Introduction to Trigonometry Test

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Introduction to Trigonometry Test - 51

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Weekly Quiz Competition

Question 1
1 / -0

In a right angled $$\triangle ABD, \angle B=60^o$$ and $$\angle A=30^o$$.
Then, $$\cot { 60^o } =$$

A
$$\sqrt { 3 } $$

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

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

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

Solution

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.
Question 2
1 / -0

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

A
$$2$$

B
$$\dfrac { 1 }{ 2 } $$

C
$$\sqrt { 3 } $$

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

Solution

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.
Question 3
1 / -0

In a right angled $$\triangle ABD$$, in which $$\angle B=60^o$$ and $$\angle A=30^o$$.
Then $$\tan { 30^o }$$ is equal to

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

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

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

D
$$\sqrt { 3 } $$

Solution

Consider an equilateral $$\triangle ABC$$, with each side $$= 2a$$.

So, each angle is $$60^o$$. From $$A$$, draw $$AD \perp BC$$.

So, $$BD=DC=a, \angle ADB=90^o, \angle BAD=30^o$$.

So, for right angled $$\triangle ADB$$, we have

$$AD=\sqrt { { AB }^{ 2 }-{ BD }^{ 2 } } =\sqrt { { \left( 2a \right) }^{ 2 }-{ a }^{ 2 } } =\sqrt { { 4a }^{ 2 }-{ a }^{ 2 } } =\sqrt { { 3a }^{ 2 } } =\sqrt { 3 } a$$.

$$\tan { 30 } =\dfrac { BD }{ AD } =\dfrac { a }{ \sqrt { 3 } a } =\dfrac { 1 }{ \sqrt { 3 } }$$.

So, $$\dfrac { 1 }{ \sqrt { 3 } } $$ is correct.
Question 4
1 / -0

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

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

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

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

D
$$\sqrt { 3 } $$

Solution

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.
Question 5
1 / -0

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

A
$$2$$

B
$$\dfrac { 1 }{ 2 } $$

C
$$3$$

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

Solution

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.
Question 6
1 / -0

In a right angle $$\triangle ABD$$, in which $$\angle B = 60^o$$ and $$\angle A = 30^o$$.
Then, $$\tan{60^o}$$ is equal to

A
$$2$$

B
$$\dfrac{1}{2}$$

C
$$\sqrt{3}$$

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

Solution

Consider an equilateral $$\triangle ABC$$ with each side $$= 2a$$.
So each angle is $$60$$.
From $$A$$, draw $$\perp AD$$ to $$BC$$,
so $$BD = DC = a$$ and $$\angle ADB = 90^o, \angle BAD = 30^o$$.
So, for $$\triangle ADB$$, we have $$AD = \sqrt{{AB}^{2} - {BD}^{2}} $$
$$= \sqrt{{\left(2a\right)}^{2}-{a}^{2}} = \sqrt{4{a}^{2}-{a}^{2}} = \sqrt{3{a}^{2}} = \sqrt{3}a$$.
$$\therefore \tan 60^o=\dfrac{AD}{BD}=\dfrac{\sqrt 3 a}{a}=\sqrt 3$$
Question 7
1 / -0

In a right angle $$\triangle ABD, \angle B=60^o$$ and $$\angle A=30^o$$.
Then $$\csc{60^o}$$ is equal to:

A
$$\sqrt { 3 } $$

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

C
$$2$$

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

Solution

Consider an equilateral $$\triangle ABC$$ with each side $$= 2a$$.
Each angle is $$60^o$$.
From $$A$$, draw $$AD \perp BC$$, then $$BD = DC = a$$.
Also, $$\angle ADB=90^o, \angle BAD=30^o$$.
From right angle $$\triangle ADB$$, we have
$$AD=\sqrt { { AB }^{ 2 }-{ BD }^{ 2 } } =\sqrt { { \left( 2a \right) }^{ 2 }-{ a }^{ 2 } }$$
$$ =\sqrt { { 4a }^{ 2 }-{ a }^{ 2 } } =\sqrt { { 3a }^{ 2 } } =\sqrt { 3 } a$$

$$\csc { 60 } =\dfrac { 1 }{ \sin { 60 } } =\dfrac { 1 }{ \dfrac { \sqrt { 3 } }{ 2 } } =\dfrac { 2 }{ \sqrt { 3 } }$$
So $$\dfrac { 2 }{ \sqrt { 3 } } $$ is correct.
Question 8
1 / -0

In a right angled $$\triangle ABD$$, in which $$\angle B=60^o$$ and $$\angle A=30^o$$.
Then $$\csc { 30 }^o$$ is equal to

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

Consider an equilateral $$\triangle ABC$$, with each side $$= 2a$$.

So, each angle is $$60^o$$.
From $$A$$, draw $$AD \perp BC$$.

So, $$BD=DC=a, \angle ADB=90^o, \angle BAD=30^o$$.

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 $$\csc { 30 } =\dfrac {AB}{BD} =\dfrac { 2a }{ a } =2$$.
Question 9
1 / -0

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

A
$$1$$

B
$$\sqrt{2}$$

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

D
None of above

Solution

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

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

$$\csc{45^o} = \dfrac{hypotenuse}{base} =\dfrac{AC}{AB}= \dfrac{\sqrt2 a}{{a}} = \sqrt{2}$$

So, $$\sqrt{2}$$ is correct.
Question 10
1 / -0

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

A
$$1$$

B
$$\sqrt{2}$$

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

D
None of above

Solution

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

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

$$\cos{45^o} = \dfrac{base}{perpendicular} =\dfrac{AB}{BC}= \dfrac{1}{1} = 1$$

So $$1$$ is correct.