Introduction to Trigonometry Test

Result

Introduction to Trigonometry Test - 50

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

$$\sec^2\theta =$$

A
$$1 - \cos^2\theta$$

B
$$1 - \tan^2\theta$$

C
$$1 + \tan^2\theta$$

D
$$1 + \cot^2\theta$$

Solution

We know that $$\sin ^2 \theta + \cos ^2 \theta = 1$$

Dividing both sides by $$\cos ^2 \theta $$,

$$ \displaystyle \frac {\sin ^2 \theta}{ \cos ^2 \theta}+ \displaystyle \frac {\cos ^2 \theta}{\cos ^2 \theta} = \frac {1}{ \cos ^2 \theta} $$

$$ \tan ^2 \theta + 1 = \sec ^2 \theta$$
Question 2
1 / -0

If $$\displaystyle X=\tan 1^{0}+\tan 2^{0}+........+\tan 45^{0}$$ and $$\displaystyle y= -(\cot 46^{0}+\cot 47^{0}+.......+\cot 89^{0})$$ then find the value of $$(x + y)$$.

A
$$1$$

B
$$0$$

C
$$-1$$

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

Solution

Since $$ \tan (90^o - x) = \cot 90^0 $$ we have
$$ x + y = \tan 1^o + \tan 2^o +....... + \tan 45^0 - (\cot 46^0 + \cot 47^0 + ... + \cot 89 ^0 ) $$
$$ =>x+y= \tan 1^o + \tan 2^o +.... + \tan 45^0 - (\cot (90^0 - 44^0) + \cot(90^0-43^0) + .. +\cot (90^0 - 1 ^0 ) $$
$$ =>x+y= \tan 1^o + \tan 2^o +.... + \tan 45^0 - (\tan 44^0 + \tan43^0 + .. + \tan 1 ^0 ) $$
$$ => x+y = 0 + 0 + .. + \tan 45^0 = 1 $$
Question 3
1 / -0

In a right angled triangle, $$\angle A = \theta$$ is an acute angle.
Then, $$\cos{\theta} \times \sec{\theta}$$ is equal to

A
$$0$$

B
$$1$$

C
$$-1$$

D
None of these

Solution

$$\cos{\theta} = \dfrac{x}{r}$$ and $$\sec{\theta} = \dfrac{r}{x}$$

$$\therefore \cos{\theta} \times \sec{\theta} = \dfrac{x}{r} \times \dfrac{r}{x} = 1$$
Question 4
1 / -0

If $$\displaystyle \cos \Theta _{1}+\cos \Theta _{2}+\cos \Theta _{3}=3,$$ find $$\displaystyle \sin \Theta _{1}+\sin \Theta _{2}+\sin \Theta _{3}.$$

A
$$0$$

B
$$1$$

C
$$2$$

D
$$3$$

Solution

Maximum value of a cosine of an angle is $$ 1 $$

Hence, when sum of three cosines $$ = 3 $$ , this means each cosine $$ = 1 $$
And each angle $$ = 0^{o} $$

So $$ \sin 0 + \sin 0 + \sin 0 = 0 $$
Question 5
1 / -0

If in a rectangular, the angle between a diagonal and a side is 30$$^o$$ and the length of the diagonal is $$8$$ cm, then the area of the rectangle is _____ .

A
$$16$$$$\sqrt3$$ cm$$^2$$

B
$$16$$ cm$$^2$$

C
$$\sqrt3$$ cm$$^2$$

D
16/$$\sqrt3$$ cm$$^2$$

Solution

Referring figure, let ABCD be a rectangular in which $$\angle$$BAC = 30$$^o$$ and AC = 8 cm

Then, cos30$$^o$$ = $$\displaystyle{\frac{AB}{AC} \Rightarrow \frac{\sqrt3}{2} = \frac{AB}{8} \Rightarrow}$$ AB = 4$$\sqrt3$$ cm

and. sin 30$$^o$$ = $$\displaystyle{\frac{BC}{AC} \Rightarrow \frac{1}{2} = \frac{BC}{8} \Rightarrow}$$ BC = 4 cm

$$\therefore$$ Area of rectangular = AB x BC
=4$$\sqrt3$$ x 4
= 16$$\sqrt3$$ cm$$^2$$

So, option A is correct.
Question 6
1 / -0

The value of $$\displaystyle \sin ^{2}5^{\circ}+ \sin ^{2}10^{\circ}+\sin ^{2}15^{\circ}......+\sin ^{2}90^{\circ}$$ is equal to

A
$$\dfrac{17}2$$

B
$$\dfrac{19}2$$

C
$$\dfrac{21}2$$

D
$$\dfrac{23}2$$

Solution

$$\displaystyle \sin ^{2}5^{\circ}+ \sin ^{2}10^{\circ}+\sin ^{2}15^{\circ}......+\sin ^{2}90^{\circ}$$

$$=(\sin^25^o+\sin^285^o)+(\sin^210^o+\sin^2{80^o})+...........+(\sin^245^o+\sin^245^o)+\sin^290^o-\sin^245^o$$

$$=(\sin^25^o+\cos^25^o)+(\sin^210^o+\cos^2{10^o})+...........+(\sin^245^o+\cos^245^o)+\sin^290^o-\sin^245^o$$

$$ = 1+1+1+.......+1-\dfrac{1}{2}$$

$$=10-\dfrac{1}{2}=\dfrac{19}{2}$$
Question 7
1 / -0

If $$\displaystyle 3\cos ^{2}A=\cos 60^{\circ}+\sin ^{2}45^{\circ} $$ then find the value of $$\displaystyle \sec ^{2}A$$.

A
$$1$$

B
$$0$$

C
$$2$$

D
$$3$$

Solution

$$ 3\cos^2 A = \cos 60^0 + \sin^2 45 $$

$$ \Rightarrow 3\cos^2 A = \dfrac {1}{2} + (\dfrac {1}{\sqrt {2}})^2 $$

$$ \Rightarrow 3\cos^2 A = \dfrac {1}{2} + \dfrac {1}{2} = 1 $$

$$ \Rightarrow \cos^2 A= \dfrac {1}{3} $$

Or $$ \sec^2 A = \dfrac {1}{\cos^2 A} = 3 $$
Question 8
1 / -0

If $$\displaystyle25\sin ^{2}\theta +10\cos ^{2}\theta =15 $$ then find $$\displaystyle\cot ^{2}\theta $$

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

$$ 25\sin ^{ 2 }{ \theta } +10\cos ^{ 2 }{ \theta } = 15 $$
$$ \Rightarrow 15\sin ^{ 2 }{ \theta } + 10\sin ^{ 2 }{ \theta } + 10\cos ^{ 2 }{ \theta } = 15 $$
$$ \Rightarrow 15\sin ^{ 2 }{ \theta } + 10(\sin ^{ 2 }{ \theta } + \cos ^{ 2 }{ \theta } ) = 15 $$
$$ \Rightarrow 15\sin ^{ 2 }{ \theta } +10 = 15 $$ (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, $$ \cot ^{ 2 }{ \theta } = \text{cosec}\, ^{ 2 }{ \theta }-1 = 3 - 1 = 2 $$
Question 9
1 / -0

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

A
$$2$$

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

C
$$\sqrt{3}$$

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

Solution

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

Each angle of $$\triangle ABC$$ is $$60^o$$.

From $$A$$, draw $$AD \perp BC$$, then $$BD = DC = a$$.

Also, $$\angle ADB = 90^o, \angle BAD = 30^o$$.

From right angled $$\triangle ADB$$, we have

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

$$= \sqrt{4{a}^{2}-{a}^{2}} = \sqrt{3{a}^{2}} = \sqrt{3}a$$.

Now, in right angled $$\triangle ADB$$, base $$BD = a, \, perpendicular = AD = \sqrt{3}a$$

and hypotenuse$$ = AB = 2a$$

$$\cos{60^o} = \dfrac{BD}{AB} = \dfrac{a}{2a} = \dfrac{1}{2}$$

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

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

A
$$\sqrt{3}$$

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

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

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

Solution

Consider an equilateral triangle with each side $$= 2a$$.
So each angle of $$\triangle ABC$$ is $$60^o$$.
From $$A$$ draw $$AD \perp BC$$ then $$BD = DC = a$$.
Also, $$\angle ADB = 90^o$$ and $$\angle BAD = 30^o$$

From right angled triangle $$\triangle ADB$$, we have
$$AD = \sqrt{{AB}^{2}-{BD}^{2}} = \sqrt{{\left(2a\right)}^{2}-{\left(a\right)}^{2}}$$
$$ = \sqrt{4{a}^{2} - {a}^{2}} = \sqrt{3{a}^{2}} = \sqrt{3}a$$

Now, in right angled $$\triangle ADB$$, base $$BD = a, AD = \sqrt{3}a$$
and hypotenuse $$= AB = 2a$$
$$\sin{60^o} = \dfrac{AD}{AB} = \dfrac{\sqrt{3} \times a}{2 \times a} = \dfrac{\sqrt{3}}{2}$$

So, $$\dfrac{\sqrt{3}}{2}$$ is correct.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Introduction to Trigonometry Test - 50

Result

Introduction to Trigonometry Test - 50

Score

-

Rank

-

SHARING IS CARING

Question 1

$$1 - \cos^2\theta$$

$$1 - \tan^2\theta$$

$$1 + \tan^2\theta$$

$$1 + \cot^2\theta$$

Solution

Question 2

$$1$$

$$0$$

$$-1$$

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

Solution

Question 3

$$0$$

$$1$$

$$-1$$

None of these

Solution

Question 4

$$0$$

$$1$$

$$2$$

$$3$$

Solution

Question 5

$$16$$$$\sqrt3$$ cm$$^2$$

$$16$$ cm$$^2$$

$$\sqrt3$$ cm$$^2$$

16/$$\sqrt3$$ cm$$^2$$

Solution

Question 6

$$\dfrac{17}2$$

$$\dfrac{19}2$$

$$\dfrac{21}2$$

$$\dfrac{23}2$$

Solution

Question 7

$$1$$

$$0$$

$$2$$

$$3$$

Solution

Question 8

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 9

$$2$$

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

$$\sqrt{3}$$

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

Solution

Question 10

$$\sqrt{3}$$

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

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

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

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.