Introduction to Trigonometry Test

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Introduction to Trigonometry Test - 24

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Weekly Quiz Competition

Question 1
1 / -0

Triangle measurement is called as _______.

A
pythagoras

B
trigonometry

C
calculus

D
area of square

Solution

Trigonometry is the study of triangle measurement.
Hence, the answer is trigonometry.
Question 2
1 / -0

Who is the founder of trigonometry?

A
Euclid

B
Issac Newton

C
William Rowan Hamilton

D
Hipparchus

Solution

Hipparchus is the founder of trigonometry.
So, option D is correct.
Question 3
1 / -0

______ mathematicians created the trigonometry system based on the sine function instead of the chords.

A
Greek

B
Indian

C
German

D
Egyptian

Solution

Indian mathematicians $$\text{Aryabhata}$$ created the trigonometry system based on the sine function instead of the chords.
.
Hence, the answer is Indian.
Question 4
1 / -0

In the early 9th century AD, _________ produced accurate sine and cosine tables, and the first table of tangents.

A
Habash al-Hasib al-Marwazi

B
Muhammad ibn Jabir al-Harrani al-Battani

C
Muhammad ibn Musa al-Khwarizmi

D
Abu al-Wafa al-Buzjani

Solution

In the early 9th century AD, Muhammad ibn Mūsā al-Khwārizmī produced accurate sine and cosine tables, and the first table of tangents. He was also a pioneer in spherical trigonometry.
Question 5
1 / -0

Which of the following is equal to $$\sin x \sec x$$?

A
$$\tan x$$

B
$$\cot x$$

C
$$\cos x \tan x$$

D
$$\cos x \text{cosec}\, x$$

E
$$\cot x \text{cosec}\, x$$

Solution

$$\sin { x }\cdot \sec { x } $$
$$=\sin { x } \times\dfrac { 1 }{ \cos x } $$
$$=\dfrac { \sin { x } }{ \cos x } $$
$$=\tan { x } $$
Question 6
1 / -0

The value of $$\tan\theta \times \cot\theta =$$ _____

A
$$2$$

B
$$0$$

C
$$1$$

D
$$-1$$

Solution

To find: Value of $$\tan\theta \times \cot\theta$$
We know that $$\cot\theta = \dfrac{1}{\tan\theta}$$
So, $$\tan\theta \times \dfrac{1}{\tan\theta} = 1$$
Question 7
1 / -0

The value of $$\sin\theta\times \text{cosec}\,\theta =$$ _____

A
$$2$$

B
$$1$$

C
$$-1$$

D
$$0$$

Solution

$$\sin\theta \times cosec\;\theta $$
$$=\sin \theta \times \dfrac{1}{\sin \theta}$$
$$=1$$
Hence, the answer is $$1.$$
Question 8
1 / -0

Who published the trigonometry in 1595?

A
William Rowan Hamilton

B
Hipparchus

C
Bartholomaeus Pitiscus

D
Newton

Solution

Trigonometry was first published by Bartholomaeus Pitiscus in 1595.
So, option C is correct.
Question 9
1 / -0

In the given figure, $$BC=15cm$$ and $$\sin{B}=\dfrac{4}{5}$$, What is the value of $$AB$$?

A
$$25cm$$

B
$$20cm$$

C
$$5cm$$

D
$$4cm$$

Solution

$$\sin B= \dfrac{AC}{AB} =\dfrac{\sqrt{AB^2 - BC^2}}{AB}$$
$$ \dfrac{4}{5} AB= \sqrt{AB^2 - 15^2}$$
$$\Rightarrow 16AB^2 = 25AB^2 - (25)(15)^2$$
$$AB^2 = \dfrac{25\times 15 \times 15}{9}$$
$$\Rightarrow \boxed {AB = 25}$$
Question 10
1 / -0

Choose the correct option. Justify your choice.
$$\displaystyle 9{ \sec }^{ 2 }A-9{ \tan }^{ 2 }A=$$

A
$$1$$

B
$$9$$

C
$$8$$

D
$$0$$

Solution

We know,
$$1= \sec ^2A-\tan^2A$$
$$\therefore 9(\sec^2A-\tan^2A)$$ $$=9\times 1=9$$

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Re-Attempt Weekly Quiz Competition

Introduction to Trigonometry Test - 24

Result

Introduction to Trigonometry Test - 24

Score

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SHARING IS CARING

Question 1

pythagoras

trigonometry

calculus

area of square

Solution

Question 2

Euclid

Issac Newton

William Rowan Hamilton

Hipparchus

Solution

Question 3

Greek

Indian

German

Egyptian

Solution

Question 4

Habash al-Hasib al-Marwazi

Muhammad ibn Jabir al-Harrani al-Battani

Muhammad ibn Musa al-Khwarizmi

Abu al-Wafa al-Buzjani

Solution

Question 5

$$\tan x$$

$$\cot x$$

$$\cos x \tan x$$

$$\cos x \text{cosec}\, x$$

$$\cot x \text{cosec}\, x$$

Solution

Question 6

$$2$$

$$0$$

$$1$$

$$-1$$

Solution

Question 7

$$2$$

$$1$$

$$-1$$

$$0$$

Solution

Question 8

William Rowan Hamilton

Hipparchus

Bartholomaeus Pitiscus

Newton

Solution

Question 9

$$25cm$$

$$20cm$$

$$5cm$$

$$4cm$$

Solution

Question 10

$$1$$

$$9$$

$$8$$

$$0$$

Solution

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