Introduction to Trigonometry Test

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Introduction to Trigonometry Test - 38

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Question 1
1 / -0

Directions For Questions

In the diagram, given below, triangle ABC is right-angled at B and BD is perpendicular to AC. Find :

...view full instructions

$$cot\, \angle DBA$$

A
$$\displaystyle \frac{3}{11}$$

B
$$\displaystyle \frac{11}{17}$$

C
$$\displaystyle \frac{7}{9}$$

D
$$\displaystyle \frac{5}{12}$$

Solution

In $$\triangle ABC$$
$$AB^2 + BC^2 = AC^2$$ (By Pythagoras theorem)
$$12^2+ 5^2 = AC^2$$
$$AC^2 = 169$$
$$AC =13$$

In $$\triangle ABC$$
$$\angle A + \angle B + \angle C = 180$$
$$\angle A + \angle C = 90$$ (As $$\angle B = 90$$).....(i)

In $$\triangle ABD$$,
$$\angle A + \angle BDA + \angle DBA = 180$$
$$\angle A + \angle DBA = 90$$ (As $$\angle BDA = 90$$).....(i)

From (i) and (ii)
$$\angle C = \angle DBA$$
Similarly, $$\angle DBC = \angle A$$

Now, $$\cot \angle DBA = \cot \angle C = \frac{B}{P} = \frac{BC}{AB}= \frac{5}{12}$$
Question 2
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If $$ \tan A + \cot A = 5$$; find the value if $$\tan^{2}\, A\, +\, \cot ^{2} A$$.

A
$$14$$

B
$$27$$

C
$$23$$

D
$$335$$

Solution

$$\tan A + \cot A = 5$$
Squaring both sides,
$$(\tan A + \cot A)^2 = 25$$
$$\tan^2 A + \cot^2 A + 2 \tan A \cot A = 25$$
$$\tan^2 A + \cot^2 A + 2 = 25$$
$$\tan^2 A + \cot^2 A = 23$$
Question 3
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Directions For Questions

$$x=co \sec^2\theta, y=\sec^2\theta, z=\displaystyle \cfrac {1}{1-\sin^2\theta \cos^2\theta}$$

...view full instructions

$$\displaystyle \frac {1}{x^2}+\frac {1}{y^2}$$ is equal to

A
$$\displaystyle \frac {2+z}{z}$$

B
$$\displaystyle \frac {2-z}{z}$$

C
$$\displaystyle \frac {z}{2+z}$$

D
$$\displaystyle \frac {z}{2-z}$$

Solution

$$\cfrac { 1 }{ x^{ 2 } } +\cfrac { 1 }{ y^{ 2 } } =\cfrac { 1 }{ co \sec^{ 4 }\theta } +\cfrac { 1 }{ \sec^{ 4 }\theta } =\sin^{ 4 }\theta +\cos^{ 4 }\theta \\ =1-2\sin^{ 2 }\theta \cos^{ 2 }\theta =1-2\left( 1-\cfrac { 1 }{ z } \right) =\cfrac { 2-z }{ z } $$
Question 4
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If $$3 \sin\theta+ 5 \cos\theta=5$$, then $$5 \sin\theta-3 \cos\theta$$ is equal to

A
$$3$$

B
$$8$$

C
$$13$$

D
none of these

Solution

Given $$3 sin\theta+ 5 cos\theta=5$$
Squaring both sides, we get
$$9\sin ^{ 2 }{ \theta } +25\cos ^{ 2 }{ \theta } +30\sin { \theta } \cos { \theta } =25$$
$$\Rightarrow 9-9\cos ^{ 2 }{ \theta } +25-25\sin ^{ 2 }{ \theta } +30\sin { \theta } \cos { \theta } =25$$
$$\Rightarrow 9\cos ^{ 2 }{ \theta } +25\sin ^{ 2 }{ \theta } -30\sin { \theta } \cos { \theta } =9$$
$$\Rightarrow { (5\sin { \theta } -3\cos { \theta } ) }^{ 2 }=9$$
$$\Rightarrow 5\sin { \theta } -3\cos { \theta } =\pm 3$$
Question 5
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$$\displaystyle \left (\frac{\sin\, 50^{\circ}}{\cos\, 40^{\circ}} \right)^{2}\, +\, \left (\frac{\cos\, 28^{\circ}}{\sin\, 62^{\circ}} \right)^{2}\, -\, 2\, \tan^{2}\, 45^{\circ}$$

A
$$0$$

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

C
$$1$$

D
$$-2$$

Solution

$$\displaystyle \left (\frac{\sin\, 50^{\circ}}{\cos\, 40^{\circ}} \right)^{2}\, +\, \left (\frac{\cos\, 28^{\circ}}{\sin\, 62^{\circ}} \right)^{2}\, -\, 2\, \tan^{2}\, 45^{\circ}$$

= $$\displaystyle \left (\frac{\sin\, (90 - 40)^{\circ}}{\cos\, 40^{\circ}} \right)^{2}\, +\, \left (\frac{\cos\, 28^{\circ}}{\sin\, (90 - 28)^{\circ}} \right)^{2}\, -\, 2\, \tan^{2}\, 45^{\circ}$$

= $$\displaystyle \left (\frac{\cos\, 40^{\circ}}{\cos\, 40^{\circ}} \right)^{2}\, +\, \left (\frac{\cos\, 28^{\circ}}{\cos\, 28^{\circ}} \right)^{2}\, -\, 2\, \tan^{2}\, 45^{\circ}$$

= $$1 + 1 - 2 (1)^2$$

= $$0$$
Question 6
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If $$x^2\tan^260^o-4x \cos^2 45^o=4 \sin 30^o \cos 60^o \tan 45^o$$ and $$x$$ is not an integer then $$x$$ is equal to:

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

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

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

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

Solution

We know $$ \tan60^\circ = \sqrt{3}, \cos45^\circ =\dfrac {1}{\sqrt{2}}, \sin30^\circ =\dfrac 12, \cos60^\circ =\dfrac 12, \tan45^\circ = 1$$

Using these given equation becomes,

$$(\sqrt3x)^2-4x\cdot \left(\dfrac {1}{\sqrt2}\right)^2=4\cdot\dfrac 12\cdot \dfrac 12 \cdot 1$$

$$ 3 x^2 - 2x = 1 \Rightarrow 3x^2 - 2x - 1 =0$$

$$\Rightarrow (3x + 1) (x - 1) = 0$$

But given x is not an integer,

Thus, $$ x = -\dfrac{1}{3}$$ is our solution.
Question 7
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If $$tan\, x^{\circ}\, =\, \dfrac{5}{12},\, tan\, y^{\circ}\, =\, \dfrac{3}{4}$$ and $$AB = 48 m$$; find the length of $$CD$$.

A
$$50m$$

B
$$45m$$

C
$$40m$$

D
$$55m$$

Solution

$$\tan x^o = \dfrac{5}{12}$$
$$\tan y^o = \dfrac{3}{4}$$
$$AB = 48m$$
Let $$CD = x\ m$$

From $$\Delta ACD,$$
$$\tan x^o = \dfrac{CD}{AC}$$
$$ \Rightarrow \dfrac{5}{12}=\dfrac{x}{AC}$$
$$\Rightarrow AC = \dfrac{12 x}{5}$$

And from $$\Delta BCD,$$
$$\tan y^o = \dfrac{CD}{BC}$$
$$\Rightarrow \dfrac{3}{4}=\dfrac{x}{BC}$$
$$\Rightarrow BC = \dfrac{4 x}{3}$$

From diagram,
$$AB = AC - BC$$
$$\Rightarrow 48 = \dfrac{12 x}{5} - \dfrac{4x}{3}$$
$$\Rightarrow48 = \dfrac{36 x - 20 x}{15}$$
$$\Rightarrow16 x = 48\times 15$$
$$\Rightarrow x = 45$$

Thus, $$CD = 45$$ m
Question 8
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If $$\sin A = \displaystyle\frac{3}{4}$$, calculate $$\cos A$$ and $$\tan A$$.

A
$$\cos A = \displaystyle\frac{\sqrt3}{4} \\ \tan A = \displaystyle\frac{2}{5}$$

B
$$\cos A = \displaystyle\frac{\sqrt7}{4} \\ \tan A = \displaystyle\frac{3}{\sqrt7}$$

C
$$\cos A = \displaystyle\frac{\sqrt3}{2} \\ \tan A = \displaystyle\frac{2}{\sqrt5}$$

D
$$\cos A = \displaystyle\frac{\sqrt5}{7} \\ \tan A = \displaystyle\frac{7}{\sqrt3}$$

Solution

Given, $$ \sin A = \displaystyle\frac{3}{4}$$
$$ \Rightarrow \displaystyle\frac{BC}{DC} = \displaystyle\frac{3}{4}$$
$$ \Rightarrow BC = 3k$$ and $$AC = 4k$$
where $$k$$ is the constant of proportionality.
By Pythagoras theorem, we have
$$AB^2 = AC^2 - BC^2 = (4k)^2 - (3k)^2 = 7k^2$$
$$ \Rightarrow AB = \sqrt7k$$
So, $$ \cos A = \displaystyle\frac{AB}{AC} = \displaystyle\frac{\sqrt7k}{4k} = \displaystyle\frac{\sqrt7}{4}$$
And $$ \tan A = \displaystyle\frac{BC}{AB} = \displaystyle\frac{3k}{\sqrt7k} = \displaystyle\frac{3}{\sqrt7}$$
Question 9
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Evaluate
$$(i)\quad \sin60^{\small\circ}\cos30^{\small\circ}+\sin30^{\small\circ}\cos60^{\small\circ}$$

$$(ii)\quad \displaystyle\frac{5\cos^260^{\small\circ}+4\sec^230^{\small\circ}-\tan^245^{\small\circ}}{\sin^230^{\small\circ}+\cos^230^{\small\circ}}$$

A
$$(i)\quad 7 \\ (ii)\quad \displaystyle\frac{5}{32}$$

B
$$(i)\quad 2 \\ (ii)\quad \displaystyle\frac{33}{8}$$

C
$$(i)\quad 1 \\ (ii)\quad \displaystyle\frac{67}{12}$$

D
$$(i)\quad 5 \\ (ii)\quad \displaystyle\frac{6}{47}$$

Solution

$$(i)\quad \sin60^{\small\circ}\cos30^{\small\circ}+\sin30^{\small\circ}\cos60^{\small\circ}$$
$$\quad = \left(\displaystyle\frac{\sqrt3}{2}\right)\left(\displaystyle\frac{\sqrt3}{2}\right) + \left(\displaystyle\frac{1}{2}\right)\left(\displaystyle\frac{1}{2}\right) = \left(\displaystyle\frac{\sqrt3}{2}\right)^2 + \left(\displaystyle\frac{1}{2}\right)^2$$
$$\quad = \displaystyle\frac{3}{4}+\displaystyle\frac{1}{4} = 1$$

$$(ii)\quad \displaystyle\frac{5\cos^260^{\small\circ}+4\sec^230^{\small\circ}-\tan^245^{\small\circ}}{\sin^230^{\small\circ}+\cos^230^{\small\circ}}$$

$$ = \quad \displaystyle\frac{5(\cos60^{\small\circ})^2+4(\sec30^{\small\circ})^2-(\tan45^{\small\circ})^2}{(\sin30^{\small\circ})^2+(\cos30^{\small\circ})^2}$$

$$\quad = \displaystyle\frac{5\left(\displaystyle\frac{1}{2}\right)^2+4\left(\displaystyle\frac{2}{\sqrt3}\right)^2 - (1)^2}{\left(\displaystyle\frac{1}{2}\right)^2 + \left(\displaystyle\frac{\sqrt3}{2}\right)^2}$$

$$\quad = \displaystyle\frac{\displaystyle\frac{5}{4}+4\times\displaystyle\frac{4}{3} - 1}{\displaystyle\frac{1}{4} + \displaystyle\frac{3}{4}} = \displaystyle\frac{\displaystyle\frac{5}{4}+\displaystyle\frac{16}{3}-1}{\displaystyle\frac{1}{4}+\displaystyle\frac{3}{4}} = \displaystyle\frac{5}{4} + \displaystyle\frac{16}{3} - 1= \displaystyle\frac{15+64-12}{12} = \displaystyle\frac{67}{12}$$
Question 10
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If, $$\displaystyle \tan \, 9^o = \frac{x}{y}$$ then, value of $$\displaystyle \frac{\sec^2\, 81^o}{1+\cot^2\, 81^o}$$ is

A
$$\displaystyle \frac{x^3}{y^3}$$

B
$$\displaystyle \frac{x^4}{y^4}$$

C
$$\displaystyle \frac{x^5}{y^5}$$

D
$$\displaystyle \frac{y^2}{x^2}$$

Solution

Given,
$$\tan\, 9^{ o }=\dfrac { x }{ y } $$

$$=\dfrac{\sec^2 \, 81^o}{1 + \cot^2 \, 81^o}$$

$$= \dfrac{\sec^2 \, 81^o}{cosec^2 \, 81^o}$$

$$=\dfrac { \dfrac { 1 }{ \cos^{ 2 }81^{ o } } }{ \dfrac { 1 }{ \sin^{ 2 }81^{ o } } \, }$$

$$=\dfrac { \sin^{ 2 }81^{ o } }{ \cos^{ 2 }\, 81^{ o } } $$

$$=\tan^2 \, 81^o$$

$$=[\tan(90^o-9^o)]^2$$

$$= (\cot\, 9^o)^2$$

$$= \cot^2\, 9^o$$

$$=\dfrac{1}{\tan^2 \, 9^o}$$

$$=\dfrac{y^2}{x^2}$$
Hence, option 'D' is correct.

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Re-Attempt Weekly Quiz Competition

Introduction to Trigonometry Test - 38

Result

Introduction to Trigonometry Test - 38

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SHARING IS CARING

Question 1

$$\displaystyle \frac{3}{11}$$

$$\displaystyle \frac{11}{17}$$

$$\displaystyle \frac{7}{9}$$

$$\displaystyle \frac{5}{12}$$

Solution

Question 2

$$14$$

$$27$$

$$23$$

$$335$$

Solution

Question 3

$$\displaystyle \frac {2+z}{z}$$

$$\displaystyle \frac {2-z}{z}$$

$$\displaystyle \frac {z}{2+z}$$

$$\displaystyle \frac {z}{2-z}$$

Solution

Question 4

$$3$$

$$8$$

$$13$$

none of these

Solution

Question 5

$$0$$

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

$$1$$

$$-2$$

Solution

Question 6

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

$$\dfrac{-1}{3}$$

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

$$-\dfrac{2}{3}$$

Solution

Question 7

$$50m$$

$$45m$$

$$40m$$

$$55m$$

Solution

Question 8

$$\cos A = \displaystyle\frac{\sqrt3}{4} \\ \tan A = \displaystyle\frac{2}{5}$$

$$\cos A = \displaystyle\frac{\sqrt7}{4} \\ \tan A = \displaystyle\frac{3}{\sqrt7}$$

$$\cos A = \displaystyle\frac{\sqrt3}{2} \\ \tan A = \displaystyle\frac{2}{\sqrt5}$$

$$\cos A = \displaystyle\frac{\sqrt5}{7} \\ \tan A = \displaystyle\frac{7}{\sqrt3}$$

Solution

Question 9

$$(i)\quad 7 \\ (ii)\quad \displaystyle\frac{5}{32}$$

$$(i)\quad 2 \\ (ii)\quad \displaystyle\frac{33}{8}$$

$$(i)\quad 1 \\ (ii)\quad \displaystyle\frac{67}{12}$$

$$(i)\quad 5 \\ (ii)\quad \displaystyle\frac{6}{47}$$

Solution

Question 10

$$\displaystyle \frac{x^3}{y^3}$$

$$\displaystyle \frac{x^4}{y^4}$$

$$\displaystyle \frac{x^5}{y^5}$$

$$\displaystyle \frac{y^2}{x^2}$$

Solution

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