o + 30 ^o)
⇒ Cos (90 ^o)
⇒ 0 [∵ cos 90 ^o = 0]
Now, RHS = Cos A Cos B – Sin A Sin B
⇒ cos(60 ^o) cos(30 ^o) – sin(60 ^o) sin (30 ^o)
$\Rightarrow\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right)-\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right)$
⇒ 0
∴ LHS = RHS
Hence Proved

---

Question 6 B 

If A=60^o and B = 30^o, verify that :

sin (A – B) = sin A cos B – cos A sin B
Sol :
Given: A=60^o and B =30^o
Now, LHS = Sin (A-B)
⇒ Sin (60 ^o - 30 ^o)
⇒ Sin (30 ^o)
$\Rightarrow\left(\frac{1}{2}\right)$

Now, RHS = Sin A Cos B – Cos A Sin B

⇒ sin(60 ^o) cos(30 ^o) – cos(60 ^o) sin (30 ^o)
$\Rightarrow\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)-\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)$
$\Rightarrow \frac{3}{4}-\frac{1}{4}$
$\Rightarrow\left(\frac{1}{2}\right)$
∴ LHS = RHS
Hence Proved

---

Question 6 C 

If A=60^o and B = 30^o, verify that :

tan (A – B) $=\frac{\tan A-\tan B}{1+\tan A \tan B}$
Sol :
Given: A=60^o and B =30^o
Now, LHS = tan (A-B)
⇒ tan (60 ^o - 30 ^o)
⇒ tan (30 ^o)
$\Rightarrow\left(\frac{1}{\sqrt{3}}\right)$

Now, RHS $=\frac{\tan A-\tan B}{1+\tan A \tan B}$
$=\frac{\tan 60^{\circ}-\tan 30^{\circ}}{1+\tan 60^{\circ} \tan 30^{\circ}}$
$\Rightarrow \frac{\sqrt{3}-\left(\frac{1}{\sqrt{3}}\right)}{1+(\sqrt{3})\left(\frac{1}{\sqrt{3}}\right)}$
$\Rightarrow \frac{\frac{3-1}{\sqrt{3}}}{1+1}$
$\Rightarrow\left(\frac{1}{\sqrt{3}}\right)$
∴ LHS = RHS
Hence Proved

---

Question 7 A 

If A = 30^o, verify that :

sin 2A = 2 sin A cos A
Sol :
Given: A =30^o
Now, LHS = sin 2(30^o)
⇒ sin 60^o
$\Rightarrow \frac{\sqrt{3}}{2}$

Now, RHS = 2 sin A cos A
⇒ 2 sin (30^o) cos (30^o)
$\Rightarrow 2\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right)$
$\Rightarrow \frac{\sqrt{3}}{2}$
∴ LHS = RHS
Hence Proved

---

Question 7 B 

If A = 30^o, verify that :

cos 2A = 1-2 sin²A=2cos² A – 1
Sol :
Given: A =30^o

Now, LHS = cos 2(30^o)
⇒ cos 60^o
$\Rightarrow \frac{1}{2}$

Now, RHS = 1- 2sin² A
⇒ 1- 2sin² (30^o)
$\Rightarrow 1-2\left(\frac{1}{2}\right)^{2}$
$\Rightarrow 1-2\left(\frac{1}{4}\right)$
$\Rightarrow \frac{2-1}{2}$
$\Rightarrow \frac{1}{2}$

Now, RHS = 2cos² A – 1
⇒ 2cos² (30^o) - 1
$\Rightarrow 2\left(\frac{\sqrt{3}}{2}\right)^{2}-1$
$\Rightarrow 2\left(\frac{3}{4}\right)-1$
$\Rightarrow \frac{3-2}{2}$
$\Rightarrow \frac{1}{2}$

∴ LHS = RHS
Hence Proved

---

Question 8 A 

If θ = 30°, verify that :
sin 3θ = 3 sinθ – 4 sin³θ
Sol :
Given: θ =30^o
Now, LHS = sin 3(30^o)
⇒ sin 90^o
= 1
Now, RHS = 3 sin θ - 4 sin³ θ
⇒ 3 sin (30^o) - 4 sin³ (30^o)
$\Rightarrow 3\left(\frac{1}{2}\right)-4\left(\frac{1}{2}\right)^{3}$
$\Rightarrow \frac{3}{2}-\frac{1}{2}$
= 1
∴ LHS = RHS
Hence Proved

---

Question 8 B 

If θ = 30°, verify that :

cos3θ = 4cos³θ – 3cosθ
Sol :
Given: θ =30^o
Now, LHS = cos 3(30^o)
⇒ cos 90^o
= 0
Now, RHS = 4 cos³ θ - 3 cos θ
⇒ 4 cos³ (30^o) - 3 cos (30^o)
$\Rightarrow 4\left(\frac{\sqrt{3}}{2}\right)^{3}-3\left(\frac{\sqrt{3}}{2}\right)$
$\Rightarrow\left(\frac{3 \sqrt{3}}{2}\right)-\left(\frac{3 \sqrt{3}}{2}\right)$
= 0
∴ LHS = RHS
Hence Proved

---

Question 9 

If sin (A + B) = 1 and cos (A – B) = $\frac{\sqrt{3}}{2}$, then find A and B.

Sol :
Given : sin (A+B) =1
⇒ Sin(A+B) = sin (90 ^o) [∵ sin (90 ^o)=1]
On equating both the sides, we get
A + B = 90 ^o …(1)
And $\cos (A-B)=\frac{\sqrt{3}}{2}$
⇒ cos(A – B) = cos (30 ^o) $\left[\because \cos \left(30^{\circ}\right)=\frac{\sqrt{3}}{2}\right]$

On equating both the sides, we get
A – B = 30 ^o …(2)

On Adding Eq. (1) and (2), we get
2A = 120 ^o
⇒ A = 60 ^o

Now, Putting the value of A in Eq.(1), we get
60 ^o + B =90 ^o
⇒ B = 30 ^o

Hence, A = 60 ^o and B = 30 ^o

---

Question 10 

If sin (A + B) = 1 and cos (A – B) = 1, find A and B.

Sol :
Given : sin (A+B) =1
⇒ Sin(A+B) = sin (90 ^o) [∵ sin (90 ^o) =1]
On equating both the sides, we get
A + B = 90 ^o …(1)
And cos (A – B) = 1
⇒ cos(A – B) = cos (0 ^o) [∵ cos(0 ^o) = 1]
On equating both the sides, we get
A – B = 0 ^o …(2)
On Adding Eq. (1) and (2), we get
2A = 90 ^o
⇒ A = 45 ^o
Now, Putting the value of A in Eq.(1), we get
45 ^o + B =90 ^o
⇒ B = 45 ^o
Hence, A = 45 ^o and B = 45 ^o

---

Question 11 

If sin (A + B) = cos (A – B) = $\frac{\sqrt{3}}{2}$, fins A and B.

Sol :
Given : $\sin (A+B)=\frac{\sqrt{3}}{2}$
⇒ Sin(A+B) = sin (60 ^o) $\left[\because \sin (60 ^{\circ})=\frac{\sqrt{3}}{2}\right]$

On equating both the sides, we get
A + B = 60 ^o …(1)
And $\cos (A-B)=\frac{\sqrt{3}}{2}$

⇒ cos(A – B) = cos (30 ^o) $\left[\because \cos (30^{\circ})=\frac{\sqrt{3}}{2}\right]$

On equating both the sides, we get
A – B = 30 ^o …(2)

On Adding Eq. (1) and (2), we get
2A = 90 ^o
⇒ A = 45 ^o

Now, Putting the value of A in Eq.(1), we get
45 ^o + B =60 ^o
⇒ B = 15 ^o
Hence, A = 45 ^o and B = 15 ^o

---

Question 12 

If sin (A – B) = 1/2, cos(A + B) = 1/2; 0^o<A+B<90^o; A > B, find A and B.

Sol :
Given : $\sin (A-B)=\frac{1}{2}$
⇒ Sin(A-B) = sin (30 ^o) $\left[\because \sin (30^{\circ})=\frac{1}{2}\right]$

On equating both the sides, we get
A - B = 30 ^o …(1)
And $\cos (A+B)=\frac{1}{2}$

⇒ cos(A + B) = cos (60 ^o) $\left[\because \cos (60^{\circ})=\frac{1}{2}\right]$

On equating both the sides, we get
A + B = 60 ^o …(2)

On Adding Eq. (1) and (2), we get
2A = 90 ^o
⇒ A = 45 ^o

Now, Putting the value of A in Eq.(2), we get
45 ^o + B =60 ^o
⇒ B = 15 ^o
Hence, A = 45 ^o and B = 15 ^o

---

Question 13 A 

Show by an example that

cos A – cos B ≠ cos (A – B)
Sol :

Let A = 60^o and B = 30^o, then
L.H.S. =cos A-cosB

=cos60°-cos30°

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

R. H. S=cos(A-B)

=cos(60°-30°)

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

 
∴ L.H.S. R.H.S

---

Question 13 B 

Show by an example that

cos C + cos D ≠ cos (C + D)
Sol :
Let C = 60^o and D = 30^o, then
L.H.S. = cos C + cos D = cos 60^o + cos 30^o
$=\frac{1}{2}+\frac{\sqrt{3}}{2}=\frac{1+\sqrt{3}}{2}$
R. H. S. = cos (C+D) = cos (60^o + 30^o) = cos 90^o= 0
∴ L.H.S. R.H.S

---

Question 13 C 

Show by an example that

sin A + sin B ≠ sin (A + B)
Sol :
Let A = 60^o and B = 30^o, then
L.H.S. = sin A + sin B = sin 60^o + sin 30^o
$=\frac{\sqrt{3}}{2}+\frac{1}{2}=\frac{\sqrt{3}+1}{2}$

R. H. S. = sin (A + B) = sin (60^o + 30^o) = sin 90^o =1
∴ L.H.S. R.H.S

---

Question 13 D 

Show by an example that

sin A – sin B ≠ sin (A – B)
Sol :
Let A = 60^o and B = 30^o, then
L.H.S. = sin A - sin B = sin 60^o - sin 30^o
$=\frac{\sqrt{3}}{2}-\frac{1}{2}=\frac{\sqrt{3}-1}{2}$

R. H. S. = sin (A - B) = sin (60^o - 30^o) = sin 30^o
$=\frac{1}{2}$
∴ L.H.S. ≠R.H.S

---

Question 14 

In a right ΔABC hypotenuse AC = 10 cm and ∠A = 60°, then find the length of the remaining sides.

Sol :

Given: ∠A = 60^o and AC = 10cm
Now, $\sin 60^{\circ}=\frac{\text { Perpendicular }}{\text { Hypotenuse }}=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{\mathrm{BC}}{10}$

Now, we know that $\sin 60^{\circ}=\frac{\sqrt{3}}{2}$
$\Rightarrow \frac{\sqrt{3}}{2}=\frac{B C}{10}$
⇒ BC = 5√3 cm
In right angled ∆ABC , we have
⇒ (AB)² + (BC)² =(AC)² [by using Pythagoras theorem]
⇒ (AB)² + (5√3)² = (10)²
⇒ (AB)² +(25×3) =100
⇒ (AB)² +75 = 100
⇒ (AB)² = 100 – 75
⇒ (AB)² = 25
⇒ AB =√25
⇒ AB = ±5
⇒ AB = 5cm [taking positive square root since, side cannot be negative]
∴ Length of the side AB = 5cm and BC =5√3 cm

---

Question 15 

In a rectangle ABCD, BD : BC = 2 : √3, then find ∠BDC in degrees.

Sol :

Given BD: BC = 2 : √3
We have to find the ∠BDC
We know that,
$\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$
$\Rightarrow \sin \theta=\frac{k \sqrt{3}}{2 k}$
$\Rightarrow \sin \theta=\frac{\sqrt{3}}{2}$
⇒ sin θ = sin 60^o
⇒ θ = 60^o

---

S.no	Chapters	Links
1	Real numbers	Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4
2	Polynomials	Exercise 2.1 Exercise 2.2 Exercise 2.3
3	Pairs of Linear Equations in Two Variables	Exercise 3.1 Exercise 3.2 Exercise 3.3 Exercise 3.4 Exercise 3.5
4	Trigonometric Ratios and Identities	Exercise 4.1 Exercise 4.2 Exercise 4.3 Exercise 4.4
5	Triangles	Exercise 5.1 Exercise 5.2 Exercise 5.3 Exercise 5.4 Exercise 5.5
6	Statistics	Exercise 6.1 Exercise 6.2 Exercise 6.3 Exercise 6.4
7	Quadratic Equations	Exercise 7.1 Exercise 7.2 Exercise 7.3 Exercise 7.4 Exercise 7.5
8	Arithmetic Progressions (AP)	Exercise 8.1 Exercise 8.2 Exercise 8.3 Exercise 8.4
9	Some Applications of Trigonometry: Height and Distances	Exercise 9.1
10	Coordinates Geometry	Exercise 10.1 Exercise 10.2 Exercise 10.3 Exercise 10.4
11	Circles	Exercise 11.1 Exercise 11.2
12	Constructions	Exercise 12.1
13	Area related to Circles	Exercise 13.1
14	Surface Area and Volumes	Exercise 14.1 Exercise 14.2 Exercise 14.3 Exercise 14.4
15	Probability	Exercise 15.1