Subject: Compulsory Maths

The word trigonometry comes from the combination of the words "tri"and goes on having the meaning of the measurement of three angles of a triangle. With the help of right-angled triangle, trigonometric ratios of an angle are found. here we will calculate the values of trigonometric ratios as form special angles 0 degrees, 30 degrees, 45, 60 and 90 degrees etc.

The word "Trigonometry" is derived from the Greek word "Tri-Gonia-Metron" where 'Tri' means 'three', 'Gonia' means 'angles' and 'Metron' mean 'measure'. So, trigonometry is a branch of mathematics which concerned with the measurement of sides, angles and their relation to a triangle.

The word trigonometry comes from the combination of the word triangle goes on having the meaning of the measurement of three angles of triangles. Hence, this is also known as the measure of the triangle. Trigonometry has wide application in the field of mathematics and science. With the help of right angles triangle, trigonometric ratios of an angle are found. Without the help of trigonometric ratios, both the development and expansion of physics and also of engineering are impossible. Hence, trigonometry is the very important branch of mathematics.

Above figure shows the shadow of the poles formed at 3 pm that stand perpendicularly on the road. For each figure, the ratio of height of pole and length of shadow and height: length are tabulated below:

Pole | height of pole | length of shadow | height:length | angle made with the ground |

a | 3m | 2m | 3:2 | 56^{o} |

b | 6m | 4m | 3:2 | 56^{o} |

Hence, the height of every pole and the length of their shadows are in proportion. The angle made by the top of the pole and with the top of shadow on the ground is also equal.

The ratio of any two sides of a right-angled triangle taking one of the side as reference are the fundamentals of trigonometric ratios.

Let know on detail about ratio with a figure. Here, in the given figure, ΔABC is a right angled triangle. \(\angle\)B = 90^{o}and \(\angle\)C =θ.

Let's take\(\angle\)C as a reference angle

The opposite side of angle C (perpendicular) (P) = AB

The adjacent side of angle C (base) (B) = BC and (hypotenuse) (H) = AC

We can make three relation with the reference angle.

**relation between perpendicular and hypotenuse**

In the above figure, the ratio of AB (perpendicular) to AC (base) with reference angle is called sine θ.

∴ sin θ = \(\frac{AB}{AC}\) = \(\frac{perpendicular}{hypotenuse}\) = \(\frac{p}{h}\)

**relation between base and hypotenuse**

In the above figure, the ratio of BC (base) to AC (base) with reference angle is called cosine θ.

∴ cos θ = \(\frac{BC}{AC}\) = \(\frac{base}{hypotenuse}\) = \(\frac{b}{h}\)

**relation between perpendicular and base**

In the above figure, the ratio of AB (perpendicular) to BC (base) with reference angle is called tangent θ.

∴ tan θ = \(\frac{AB}{BC}\) = \(\frac{perpendicular}{base}\) = \(\frac{p}{b}\)

The relationship between the three sides of a triangle is simply known as Pythagoras Theorem. The relation was given by the popular Mathematician Pythagoras so it is called as Pythagorean theorem.

In mathematics, the Pythagorean theorem, also known as Pythagoras's theorem, is a relation in Euclidean geometry among the three sides of a right triangle. It states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

According to this theorem "In any right-angled triangle, the area of the square on the hypotenuse is equal to the sum of the areas of squares of perpendicular and base".

By Pythagoras Theorem

Hypotenuse (h^{2}) = Perpendicular (p^{2}) + Base (b^{2})

or, h^{2}= p^{2}+ b^{2}

From this theory we can derive,

h = \(\sqrt{p^{2} + b^{2}}\)

p = \(\sqrt{h^{2} - b^{2}}\)

b = \(\sqrt{h^{2} - p^{2}}\)

**Theoretical proof:**

**Given:** ΔABC is a right angled triangle in which \(\angle\)ABC = 90^{o}.

**To prove:** CA^{2} = AB^{2} + BC^{2}

**Construction:** BD ⊥AC is draawn

**Proof:**

S.N. | Statements | Reasons |

1. | In ΔABC and ΔBCD | |

i. | \(\angle\)ABC =\(\angle\)BDC (A) | Both of them are right angles |

ii. | \(\angle\)BCA =\(\angle\)BCD (A) | ommon angles |

iii. | \(\angle\)BAC = \(\angle\)DBC (A) | Remaining angles of the triangles |

2. |
\(\frac{CA}{BC}\) = \(\frac{BC}{CD}\) or, BC |
A.A.A. axiom |

3. | ΔABC∼ΔABD | Same as above |

4. |
\(\frac{CA}{AB}\) = \(\frac{AB}{AD}\) or, AB |
Corresponding sides of similar triangles |

5. |
AB or, AB or, AB or, AB |
Adding the statments (2) and (4) |

Angle | Degree | Radinas |

Right Angle | 90^{o} |
π/2 |

Straight Angle | 180^{o} |
π |

Full Rotation | 360^{o} |
2 π |

Trigonometry is all about finding triangles. The terms like Sin, Cos and Tan helps us in trigonometry.

- It includes every relationship which established among the people.
- There can be more than one community in a society. Community smaller than society.
- It is a network of social relationships which cannot see or touched.
- common interests and common objectives are not necessary for society.

Calculate the three fundamental trigonometric ratios for the acute angle x of each of the folllowing triangles expressing them in the lowest terms and correct to two decimal places.

Given , in BCA

Opposite side of x (AB) = p = 24

Adjacent side of x (AC) = b = 1 0

Opposite side of (<A = 90^{o}) h = 26

Here , sin x = \(\frac{p}{h}\) = \(\frac{24}{26}\) = \(\frac{12}{13}\) = 0.92

cos x = \(\frac{b}{h}\) = \(\frac{10}{26}\) = \(\frac{5}{13}\) = 0.38

and

tan x = \(\frac{p}{b}\) = \(\frac{24}{10}\) = \(\frac{12}{5}\) = 2.40

Find out the third side by using Pythagoras theroem and then calculate sin x , cos x and tna x.

In given ACB ,

Opposite side of x = p = 1.2 cm

Adjacent side of x = b = 0.9 cm

Oppsite side of B = h = ?

Here , by pythagoras theroem ,

h^{2} =p^{2} + b^{2}

h = \(\sqrt{p^{2} + b^{2}}\)

= \(\sqrt{1.2^{2} + 0.9 ^{2}}\)

= \(\sqrt{1.44 + 0.81}\)

= \(\sqrt{2.25}\)

= 1.5 cm.

Now ,

sinx = \(\frac{p}{h}\) = \(\frac{1.2 cm}{1.5 cm}\) = \(\frac{12}{15}\) = \(\frac{4}{5}\)

cos x = \(\frac{b}{h}\) = \(\frac{0.9 cm}{1.5 cm}\) = \(\frac{09}{15}\) = \(\frac{3}{5}\)

and

tanx = \(\frac{p}{b}\) = \(\frac{1.2cm}{0.9 cm}\) = \(\frac{12}{9}\) = \(\frac{4}{3}\)

Find out the third side by using Pythagoras theorem :

In given triangle EFG ,

p = EF = 9 cm , b = FG = 40cm , h = EG = ?

h^{2} = p^{2} + b^{2}

or , h = \(\sqrt{p^{2} + b^{2}}\)

or , \(\sqrt{9^{2} + 40^{2}}\) = \(\sqrt{81 + 1600}\) = \(\sqrt{1681}\) = 41 cm.

Now ,

sinx =\(\frac{p}{h}\) = \(\frac{9}{41}\) = \(\frac{9}{41}\)

cosx = \(\frac{b}{h}\) = \(\frac{40}{41}\) = \(\frac{40}{41}\)

and tanx = \(\frac{p}{b}\) = \(\frac{9}{40}\) = \(\frac{9}{40}\)

Find out the third side by using Pythagoras theorem.

In given PQR , h = QR = 25cm , b = QP = 7cm ,

p = PR = ?

By Pythagoras theorem ,

p^{2} + b^{2} = h^{2 }

or , p_{2} = h^{2} - h^{2}

\(\therefore\) p = \(\sqrt{h ^{2} - b^{2}}\) = \(\sqrt{25 ^{2} - 7^{2}}\)

= \(\sqrt{576}\) = 24 cm.

Here ,

sinx = \(\frac{p}{h}\) = \(\frac{24}{25}\) = \(frac{24}{25}\)

cosx = \(\frac{b}{h}\) = \(\frac{7}{25}\) = \(\frac{7}{25}\)

and

tanx = \(\frac{p}{b}\) = \(\frac{24}{7}\) = \(=frac{24}{7}\)

Find the third side by using Pythagoras theorem,

In given XYZ ,

p = XY = 16 , h = XZ = 65 , b = YZ = ?

p^{2} + b^{2} = h^{2 }

or , b^{2} = h^{2} - p^{2}

b = \(\sqrt{h^{2} - p^{2}}\) =\(\sqrt{65^{2} - 16^{2}}\)

= \(\sqrt{3969}\) = 63.

Here ,

sinx = \(\frac{p}{h}\) = \(\frac{16}{65}\) = \(\frac{16}{65}\)

cosx = \(\frac{b}{h}\) = \(\frac{63}{65}\) = \(\frac{63}{65}\)

and

tanx = \(\frac{p}{b}\) = \(\frac{16}{63}\) = \(\frac{16}{63}\)

Calculate the value of given equations :

1. cos60^{o} \(\times\) sin 60^{o }

= \(\frac{1}{2}\) \(\times\) \(\frac{\sqrt{3}}{2}\) = \(\frac{\sqrt{3}}{4}\).

2cos60^{o} \(\times\) sin30^{o }

= 2 \(\frac{1}{2}\) \(\times\) \(\frac{1}{2}\) = \(\frac{1}{2}\) Ans.

Calculate the value of given equation:

sin45^{o} \(\times\) 2cos45^{o }

= \(\frac{1}{\sqrt{2}}\) \(\times\) 2 \(\times\)\(\frac{1}{\sqrt{2}}\)

= \(\frac{2}{2}\)

= 1.

Calculate the value of given equations :

tan45^{o} + sin 30^{o }

= 1 +\(\frac{1}{2}\)

= \(\frac{3}{2}\)

= 1 \(\frac{1}{2}\).

Find the values of the given equation :

sin^{2}30^{o }+ sin^{2}45^{o} + sin^{2}60^{o }

= ( \(\frac{1}{2}\) )^{2} + ( \(\frac{1}{\sqrt {2}}\) )^{2} + ( \(\frac{\sqrt{3}}{2}\) )^{2}

= \(\frac{1}{4}\)+ \(\frac{1}{2}\) +\(\frac{3}{4}\)

= \(\frac{1+2+3}{4}\)

= \(\frac{3}{2}\)

= 1\(\frac{1}{2}\)

Solve:

cos\(\beta\) = \(\sqrt{3}\) cos60^{o }

cos\(\beta\) = \(\sqrt{3}\) \(\times\) \(\frac{1}{2}\)

or ,cos\(\beta\) = \(\frac{\sqrt{3}}{2}\)

or ,cos\(\beta\) = cos 30^{o }

\(\therefore\) \(\beta\) = 30^{o} Ans.

In a right angle ABC , AC = 30 , A = 30^{o } , B = 60^{o }. What is the length of BC ?

Here , A + B = 30^{o}+ 60^{o} = 90^{o}

Remaning angle C=90^{o}

In rt. angled triangle ABC ,

tan60^{o} = \(\frac{AC}{BC}\)

\(\sqrt{3}\) = \(\frac{30}{BC}\)

BC \(\times\) \(\sqrt{30}\) = 30

BC = \(\frac{30}{\sqrt{30}}\) = \(\frac{3 \times 10}{\sqrt{30}}\) = 1.732 \(\times\) = 17. 32 Ans.

In the adjoining figure , AB = 6 cm , BC = 8 cm , ABC = 90^{o } , BD⊥AC and ABD = \(\theta\). Calculate the value of sin\(\theta\).

Here , AC^{2} = AB^{2} + AB^{2}

AC = \(\sqrt{AB^{2} + BC^{2}}\)

= \(\sqrt{6^{2} + 8^{2}}\)

= 10.

In ABC and ABD ,

ABC = ADB [both are right angles]

BAC = BAD [common angles]

ACB = ABD = \(\theta\) [remaining angle]

InΔABC ,

sin ACB = \(\frac{AB}{AC}\)

\(\therefore\) sin \(\theta\) = \(\frac{6}{10}\) = \(\frac{3}{5}\)

In the adjoining DEF , DG⊥ EF. If DE = EF = FD = x,

prove that : cos\(\theta\) = \(\frac{\sqrt{3}}{2}\).

EG = GF [perpendicular drawn from any vertex to the opposite side of equilateral triangle bisects that side]

EG = \(\frac{EF}{2}\) = \(\frac{x}{2}\)

In rt. angled triangle DGE ,

or , DG^{2} + GE^{2} = DE^{2}

or , DG^{2} + \(\frac{x}{2}\)^{2} = x^{2}

or , DG = \(\frac{\sqrt{4x^{2} - x^{2}}}{4}\) = \(\frac{\sqrt{3x^{2}}}{4}\) = \(\frac{\sqrt{3x}}{2}\)

cos \(\theta\) = \(\frac{DG}{DE}\) = \(\frac{\sqrt{3}}{2}\) Proved.

Solve :

4 sin^{2}\(\beta\) = 1.

or , sin^{2}\(\beta\) = \(\frac{1}{4}\)

or , sin^{2} \(\beta\) =( \(\frac{1}{2}\) )^{2}

or , sin^{2} \(\beta\) = (sin30^{o})^{2}

or ,sin \(\beta\) = sin30^{o}

\(\therefore\) \(\beta\) = 30^{o} Ans.

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