Introduction to Trigonometry Test

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Introduction to Trigonometry Test - 52

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Weekly Quiz Competition

Question 1
1 / -0

In right angle $$\triangle ABC, \angle B=90^o, \angle A=45^o$$, then $$\sin { 45^o } =$$

A
$$1$$

B
$$\sqrt{2}$$

C
$$\dfrac{1}{\sqrt{2}}$$

D
None of above

Solution

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.
Question 2
1 / -0

In right angle $$\triangle ABC, \angle B=90^o, \angle A=45^o$$, then $$\tan { 45^o } =$$

A
$$1$$

B
$$\sqrt{2}$$

C
$$\dfrac{1}{\sqrt{2}}$$

D
None of above

Solution

$$\angle A = \angle C \Rightarrow AB = BC = a$$

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.
Question 3
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In right angle $$\triangle ABC, \angle B=90^o, \angle A=45^o=\angle C$$, then $$\sec { 45^o} =$$

A
$$1$$

B
$$\sqrt{2}$$

C
$$\dfrac{1}{\sqrt{2}}$$

D
None of above

Solution

$$\angle A = \angle C \Rightarrow AB = BC = a$$

$$AC = \sqrt{{AB}^{2}+{BC}^{2}} = \sqrt{{a}^{2}+{a}^{2}} = \sqrt{2{a}^{2}} = \sqrt{2} a$$

$$\sec{45^o} = \dfrac{hypotenuse}{base} =\dfrac{AC}{AB}= \dfrac{\sqrt2a}{a} = \sqrt{2}$$
So, $$\sqrt{2}$$ is correct.
Question 4
1 / -0

In right angled $$\triangle ABD, \angle B=60^o, \angle A=30^o$$.
Then $$\sec { 30^o}$$ is equal to

A
$$\dfrac { 1 }{ \sqrt { 3 } } $$

B
$$\dfrac { 2 }{ \sqrt { 3 } } $$

C
$$\dfrac { 3 }{ \sqrt { 3 } } $$

D
$$\dfrac { 4 }{ \sqrt { 3 } } $$

Solution

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.
Question 5
1 / -0

In right angle $$\triangle ABD, \angle B={ 60 }^{ o }, \angle A={ 30 }^{ o }$$, then $$\cot { { 60 }^{ o } } =$$

A
$$\sqrt { 3 } $$

B
$$\dfrac { 1 }{ \sqrt { 3 } } $$

C
$$\dfrac { 2 }{ \sqrt { 3 } } $$

D
$$\dfrac{1}{3}$$

Solution

in $$\triangle ABD$$
$$\cot60^o=\cot B=\dfrac{base}{perpendicular}=\dfrac{a}{\sqrt3a}=\dfrac{1}{\sqrt3}$$

Therefore, Answer is $$\dfrac{1}{\sqrt3}$$
Question 6
1 / -0

In $$\triangle ABC, \angle B=90^o, AB=3\ cm, AC=6\ cm, \angle A=$$

A
$$30^o$$

B
$$45^o$$

C
$$60^o$$

D
None of these

Solution

$$\sin{C} = \dfrac{AB}{AC} = \dfrac{3}{6} = \dfrac{1}{2}$$

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.
Question 7
1 / -0

In right angle $$\triangle ABC, \angle B=90^o, \angle A=45^o$$, then $$\cos { 45^o } =$$

A
$$1$$

B
$$\sqrt{2}$$

C
$$\dfrac{1}{\sqrt{2}}$$

D
None of above

Solution

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.
Question 8
1 / -0

In $$\triangle ABC, \angle B=90^o, \angle A=30^o, AB =9 cm\ $$, then $$BC =$$

A
$$3 cm$$

B
$$2\sqrt{3} cm$$

C
$$3\sqrt{3} cm$$

D
$$6 cm$$

Solution

$$\tan{30^0} = \dfrac{BC}{AB}$$ 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.
Question 9
1 / -0

Value of $$\left(\sin{60^o}\cos{30^o}-\cos{60^o}\sin{30^o}\right)$$ is

A
$$0$$

B
$$1$$

C
$$\dfrac{1}{2}$$

D
$$\dfrac{\sqrt{3}}{2}$$

Solution

Given that: $$\left( \sin { 60^o } \cos { 30^o } -\cos { 60^o } \sin { 30^o } \right) $$

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

$$=\left( \dfrac { 3 }{ 4 } -\dfrac { 1 }{ 4 } \right) $$

$$=\dfrac { 2 }{ 4 } =\dfrac { 1 }{ 2 } $$
Question 10
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Value of $$\left(\tan{30^0}\csc{60^0} + \tan{60^0}\sec{30^0}\right)$$ is

A
$$2\dfrac{1}{3}$$

B
$$2\dfrac{2}{3}$$

C
$$3\dfrac{1}{3}$$

D
$$\dfrac{7}{4}$$

Solution

Taking
$$\left( \tan { 30^0 } \csc { 60^0+\tan { 60^0 } \sec { 30^0 } } \right) $$
Substituting the values of tan $$30^0$$ ,csc $$60^0$$, tan $$60^0$$,sec $$30^0$$
$$=\left( \dfrac { 1 }{ \sqrt { 3 } } \times \dfrac { 2 }{ \sqrt { 3 } } \right) +\left( \sqrt { 3 } \times \dfrac { 2 }{ \sqrt { 3 } } \right) $$
$$=\dfrac { 2 }{ 3 } +2$$ taking L.CM we get
$$=2\dfrac { 2 }{ 3 } $$