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CBSE Questions for Class 10 Maths Introduction To Trigonometry Quiz 9 - MCQExams.com
CBSE
Class 10 Maths
Introduction To Trigonometry
Quiz 9
If $$\displaystyle25\sin ^{2}\theta +10\cos ^{2}\theta =15 $$ then find $$\displaystyle\cot ^{2}\theta $$
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$$1$$
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$$2$$
0%
$$3$$
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$$4$$
Explanation
$$ 25\sin ^{ 2 }{ \theta } +10\cos ^{ 2 }{ \theta } = 15 $$
$$ \Rightarrow 15\sin ^{ 2 }{ \theta } + 10\sin ^{ 2 }{ \theta } + 10\cos ^{ 2 }{ \theta } = 15 $$
$$ \Rightarrow 15\sin ^{ 2 }{ \theta } + 10(\sin ^{ 2 }{ \theta } + \cos ^{ 2 }{ \theta } ) = 15 $$
$$ \Rightarrow 15\sin ^{ 2 }{ \theta } +10 = 15 $$ (Since, $$ \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } = 1 $$)
$$ \Rightarrow 15\sin ^{ 2 }{ \theta } = 5 $$
$$ \Rightarrow \sin ^{ 2 }{ \theta } = \dfrac {5}{15} = \dfrac {1}{3} $$
And, $$ \text{cosec}\, ^{ 2 }{ \theta } = \dfrac {1}{ \sin ^{ 2 }{ \theta }} = 3 $$
Also, $$ \cot ^{ 2 }{ \theta } = \text{cosec}\, ^{ 2 }{ \theta }-1 = 3 - 1 = 2 $$
In a right angled $$\triangle ABD, \angle B=60^o$$ and $$\angle A=30^o$$.
Then, $$\cot { 60^o } =$$
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$$\sqrt { 3 } $$
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$$\dfrac { 1 }{ \sqrt { 3 } } $$
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$$\dfrac { \sqrt { 3 } }{ 2 } $$
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$$\dfrac { 2 }{ \sqrt { 3 } } $$
Explanation
Consider an equilateral $$\triangle ABC$$, with each side $$= 2a$$.
So each angle of $$\triangle ABD$$ is $$60^o$$.
From $$A$$, draw $$AD \perp BC$$, so $$BD = DC = a$$.
Also, $$\angle BAD=30^o$$ and $$\angle ADB=90^o$$.
From right angled $$\triangle ADB, AD=\sqrt { { AB }^{ 2 }-{ BD }^{ 2 } } $$
$$=\sqrt { { \left( 2a \right) }^{ 2 }-{ a }^{ 2 } } =\sqrt { { 4a }^{ 2 }-{ a }^{ 2 } } =\sqrt { { 3a }^{ 2 } } =\sqrt { 3 } a$$.
$$\cot { 60^o } =\dfrac { base}{ perpendicular } =\dfrac{BD}{AD}=\dfrac { a }{ \sqrt { 3 } a }=\dfrac{1}{\sqrt3}$$.
So $$\dfrac { 1 }{ \sqrt { 3 } }$$ is correct.
In a right angle $$\triangle ABD, \angle B=60^o$$ and $$\angle A=30^o$$.
Then $$\sec { 60 }$$ is equal to:
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$$2$$
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$$\dfrac { 1 }{ 2 } $$
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$$\sqrt { 3 } $$
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$$\dfrac { 1 }{ \sqrt { 3 } } $$
Explanation
Consider an equilateral $$\triangle ABC$$, where each side $$= 2a$$.
From $$A$$, draw $$AD \perp BC$$,
so $$BD = DC = a$$, and $$\angle ADB=90^o, \angle BAD=30^o$$.
From right angle $$\triangle ADB$$, we have $$AD=\sqrt { { AB }^{ 2 }-{ BD }^{ 2 } } =\sqrt { { \left( 2a \right) }^{ 2 }-{ a }^{ 2 } } $$
$$=\sqrt { { 4a }^{ 2 }-{ a }^{ 2 } } =\sqrt { { 3a }^{ 2 } } =\sqrt { 3 } a$$.
$$\sec { 60^o } =\dfrac{hypotenuse}{base}=\dfrac{AB}{BD}=\dfrac{2a}{a}=2$$.
So, option A that is $$2$$ is correct.
In right angle $$\triangle ABC, \angle B=90^o, \angle A=45^o$$, then $$\sin { 45^o } =$$
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$$1$$
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$$\sqrt{2}$$
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$$\dfrac{1}{\sqrt{2}}$$
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None of above
Explanation
Given $$\angle A=45^o, \angle B=90^o$$
$$\therefore \angle C=90^o$$
$$\angle A = \angle C \Rightarrow AB = BC = a$$
$$AC = \sqrt{{AB}^{2}+{BC}^{2}} = \sqrt{{a}^{2}+{a}^{2}} = \sqrt{2{a}^{2}} = \sqrt{2} a$$
$$\sin{45^o} = \dfrac{BC}{AC} = \dfrac{a}{\sqrt{2}a} = \dfrac{1}{\sqrt{2}}$$
So, $$\dfrac{1}{\sqrt{2}}$$ is correct.
In a right angled $$\triangle ABD, \angle B=60^o$$ and $$\angle A=30^o$$.
Then $$\cos { 30^o }$$ is equal to:
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$$\dfrac { \sqrt { 3 } }{ 2 } $$
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$$\dfrac { 1 }{ \sqrt { 3 } } $$
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$$\dfrac { 2 }{ \sqrt { 3 } } $$
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$$\sqrt { 3 } $$
Explanation
Consider an equilateral $$\triangle ABC$$ with each side $$= 2a$$.
So, each angle of $$\triangle ABC=60^o$$.
From $$A$$ draw $$AD \perp BC$$, so $$BD=DC=a, \angle ADB=90^o, \angle BAD=30^o$$.
From right angled $$\triangle ADB$$, we have
$$AD=\sqrt { { AB }^{ 2 }-{ BD }^{ 2 } } =\sqrt { { \left( 2a \right) }^{ 2 }-{ a }^{ 2 } } =\sqrt { { 4a }^{ 2 }-{ a }^{ 2 } } =\sqrt { { 3a }^{ 2 } } =\sqrt { 3 } a$$.
$$\cos { 30^o } =\dfrac { AD }{ AB } =\dfrac { \sqrt { 3 } a }{ 2a } =\dfrac { \sqrt { 3 } }{ 2 }$$.
So, $$\dfrac { \sqrt { 3 } }{ 2 } $$ is correct.
In a right angled $$\triangle ABD, \angle B=60^o, \angle A=30^o$$.
Then $$\sin { 30^o }$$ is equal to
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$$2$$
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$$\dfrac { 1 }{ 2 } $$
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$$3$$
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$$\dfrac { 1 }{ 3 } $$
Explanation
Consider an equilateral $$\triangle ABC$$, with each side $$= 2a$$.
So, each angle of $$\triangle ABC=60^o$$.
From $$A$$ draw $$AD \perp BC$$. So $$BD = DC = a$$, $$\angle ADB=90^o, \angle BAD=30^o$$.
From right angled $$\triangle ADB$$, we have $$AD=\sqrt { { AB }^{ 2 }-{ BD }^{ 2 } } =\sqrt { { \left( 2a \right) }^{ 2 }-{ a }^{ 2 } }$$
$$ =\sqrt { { 4a }^{ 2 }-{ a }^{ 2 } } =\sqrt { { 3a }^{ 2 } } =\sqrt { 3 } a$$.
$$\sin { 30^o } =\dfrac { BD }{ AB } =\dfrac { a }{ 2a } =\dfrac { 1 }{ 2 }$$.
So, $$\dfrac { 1 }{ 2 } $$ is correct.
In right angled $$\triangle ABD, \angle B=60^o, \angle A=30^o$$.
Then $$\sec { 30^o}$$ is equal to
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$$\dfrac { 1 }{ \sqrt { 3 } } $$
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$$\dfrac { 2 }{ \sqrt { 3 } } $$
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$$\dfrac { 3 }{ \sqrt { 3 } } $$
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$$\dfrac { 4 }{ \sqrt { 3 } } $$
Explanation
Consider an equilateral $$\triangle ABC$$, with each side $$= 2a$$.
So, each angle is $$60^o$$.
From $$A$$, draw $$AD \perp BC$$.
So, $$BD=DC=a, \angle BAD=30^o$$ and $$\angle ADB=90^o$$.
So, for right angled $$\triangle ADB$$, we have
$$AD=\sqrt { { AB }^{ 2 }-{ BD }^{ 2 } } =\sqrt { { \left( 2a \right) }^{ 2 }-{ a }^{ 2 } } =\sqrt { { 4a }^{ 2 }-{ a }^{ 2 } } =\sqrt { { 3a }^{ 2 } } =\sqrt { 3 } a$$.
$$\sec { 30^o } =\dfrac { 1 }{ \cos { 30^o } } =\dfrac { 1 }{ \dfrac { \sqrt { 3 } }{ 2 } } =\dfrac{AB}{AD}=\dfrac { 2 }{ \sqrt { 3 } }$$.
So $$ \dfrac { 2 }{ \sqrt { 3 } } $$ is correct.
In a right angled $$\triangle ABD$$, in which $$\angle B=60^o$$ and $$\angle A=30^o$$.
Then $$\csc { 30 }^o$$ is equal to
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$$1$$
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$$2$$
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$$3$$
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$$4$$
Explanation
Consider an equilateral $$\triangle ABC$$, with each side $$= 2a$$.
So, each angle is $$60^o$$.
From $$A$$, draw $$AD \perp BC$$.
So, $$BD=DC=a, \angle ADB=90^o, \angle BAD=30^o$$.
So, for right angled $$\triangle ADB$$, we have
$$AD=\sqrt { { AB }^{ 2 }-{ BD }^{ 2 } } =\sqrt { { \left( 2a \right) }^{ 2 }-{ a }^{ 2 } } =\sqrt { { 4a }^{ 2 }-{ a }^{ 2 } } =\sqrt { { 3a }^{ 2 } } =\sqrt { 3 } a$$.
Now $$\csc { 30 } =\dfrac {AB}{BD} =\dfrac { 2a }{ a } =2$$.
In right angle $$\triangle ABD, \angle B={ 60 }^{ o }, \angle A={ 30 }^{ o }$$, then $$\cot { { 60 }^{ o } } =$$
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$$\sqrt { 3 } $$
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$$\dfrac { 1 }{ \sqrt { 3 } } $$
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$$\dfrac { 2 }{ \sqrt { 3 } } $$
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$$\dfrac{1}{3}$$
Explanation
in $$\triangle ABD$$
$$\cot60^o=\cot B=\dfrac{base}{perpendicular}=\dfrac{a}{\sqrt3a}=\dfrac{1}{\sqrt3}$$
Therefore, Answer is $$\dfrac{1}{\sqrt3}$$
In right angle $$\triangle ABC, \angle B=90^o, \angle A=45^o$$, then $$\cos { 45^o } =$$
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$$1$$
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$$\sqrt{2}$$
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$$\dfrac{1}{\sqrt{2}}$$
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None of above
Explanation
Given $$\angle A=45^o, \angle B=90^o$$
$$\therefore \angle C=45^o$$
$$\angle A = \angle C \Rightarrow AB = BC $$
Let $$AB=BC=a$$
$$AC = \sqrt{{AB}^{2}+{BC}^{2}} = \sqrt{{a}^{2}+{a}^{2}} = \sqrt{2{a}^{2}} = \sqrt{2} a$$
$$\cos{45} = \dfrac{AB}{AC} = \dfrac{a}{\sqrt{2}a} = \dfrac{1}{\sqrt{2}}$$
So $$\dfrac{1}{\sqrt{2}}$$ is correct.
Value of $$\left(\sin{60^o}\cos{30^o}-\cos{60^o}\sin{30^o}\right)$$ is
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$$0$$
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$$1$$
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$$\dfrac{1}{2}$$
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$$\dfrac{\sqrt{3}}{2}$$
Explanation
Given that: $$\left( \sin { 60^o } \cos { 30^o } -\cos { 60^o } \sin { 30^o } \right) $$
$$=\left( \dfrac { \sqrt { 3 } }{ 2 } \times \dfrac { \sqrt { 3 } }{ 2 } -\dfrac { 1 }{ 2 } \times \dfrac { 1 }{ 2 } \right) $$
$$=\left( \dfrac { 3 }{ 4 } -\dfrac { 1 }{ 4 } \right) $$
$$=\dfrac { 2 }{ 4 } =\dfrac { 1 }{ 2 } $$
In right angle $$\triangle ABC, \angle B=90^o, \angle A=45^o=\angle C$$, then $$cosec { 45^o } =$$
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$$1$$
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$$\sqrt{2}$$
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$$\dfrac{1}{\sqrt{2}}$$
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None of above
Explanation
$$\angle A = \angle C \Rightarrow AB = BC = a$$
$$AC = \sqrt{{AB}^{2}+{BC}^{2}} = \sqrt{{a}^{2}+{a}^{2}} = \sqrt{2{a}^{2}} = \sqrt{2} a$$
$$\csc{45^o} = \dfrac{hypotenuse}{base} =\dfrac{AC}{AB}= \dfrac{\sqrt2 a}{{a}} = \sqrt{2}$$
So, $$\sqrt{2}$$ is correct.
In right angle $$\triangle ABC, \angle B=90^o, \angle A=45^o=\angle C$$, then $$\cot { 45^o } =$$
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$$1$$
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$$\sqrt{2}$$
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$$\dfrac{1}{\sqrt{2}}$$
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None of above
Explanation
$$\angle A = \angle C \Rightarrow AB = BC = a$$
$$AC = \sqrt{{AB}^{2}+{BC}^{2}} = \sqrt{{a}^{2}+{a}^{2}} = \sqrt{2{a}^{2}} = \sqrt{2} a$$
$$\cos{45^o} = \dfrac{base}{perpendicular} =\dfrac{AB}{BC}= \dfrac{1}{1} = 1$$
So $$1$$ is correct.
In right angle $$\triangle ABC, \angle B=90^o, \angle A=45^o$$, then $$\tan { 45^o } =$$
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$$1$$
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$$\sqrt{2}$$
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$$\dfrac{1}{\sqrt{2}}$$
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None of above
Explanation
$$\angle A = \angle C \Rightarrow AB = BC = a$$
So $$AC = \sqrt{{AB}^{2}+{BC}^{2}} = \sqrt{{a}^{2}+{a}^{2}} = \sqrt{2{a}^{2}} = \sqrt{2} a$$
$$\tan{45} = \dfrac{BC}{AB} = \dfrac{a}{a} =1$$
So $$1$$ is correct.
In right angle $$\triangle ABC, \angle B=90^o, \angle A=45^o=\angle C$$, then $$\sec { 45^o} =$$
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$$1$$
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$$\sqrt{2}$$
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$$\dfrac{1}{\sqrt{2}}$$
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None of above
Explanation
$$\angle A = \angle C \Rightarrow AB = BC = a$$
$$AC = \sqrt{{AB}^{2}+{BC}^{2}} = \sqrt{{a}^{2}+{a}^{2}} = \sqrt{2{a}^{2}} = \sqrt{2} a$$
$$\sec{45^o} = \dfrac{hypotenuse}{base} =\dfrac{AC}{AB}= \dfrac{\sqrt2a}{a} = \sqrt{2}$$
So, $$\sqrt{2}$$ is correct.
If $${\tan}^{2}{45^o}- {\cos}^{2}{30^o} = x \sin{45^o} \cos{45^o}$$, then $$x =$$
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$$2$$
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$$-2$$
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$$\dfrac{1}{2}$$
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$$-\dfrac{1}{2}$$
Explanation
Given that: $$\tan ^{ 2 }{ 45^o } -\cos ^{ 2 }{ 30^o } =x\sin { 45^o } \cos { 45^o } $$
$$x=\dfrac { \tan ^{ 2 }{ 45^o } -\cos ^{ 2 }{ 30^o } }{ \sin { 45^o } \cos { 45^o } } $$
$$x=\dfrac { 1-{ \left( \dfrac { \sqrt { 3 } }{ 2 } \right) }^{ 2 } }{ \dfrac { 1 }{ \sqrt { 2 } } \times \dfrac { 1 }{ \sqrt { 2 } } } =\dfrac { 1-\dfrac { 3 }{ 4 } }{ \dfrac { 1 }{ 2 } } =\frac { \dfrac { 1 }{ 4 } }{ \dfrac { 1 }{ 2 } } =\dfrac { 1 }{ 4 } \times \dfrac { 2 }{ 1 } =\dfrac { 1 }{ 2 } $$
Value of $$\left(\cos{60^o}\cos{30^o}-\sin{60^o}\sin{30^o}\right)$$ is
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$$0$$
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$$\dfrac{\sqrt{3}}{2}$$
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$$\dfrac{1}{2}$$
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$$1$$
Explanation
$$\left( \cos { 60^0 } \cos { 30^0-\sin { 60^0 } \sin { 30^0 } } \right) $$
$$=\dfrac { 1 }{ 2 } \times \dfrac { \sqrt { 3 } }{ 2 } -\dfrac { \sqrt { 3 } }{ 2 } \times \dfrac { 1 }{ 2 } $$
$$=\dfrac { \sqrt { 3 } }{ 4 } -\dfrac { \sqrt { 3 } }{ 4 } $$
$$=0$$
In $$\triangle ABC, \angle B=90^o, AB=3\ cm, AC=6\ cm, \angle A=$$
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$$30^o$$
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$$45^o$$
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$$60^o$$
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None of these
Explanation
$$\sin{C} = \dfrac{AB}{AC} = \dfrac{3}{6} = \dfrac{1}{2}$$
But $$\sin{30^o} = \dfrac{1}{2}$$, so, $$\angle C = 30^o$$
$$\angle A = \left(90-\angle C\right)^o = \left(90-30\right)^o = 60^o$$.
So $$60^o$$ is correct.
In $$\triangle ABC, \angle B=90^o, \angle A=30^o, AB =9 cm\ $$, then $$BC =$$
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$$3 cm$$
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$$2\sqrt{3} cm$$
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$$3\sqrt{3} cm$$
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$$6 cm$$
Explanation
$$\tan{30^0} = \dfrac{BC}{AB}$$ and $$\tan{30^0} = \dfrac{1}{\sqrt{3}}$$
$$\therefore \dfrac{BC}{9} = \dfrac{1}{\sqrt{3}}$$, so $$BC = \dfrac{9}{\sqrt{3}} = 3\sqrt{3}$$.
So, $$3\sqrt{3}$$ is correct.
$$\sec (90 - \theta) = $$
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$$cosec \theta$$
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$$\sin \theta$$
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$$\cos \theta$$
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None of these
Explanation
In the figure,
$$\sec (90 - \theta) = \dfrac {r}{y} = \csc \theta$$
So, $$\sec (90 - \theta) = \csc \theta$$
So, $$\csc \theta$$ is correct.
$$(\cos 0^{\circ} + \sin 45^{\circ} + \sin 30^{\circ}) (\sin 90^{\circ} - \cos 45^{\circ} + \cos 60^{\circ}) = $$
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$$\dfrac {5}{4}$$
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$$\dfrac {9}{4}$$
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$$\dfrac {7}{4}$$
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None of these
Explanation
$$\left( \cos { { 0^o } + } \sin { { 45^o } + } \sin { { 30^o } } \right) \left( \sin { { 90^o- } } \cos { { 45^o+ } } \cos {{ 60^o } } \right) $$
$$=\left( 1+\dfrac { 1 }{ \sqrt { 2 } } +\dfrac { 1 }{ 2 } \right) \left( 1-\dfrac { 1 }{ \sqrt { 2 } } +\dfrac { 1 }{ 2 } \right) $$
$$=\left( 1+\dfrac { 1 }{ 2 } +\dfrac { 1 }{ \sqrt { 2 } } \right) \left( 1+\dfrac { 1 }{ 2 } -\dfrac { 1 }{ \sqrt { 2 } } \right) $$
$$=\left( \dfrac { 3 }{ 2 } +\dfrac { 1 }{ \sqrt { 2 } } \right) \left( \dfrac { 3 }{ 2 } -\dfrac { 1 }{ \sqrt { 2 } } \right) $$
$$={ \left( \dfrac { 3 }{ 2 } \right) }^{ 2 }-{ \left( \dfrac { 1 }{ \sqrt { 2 } } \right) }^{ 2 }$$
$$=\dfrac { 9 }{ 4 } -\dfrac { 1 }{ 2 } =\dfrac { 7 }{ 4 } $$
$$\cot (90 - \theta) = $$
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$$\sin \theta$$
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$$\cos \theta$$
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$$\tan \theta$$
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None of these
Explanation
$$\cot (90 - \theta) = \dfrac {y}{x} = \tan \theta$$
So, $$\cot (90 - \theta) = \tan \theta$$
So, $$\tan \theta$$ is correct.
If $$A$$ and $$B$$ are acute angle such that $$\sin A = \cos B$$, then $$(A + B) = .....$$
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$$45^{\circ}$$
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$$60^{\circ}$$
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$$90^{\circ}$$
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$$180^{\circ}$$
Explanation
$$\sin A=\cos B$$
$$\Rightarrow \sin A=\sin(90^o-B)$$
$$\Rightarrow A=90^o-B$$
$$\Rightarrow A+B=90^o$$
Therefore, Answer is $$90^o$$
$$4\left( \sin ^{ 4 }{ 30^0} +\cos ^{ 4 }{ 60^0 } \right) -\dfrac { 2 }{ 3 } \left( \sin ^{ 2 }{ 60^0 } -\cos ^{ 2 }{ 45^0 } \right) +\dfrac { 1 }{ 2 } \tan ^{ 2 }{ 60^0 } =$$
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$$\dfrac { 8 }{ 3 } $$
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$$\dfrac { 11 }{ 6 } $$
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$$\dfrac { 13 }{ 6 } $$
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$$\dfrac { 13 }{ 8 } $$
Explanation
$$4\left( \sin ^{ 4 }{ 30^0} +\cos ^{ 4 }{ 60^0 } \right) -\dfrac { 2 }{ 3 } \left( \sin ^{ 2 }{ 60^0 } -\cos ^{ 2 }{ 45^0 } \right) +\dfrac { 1 }{ 2 } \tan ^{ 2 }{ 60^0 } $$
$$ =4\left[ { \left( \dfrac { 1 }{ 2 } \right) }^{ 4 }+{ \left( \dfrac { 1 }{ 2 } \right) }^{ 4 } \right] -\dfrac { 2 }{ 3 } \left[ { \left( \dfrac { \sqrt { 3 } }{ 2 } \right) }^{ 2 }-{ \left( \dfrac { 1 }{ \sqrt { 2 } } \right) }^{ 2 } \right] +\dfrac { 1 }{ 2 } \times { \left( \sqrt { 3 } \right) }^{ 2 }$$
$$ =4\left( \dfrac { 1 }{ 16 } +\dfrac { 1 }{ 16 } \right) +\dfrac { 2 }{ 3 } \left( \dfrac { 3 }{ 4 } -\dfrac { 1 }{ 2 } \right) +\dfrac { 3 }{ 2 }$$
$$ =4\times \dfrac { 1 }{ 8 } -\dfrac { 2 }{ 3 } \times \dfrac { 1 }{ 4 } +\dfrac { 3 }{ 2 }$$
$$ =\dfrac { 1 }{ 2 } -\dfrac { 1 }{ 6 } +\dfrac { 3 }{ 2 }$$
$$ =\dfrac { 11 }{ 6 } $$
$$2\sin 30˚ + 2\tan 45˚ - 3\cos 60˚ - 2\cos^{2} 30˚ =$$
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0%
$$0$$
0%
$$1$$
0%
$$-1$$
0%
None of these
Explanation
$$2\sin 30˚ + 2\tan 45˚ - 3\cos 60˚ - 2\cos^{2} 30˚$$
$$= 2\times \dfrac {1}{2} + 2\times 1 - 3\times \dfrac {1}{2} - 2\left (\dfrac {\sqrt {3}}{2}\right )^{2}$$
$$= 1 + 2 - \dfrac {3}{2} - \dfrac {3}{2}$$
$$= 0$$
$$\sin 30^{\circ} \cos 60^{\circ} + \sin 60^{\circ} \cos 30^{\circ} $$ is equal to
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0%
$$\dfrac {1}{2}$$
0%
$$\dfrac {\sqrt {3}}{2}$$
0%
$$1$$
0%
$$\dfrac {1}{4}$$
Explanation
Given: $$\sin { { 30^o } } \cos { { 60^o } } +\sin { { 60 ^o } } \cos { { 30^o } } $$
(Putting the values as per trigonometry table)
$$=\dfrac { 1 }{ 2 } \times \dfrac { 1 }{ 2 } +\dfrac { \sqrt { 3 } }{ 2 } \times \dfrac { \sqrt { 3 } }{ 2 } $$
$$=\dfrac { 1 }{ 4 } +\dfrac { 3 }{ 4 } =\dfrac { 4 }{ 4 } =1$$
$$\dfrac { 1 }{ 4 } \left[ \cot ^{ 4 }{ 30^0 } -\csc ^{ 4 }{ 60^0 } \right] +\dfrac { 3 }{ 2 } \left[ \sec ^{ 2 }{ 45^0 } -\tan ^{ 2 }{ 30^0 } \right] -5\cos ^{ 2 }{ 60^0 } =$$
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$$\dfrac { 35 }{ 9 } $$
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$$\dfrac { 45 }{ 8 } $$
0%
$$\dfrac { 55 }{ 18 } $$
0%
$$\dfrac { 49 }{ 36 } $$
Explanation
$$\dfrac { 1 }{ 4 } \left[ \cot ^{ 4 }{ 30^0 } -\csc ^{ 4 }{ 60^0 } \right] +\dfrac { 3 }{ 2 } \left[ \sec ^{ 2 }{ 45^0 } -\tan ^{ 2 }{ 30^0 } \right] -5\cos ^{ 2 }{ 60^0 } =$$
$$ =\dfrac { 1 }{ 4 } \left[ { \left( \sqrt { 3 } \right) }^{ 4 }-{ \left( \dfrac { 2 }{ \sqrt { 3 } } \right) }^{ 4 } \right] +\dfrac { 3 }{ 2 } \left[ { \left( \sqrt { 2 } \right) }^{ 2 }-{ \left( \dfrac { 1 }{ \sqrt { 3 } } \right) }^{ 2 } \right] -5\times { \left( \dfrac { 1 }{ 2 } \right) }^{ 2 }$$, substitute the values from the table
$$ =\dfrac { 1 }{ 4 } \left( 9-\dfrac { 16 }{ 9 } \right) +\dfrac { 3 }{ 2 } \left( 2-\dfrac { 1 }{ 3 } \right) -\dfrac { 5 }{ 4 } $$
$$=\dfrac { 65 }{ 36 } +\dfrac { 5 }{ 2 } -\dfrac { 5 }{ 4 } =\dfrac { 110 }{ 36 } =\dfrac { 55 }{ 18 } $$
$$\tan (90 - \theta) $$ is equivalent to
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0%
$$\cos \theta$$
0%
$$\sin \theta$$
0%
$$(\sin \theta + \cos \theta)$$
0%
$$\cot \theta$$
Explanation
In the figure,
$$\tan (90^{\circ} - \theta) = \dfrac {x}{y}$$ and $$\cot \theta = \dfrac {x}{y}$$
So $$\tan (90 - \theta) = \cot \theta$$
So $$\cot \theta$$ is correct.
In $$\triangle ABC$$, if $$AD\perp BC$$ and $$BD = 10\ cm; \angle B = 60^{\circ}$$ and $$\angle C = 30^{\circ}$$, then $$CD = .........$$
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$$10\ cm$$
0%
$$20\ cm$$
0%
$$40\ cm$$
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$$30\ cm$$
Explanation
In right angled $$\triangle ABD$$,
$$\tan B = \dfrac {AD}{BD} = \dfrac {AD}{10}$$
$$\Rightarrow \tan 60^{\circ} = \dfrac {AD}{10}\Rightarrow \sqrt {3} = \dfrac {AD}{10}$$
$$\Rightarrow AD = 10\sqrt {3}$$
Also, in $$\triangle ACD$$,
$$\tan 30^{\circ} = \dfrac {AD}{DC} = \dfrac {1}{\sqrt {3}} \Rightarrow \dfrac {10\sqrt {3}}{DC} = \dfrac {1}{\sqrt {3}}$$
$$\Rightarrow DC = 30\ cm$$.
$$\cos^{2}45^0 + \sin^{2}60^0 + \sin^{2} 30^0 = $$
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0%
$$\dfrac {1}{2}$$
0%
$$\dfrac {3}{2}$$
0%
$$\dfrac {5}{2}$$
0%
$$\dfrac {7}{2}$$
Explanation
Taking
$$\cos^{2}45^0 + \sin^{2}60^0 + \sin^{2} 30^0 $$
Substituting the values of sin $$30 ^0$$ , sin $$60 ^0$$ , cos $$45 ^0$$
$$={ \left( \dfrac { 1 }{ \sqrt { 2 } } \right) }^{ 2 }+{ \left( \dfrac { \sqrt { 3 } }{ 2 } \right) }^{ 2 }+{ \left( \dfrac { 1 }{ 2 } \right) }^{ 2 }$$
$$=\dfrac { 1 }{ 2 } +\dfrac { 3 }{ 4 } +\dfrac { 1 }{ 4 } $$
$$=\dfrac { 6 }{ 4} $$
$$=\dfrac { 3 }{ 2 } $$
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