Introduction to Trigonometry Test

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Introduction to Trigonometry Test - 59

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Weekly Quiz Competition

Question 1
1 / -0

If $$\sin\theta = \dfrac{6}{10}$$ and $$\theta$$ is acute, what is the value of $$\sec\theta$$?

A
$$\dfrac{10}{8}$$

B
$$\dfrac{8}{10}$$

C
$$\dfrac{1}{8}$$

D
$$\dfrac{10}{7}$$

Solution

First use the identity: $$\sin^{2}\theta + \cos^{2}\theta = 1$$ and substitute the value of $$\sin\theta = \dfrac{6}{10}$$

Therefore, $$\left(\dfrac{6}{10}\right)^{2}+\cos^{2}\theta = 1$$

$$\cos^{2}\theta = 1 - \dfrac{36}{100} = \dfrac{64}{100}$$

Square root on both the sides, we get
$$\cos^{}\theta = \sqrt{\dfrac{64}{100}} = \dfrac{8}{10}$$

Now we know the value of $$\cos^{2}\theta$$, to find $$\sec^{2}\theta$$ we just take the reciprocal.

Therefore, $$\sec\theta = \dfrac{10}{8}$$

So, option $$A$$ is correct.
Question 2
1 / -0

What is the meaning of trigonometry in Greek language?

A
Measurement

B
Triangle Measure

C
Angle Measure

D
Degree Measure

Solution

Trigonometry in the Greek language means triangle measure.
Trigono $$\Rightarrow $$ triangle $$+$$ metron $$\Rightarrow $$ measure

So, option $$B$$ is correct.
Question 3
1 / -0

If $$\cos\theta = \dfrac{12}{13}$$ and $$\theta$$ is acute, what is the value of $$\text{cosec}\,\theta$$?

A
$$\dfrac{12}{5}$$

B
$$\dfrac{5}{13}$$

C
$$\dfrac{13}{5}$$

D
$$\dfrac{11}{5}$$

Solution

First use the identity: $$\sin ^{2}\theta + \cos ^{2}\theta = 1$$ and substitute the value of $$\cos \theta = \dfrac{12}{13}$$

Therefore $$\left(\dfrac{12}{13}\right)^{2}+\sin ^{2}\theta = 1$$

$$\sin ^{2}\theta = 1 - \dfrac{144}{169} = \dfrac{25}{169}$$

Squaring on both the sides, we get
$$\sin ^{2}\theta = \sqrt{\dfrac{25}{169}} = \dfrac{5}{13}$$

Now we know the value of $$\sin ^{2}\theta$$, to find $$\text{cosec}\,^{2}\theta$$ we just take the reciprocal.

Therefore, $$\text{cosec}\,^{2}\theta = \dfrac{13}{5}$$

So, option $$C$$ is correct.
Question 4
1 / -0

If $$\sin \theta = \dfrac{5}{13}$$ and $$\theta$$ is acute, what is the value of $$\sec \theta$$?

A
$$\dfrac{2}{13}$$

B
$$\dfrac{13}{12}$$

C
$$\dfrac{12}{13}$$

D
$$\dfrac{1}{13}$$

Solution

First use the identity: $$\sin ^{2}\theta + \cos ^{2}\theta = 1$$ and substitute the value of $$\sin \theta = \dfrac{5}{13}$$
Therefore, $$(\dfrac{5}{13})^{2}+\cos ^{2}\theta = 1$$

$$\cos ^{2}\theta = 1 - \dfrac{25}{169} = \dfrac{144}{169}$$

Squaring on both the sides, we get
$$\cos ^{2}\theta = \sqrt{\dfrac{144}{169}} = \dfrac{12}{13}$$

Now we know the value of $$\cos ^{2}\theta$$, to find $$\sec^{2}\theta$$

Therefore, $$\sec^{2}\theta= \dfrac{13}{12}$$

So, option $$B$$ is correct.
Question 5
1 / -0

In the 5th century who created the table of chords with increasing 1 degree?

A
Hipparchus

B
William Rowan Hamilton

C
Euclid

D
Ptolemy

Solution

Ptolemy used length of chords to define his trigonometric functions
Question 6
1 / -0

If $$\cos\theta = \dfrac{8}{10}$$ and $$\theta$$ is acute, what is the value of $$\text{cosec}\,\theta$$?

A
$$\dfrac{10}{6}$$

B
$$\dfrac{8}{10}$$

C
$$\dfrac{1}{8}$$

D
$$\dfrac{10}{7}$$

Solution

First use the identity: $$\sin^{2}\theta + \cos^{2}\theta = 1$$ and substitute the value of $$\cos\theta = \dfrac{8}{10}$$

Therefore,

$$\left(\dfrac{8}{10}\right)^{2}+\sin^{2}\theta = 1$$

$$\sin^{2}\theta = 1 - \dfrac{64}{100} = \dfrac{36}{100}$$

$$\sin\theta = \sqrt{\dfrac{36}{100}} = \dfrac{6}{10}$$

Now that we know the value of $$\sin\theta$$, therefore to find $$\text{cosec}\,\theta$$ we just have to take the reciprocal.

Therefore, $$\text{cosec}\,\theta =\dfrac{1}{\sin\theta}= \dfrac{10}{6}$$

So, option $$A$$ is correct.
Question 7
1 / -0

If $$\cos\theta = \dfrac{14}{4}$$ and $$\sin\theta$$ $$=$$ $$\dfrac{8}{3}$$, what is the value of $$\cot\theta$$?

A
$$\dfrac{11}{16}$$

B
$$\dfrac{13}{16}$$

C
$$\dfrac{16}{21}$$

D
$$\dfrac{21}{16}$$

Solution

$$\cot \theta$$ $$=$$ $$\dfrac{\cos\theta}{\sin\theta}$$ $$=$$ $$\dfrac{\dfrac{14}{4}}{\dfrac{8}{3}}$$

$$=\dfrac{14}{4}\times\dfrac{3}{4}$$

$$\cot\theta = \dfrac{21}{16}$$

So, option $$D$$ is correct.
Question 8
1 / -0

If $$\cos\theta = \dfrac{2}{21}$$ and $$\sin\theta = \dfrac{6}{7}$$, what is the value of $$\tan\theta$$?

A
$$4$$

B
$$9$$

C
$$6$$

D
$$7$$

Solution

$$\tan\theta = \dfrac{\sin\theta}{\cos\theta}$$ $$=$$ $$\dfrac{\dfrac{6}{7}}{\dfrac{2}{21}}$$

$$=\dfrac{6}{7}\times\dfrac{21}{2}$$

$$\tan\theta = 9$$

So, option $$B$$ is correct.
Question 9
1 / -0

Choose the correct option. Justify your choice
$$\displaystyle \left( 1+\tan { \theta +\sec { \theta } } \right) \left( 1+\cot { \theta } -\text{cosec }\theta \right) =$$

A
$$0$$

B
$$1$$

C
$$2$$

D
$$-1$$

Solution

The value of $$(1+\tan \theta+ \sec \theta)(1
+\cot \theta - \text{cosec }\theta)$$
$$=\displaystyle \left(1+\frac{\sin \theta}{\cos \theta}+\frac{1}{\cos \theta}\right)\left(1+\frac{\cos\theta}{\sin\theta}-\frac{1}{\sin\theta}\right)$$

$$\displaystyle =\frac{(\sin\theta+\cos\theta)^2-(1)^2}{\sin\theta \,\cos\theta}$$
$$\displaystyle =\frac{(\sin\theta)^2+(\cos\theta)^2+2\sin\theta \cos\theta-(1)^2}{\sin\theta \,\cos\theta}$$

$$\displaystyle =\frac{2\sin\theta\,\cos\theta}{\sin\theta\,\cos\theta}=2$$
Question 10
1 / -0

Choose the correct option and justify your choice :
$$\displaystyle \frac { 2\tan { { 30 }^{ \circ } } }{ 1-{ \tan }^{ 2 }{ 30 }^{ \circ } } =$$

A
$$\sqrt 3 $$

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

C
$$2 $$

D
$$\sqrt 2 $$

Solution

We know that $$\tan30^0 = \dfrac1{\sqrt3}$$

The value of $$\dfrac {2\tan 30^o}{1-\tan^2 30^o}$$

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

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Re-Attempt Weekly Quiz Competition

Introduction to Trigonometry Test - 59

Result

Introduction to Trigonometry Test - 59

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SHARING IS CARING

Question 1

$$\dfrac{10}{8}$$

$$\dfrac{8}{10}$$

$$\dfrac{1}{8}$$

$$\dfrac{10}{7}$$

Solution

Question 2

Measurement

Triangle Measure

Angle Measure

Degree Measure

Solution

Question 3

$$\dfrac{12}{5}$$

$$\dfrac{5}{13}$$

$$\dfrac{13}{5}$$

$$\dfrac{11}{5}$$

Solution

Question 4

$$\dfrac{2}{13}$$

$$\dfrac{13}{12}$$

$$\dfrac{12}{13}$$

$$\dfrac{1}{13}$$

Solution

Question 5

Hipparchus

William Rowan Hamilton

Euclid

Ptolemy

Solution

Question 6

$$\dfrac{10}{6}$$

$$\dfrac{8}{10}$$

$$\dfrac{1}{8}$$

$$\dfrac{10}{7}$$

Solution

Question 7

$$\dfrac{11}{16}$$

$$\dfrac{13}{16}$$

$$\dfrac{16}{21}$$

$$\dfrac{21}{16}$$

Solution

Question 8

$$4$$

$$9$$

$$6$$

$$7$$

Solution

Question 9

$$0$$

$$1$$

$$2$$

$$-1$$

Solution

Question 10

$$\sqrt 3 $$

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

$$2 $$

$$\sqrt 2 $$

Solution

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