Introduction to Trigonometry Test

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Introduction to Trigonometry Test - 53

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Weekly Quiz Competition

Question 1
1 / -0

$${\cos}^{2}{30^\circ}\ {\cos}^{2}{45^\circ}+4\ {\sec}^{2}{60^\circ}+\dfrac{1}{2}\ {\cos}^{2}{90^\circ}-2{\tan}^{2}{60^\circ} =$$

A
$$\dfrac{73}{8}$$

B
$$\dfrac{75}{8}$$

C
$$\dfrac{81}{8}$$

D
$$\dfrac{83}{8}$$

Solution

Taking $$\cos ^{ 2 }{ 30^\circ } \cos ^{ 2 }{ 45^\circ } +4\sec ^{ 2 }{ 60^\circ } +\frac { 1 }{ 2 } \cos ^{ 2 }{ 90^\circ } -2\tan ^{ 2 }{ 60^\circ } $$

substitute the values of $$\cos 30^\circ$$, $$\cos 45^\circ$$, $$\cos 90^\circ$$, $$\sec 60^\circ$$, $$\tan 60^\circ$$

$$={ \left( \dfrac { \sqrt { 3 } }{ 2 } \right) }^{ 2 }\times { \left( \dfrac { 1 }{ \sqrt { 2 } } \right) }^{ 2 }+4\times { \left( 2 \right) }^{ 2 }+\dfrac { 1 }{ 2 } \times { 0 }^{ 2 }-[2\times { \left( \sqrt { 3 } \right) }^{ 2 }]$$

$$=\bigg(\dfrac { 3 }{ 4 } \times \dfrac { 1 }{ 2 }\bigg) +16-6$$

$$=\dfrac { 3 }{ 8 } +10$$

$$=\dfrac { 3+80 }{ 8 } $$

$$=\dfrac { 83 }{ 8 } $$
Question 2
1 / -0

Value of $$\left(3{\cos}^{2}{60^0}+2{\cot}^{2}{30^0}-5{\sin}^{2}{45^0}\right)$$ is:

A
$$\dfrac{13}{6}$$

B
$$\dfrac{17}{4}$$

C
$$1$$

D
$$4$$

Solution

Given:
$$\left( 3\cos ^{ 2 }{ 60^0} +2\cot ^{ 2 }{ 30^0 } -5\sin ^{ 2 }{ 45^0 } \right) $$

Substituting the value of sin $$45^ 0$$ , cos $$60^ 0$$, cot $$30^ 0$$ we get,

$$=3\times { \left( \dfrac { 1 }{ 2 } \right) }^{ 2 }+2\times { \left( \sqrt { 3 } \right) }^{ 2 }-5\times { \left( \dfrac { 1 }{ \sqrt { 2 } } \right) }^{ 2 }$$

$$=\left( 3\times \dfrac { 1 }{ 4 } \right) +\left( 2\times 3 \right) -\left( 5\times \dfrac { 1 }{ 2 } \right) $$

$$=\dfrac { 3 }{ 4 } +6-\dfrac { 5 }{ 2 } $$

$$=\dfrac { 3+24-10 }{ 4 } =\dfrac { 17 }{ 4 } $$
Question 3
1 / -0

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

A
$$0$$

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

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

D
$$1$$

Solution

$$\left( \cos { 60^0 } \cos { 30^0-\sin { 60^0 } \sin { 30^0 } } \right) $$

$$=\dfrac { 1 }{ 2 } \times \dfrac { \sqrt { 3 } }{ 2 } -\dfrac { \sqrt { 3 } }{ 2 } \times \dfrac { 1 }{ 2 } $$

$$=\dfrac { \sqrt { 3 } }{ 4 } -\dfrac { \sqrt { 3 } }{ 4 } $$

$$=0$$
Question 4
1 / -0

$$\dfrac {2}{3}(\cos^{4} 30^0 - \sin^{4} 45^0) - 3 (\sin^{2}60^0 - \sec^{2} 45^0) + \dfrac {1}{4} \cot^{2} 30^0 = $$

A
$$\dfrac {53}{12}$$

B
$$\dfrac {73}{24}$$

C
$$\dfrac {113}{24}$$

D
$$\dfrac {83}{12}$$

Solution

Given
$$\dfrac { 2 }{ 3 } \left( \cos ^{ 4 }{ { 30^o-\sin ^{ 4 }{ { 45^o } } } } \right) -3\left( \sin ^{ 2 }{ { 60^o } } -\sec ^{ 2 }{ { 45^o } } \right) +\dfrac { 1 }{ 4 } \cot ^{ 2 }{ 30^0 } $$
$$=\dfrac { 2 }{ 3 } \left[ { \left( \frac { \sqrt { 3 } }{ 2 } \right) }^{ 4 }-{ \left( \dfrac { 1 }{ \sqrt { 2 } } \right) }^{ 4 } \right] -3\left[ { \left( \dfrac { \sqrt { 3 } }{ 2 } \right) }^{ 2 }-{ \left( \sqrt { 2 } \right) }^{ 2 } \right] +\dfrac { 1 }{ 4 } \times { \left( \sqrt { 3 } \right) }^{ 2 }$$
$$=\dfrac { 2 }{ 3 } \left( \dfrac { 9 }{ 16 } -\dfrac { 1 }{ 4 } \right) -3\left( \dfrac { 3 }{ 4 } -2 \right) +\dfrac { 3 }{ 4 } $$
$$=\dfrac { 2 }{ 3 } \left( \dfrac { 9-4 }{ 16 } \right) -3\left( \dfrac { 3-8 }{ 4 } \right) +\dfrac { 3 }{ 4 } $$
$$=\dfrac { 2 }{ 3 } \times \dfrac { 5 }{ 16 } -3\left( -\dfrac { 5 }{ 4 } \right) +\dfrac { 3 }{ 4 } $$
$$=\dfrac { 5 }{ 24 } +\dfrac { 15 }{ 4 } +\dfrac { 3 }{ 4 } $$

$$=\dfrac { 5+90+18 }{ 24 } $$
$$=\dfrac { 113 }{ 24 } $$
Question 5
1 / -0

$$\left( \dfrac { 4 }{ 3 } \cot ^{ 2 }{ 30^0 } +3\sin ^{ 2 }{ 60^0 } -2\csc ^{ 2 }{ 60^0 } -\dfrac { 3 }{ 4 } \tan ^{ 2 }{ 30^0 } \right) =$$

A
$$3$$

B
$$\dfrac{8}{3}$$

C
$$\dfrac{10}{3}$$

D
$$\dfrac{9}{4}$$

Solution

$$\dfrac { 4 }{ 3 } \cot ^{ 2 }{ 30 } +3\sin ^{ 2 }{ 60 } -2\csc ^{ 2 }{ 60 } -\dfrac { 3 }{ 4 } \tan ^{ 2 }{ 30 } $$
$$=\dfrac { 4 }{ 3 } \times { \left( \sqrt { 3 } \right) }^{ 2 }+{ 3\left( \dfrac { \sqrt { 3 } }{ 2 } \right) }^{ 2 }-2{ \left( \dfrac { 2 }{ \sqrt { 3 } } \right) }^{ 2 }-\dfrac { 3 }{ 4 } { \left( \dfrac { 1 }{ \sqrt { 3 } } \right) }^{ 2 }$$, substitute the values from the table
$$=\dfrac { 4 }{ 3 } \times 3+3\times \dfrac { 3 }{ 4 } -2\times \dfrac { 4 }{ 3 } -\dfrac { 3 }{ 4 } \times \dfrac { 1 }{ 3 } $$
$$=4+\dfrac { 9 }{ 4 } -\dfrac { 8 }{ 3 } -\dfrac { 1 }{ 4 } $$
$$=\dfrac { 40 }{ 12 } = \dfrac { 10 }{ 3 } $$.
Question 6
1 / -0

Value of $$\left(\cos{0^0}+\sin{30^0}+\sin{45^0}\right)\left(\sin{90^0}+\cos{60^0}-\cos{45^0}\right)$$ is

A
$$\dfrac{5}{6}$$

B
$$\dfrac{5}{8}$$

C
$$\dfrac{3}{5}$$

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

Solution

Taking
$$\left( \cos { 0^0 } +\sin { 30^0 } +\sin { 45 ^0} \right) \left( \sin { 90^0+ } \cos { 60^0-\cos { 45^0 } } \right) $$
Substituting the values of $$cos 0^0,cos 45^0,cos 60^0, sin 30^0,sin 45^0,sin 90^0$$ we get
$$=\left( 1+\dfrac { 1 }{ 2 } +\dfrac { 1 }{ \sqrt { 2 } } \right) \left( 1+\dfrac { 1 }{ 2 } -\dfrac { 1 }{ \sqrt { 2 } } \right) $$ taking L.C.M
$$=\left( \dfrac { 3 }{ 2 } +\dfrac { 1 }{ \sqrt { 2 } } \right) \left( \dfrac { 3 }{ 2 } -\dfrac { 1 }{ \sqrt { 2 } } \right) $$ using $$(a^2-b^2)=(a-b)(a+b)$$
$$={ \left( \dfrac { 3 }{ 2 } \right) }^{ 2 }-{ \left( \dfrac { 1 }{ \sqrt { 2 } } \right) }^{ 2 }$$
$$=\dfrac { 9 }{ 4 } -\dfrac { 1 }{ 2 } =\dfrac { 7 }{ 4 } $$
Question 7
1 / -0

Value of $$\left({\sin}^{2}{30}+4{\cot}^{2}{45}-{\sec}^{2}{60}\right)$$ is

A
$$0$$

B
$$\dfrac{1}{4}$$

C
$$4$$

D
$$1$$

Solution

Taking
$$\left( \sin ^{ 2 }{ 30^0 } +4\cot ^{ 2 }{ 45^0-\sec ^{ 2 }{ 60^0 } } \right) $$
substituting the value of sin $$30^ 0$$, cot$$45^ 0$$, $$sec60^ 0$$
$$={ \left( \dfrac { 1 }{ 2 } \right) }^{ 2 }+4\times { \left( 1 \right) }^{ 2 }-{ 2 }^{ 2 }$$
$$=\dfrac { 1 }{ 4 } +4-4$$

$$=\dfrac { 1 }{ 4 } $$
Question 8
1 / -0

Find the value of:
$$\sin ^{ 2 }{ 30^0 } \cos ^{ 2 }{ 45^0 } +4\tan ^{ 2 }{ 30^0 } +\cfrac { 1 }{ 2 } \sin ^{ 2 }{ 90^0 } +\cfrac { 1 }{ 2 } \cot ^{ 2 }{ 60^0 } $$

A
$$2$$

B
$$3$$

C
$$4$$

D
$$\dfrac{17}{8}$$

Solution

$$\sin ^{ 2 }{ 30^0 } \cos ^{ 2 }{ 45^0 } +4\tan ^{ 2 }{ 30^0 } +\dfrac { 1 }{ 2 } \sin ^{ 2 }{ 90^0 } +\dfrac { 1 }{ 2 } \cot ^{ 2 }{ 60^0 } $$

$$={ \left( \dfrac { 1 }{ 2 } \right) }^{ 2 }\times { \left( \dfrac { 1 }{ \sqrt { 2 } } \right) }^{ 2 }+4\times { \left( \dfrac { 1 }{ \sqrt { 3 } } \right) }^{ 2 }+\dfrac { 1 }{ 2 } \times { \left( 1 \right) }^{ 2 }+\dfrac { 1 }{ 2 } \times { \left( \dfrac { 1 }{ \sqrt { 3 } } \right) }^{ 2 }$$,

$$=\dfrac { 1 }{ 4 } \times \dfrac { 1 }{ 2 } +4\times \dfrac { 1 }{ 3 } +\dfrac { 1 }{ 2 } +\dfrac { 1 }{ 2 } \times \dfrac { 1 }{ 3 } $$

$$=\dfrac { 1 }{ 8 } +\dfrac { 4 }{ 3 } +\dfrac { 1 }{ 2 } +\dfrac { 1 }{6 } $$

$$=\dfrac { 3+32+12+4 }{ 24 } $$

$$=\dfrac { 17 }{ 8 }$$
Question 9
1 / -0

$$(\cos 0^{\circ} + \sin 45^{\circ} + \sin 30^{\circ}) (\sin 90^{\circ} - \cos 45^{\circ} + \cos 60^{\circ}) = $$

A
$$\dfrac {5}{4}$$

B
$$\dfrac {9}{4}$$

C
$$\dfrac {7}{4}$$

D
None of these

Solution

$$\left( \cos { { 0^o } + } \sin { { 45^o } + } \sin { { 30^o } } \right) \left( \sin { { 90^o- } } \cos { { 45^o+ } } \cos {{ 60^o } } \right) $$

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

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

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

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

$$=\dfrac { 9 }{ 4 } -\dfrac { 1 }{ 2 } =\dfrac { 7 }{ 4 } $$
Question 10
1 / -0

In $$\triangle ABC$$, if $$AD\perp BC$$ and $$BD = 10\ cm; \angle B = 60^{\circ}$$ and $$\angle C = 30^{\circ}$$, then $$CD = .........$$

A
$$10\ cm$$

B
$$20\ cm$$

C
$$40\ cm$$

D
$$30\ cm$$

Solution

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 $$\triangle ACD$$,
$$\tan 30^{\circ} = \dfrac {AD}{DC} = \dfrac {1}{\sqrt {3}} \Rightarrow \dfrac {10\sqrt {3}}{DC} = \dfrac {1}{\sqrt {3}}$$

$$\Rightarrow DC = 30\ cm$$.

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Re-Attempt Weekly Quiz Competition

Introduction to Trigonometry Test - 53

Result

Introduction to Trigonometry Test - 53

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SHARING IS CARING

Question 1

$$\dfrac{73}{8}$$

$$\dfrac{75}{8}$$

$$\dfrac{81}{8}$$

$$\dfrac{83}{8}$$

Solution

Question 2

$$\dfrac{13}{6}$$

$$\dfrac{17}{4}$$

$$1$$

$$4$$

Solution

Question 3

$$0$$

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

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

$$1$$

Solution

Question 4

$$\dfrac {53}{12}$$

$$\dfrac {73}{24}$$

$$\dfrac {113}{24}$$

$$\dfrac {83}{12}$$

Solution

Question 5

$$3$$

$$\dfrac{8}{3}$$

$$\dfrac{10}{3}$$

$$\dfrac{9}{4}$$

Solution

Question 6

$$\dfrac{5}{6}$$

$$\dfrac{5}{8}$$

$$\dfrac{3}{5}$$

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

Solution

Question 7

$$0$$

$$\dfrac{1}{4}$$

$$4$$

$$1$$

Solution

Question 8

$$2$$

$$3$$

$$4$$

$$\dfrac{17}{8}$$

Solution

Question 9

$$\dfrac {5}{4}$$

$$\dfrac {9}{4}$$

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

None of these

Solution

Question 10

$$10\ cm$$

$$20\ cm$$

$$40\ cm$$

$$30\ cm$$

Solution

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