Note on Trigonometry

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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.

Trigonometric Ratios

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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 56o
b 6m 4m 3:2 56o

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.

 Trigonometric Ratios of some special angles
Trigonometric Ratios of some special angles

 

 

 

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Fundamentals of Trigonometric Ratios

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 = 90oand \(\angle\)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 = \(\frac{AB}{BC}\) = \(\frac{perpendicular}{base}\) = \(\frac{p}{b}\)

 

 

Pythagoras Theorem

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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 (h2) = Perpendicular (p2) + Base (b2)
or, h2= p2+ b2
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 = 90o.

To prove: CA2 = AB2 + BC2

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, BC2 = CA.CD

A.A.A. axiom
3. ΔABC∼ΔABD Same as above
4.

\(\frac{CA}{AB}\) = \(\frac{AB}{AD}\)

or, AB2 = CA.AD

Corresponding sides of similar triangles
5.

AB2 + BC2 = CA.AD + CA.CD

or, AB2 + BC2 = CA (AD + CD)

or, AB2 + BC2 = CA.CA

or, AB2 + BC2 = CA2

Adding the statments (2) and (4)

 

 

 

Angle Degree Radinas
Right Angle 90o  π/2
Straight Angle 180o  π
Full Rotation 360o 2 π

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

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Very Short Questions

Given , in BCA
Opposite side of x (AB) = p = 24
Adjacent side of x (AC) = b = 1 0
Opposite side of (<A = 90o) 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

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 ,
h2 =p2 + b2
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 = \(\frac{p}{b}\) = \(\frac{1.2cm}{0.9 cm}\) = \(\frac{12}{9}\) = \(\frac{4}{3}\)

In given triangle EFG ,
p = EF = 9 cm , b = FG = 40cm , h = EG = ?
h2 = p2 + b2
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 = \(\frac{p}{b}\) = \(\frac{9}{40}\) = \(\frac{9}{40}\)

In given PQR , h = QR = 25cm , b = QP = 7cm ,
p = PR = ?
By Pythagoras theorem ,
p2 + b2 = h2
or , p2 = h2 - h2
\(\therefore\) 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 = \(\frac{p}{b}\) = \(\frac{24}{7}\) = \(=frac{24}{7}\)

In given XYZ ,
p = XY = 16 , h = XZ = 65 , b = YZ = ?
p2 + b2 = h2
or , b2 = h2 - p2
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 = \(\frac{p}{b}\) = \(\frac{16}{63}\) = \(\frac{16}{63}\)

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

cos\(\beta\) = \(\sqrt{3}\) \(\times\) \(\frac{1}{2}\)
or ,cos\(\beta\) = \(\frac{\sqrt{3}}{2}\)
or ,cos\(\beta\) = cos 30o
\(\therefore\) \(\beta\) = 30o Ans.

or , sin2\(\beta\) = \(\frac{1}{4}\)
or , sin2 \(\beta\) =( \(\frac{1}{2}\) )2
or , sin2 \(\beta\) = (sin30o)2
or ,sin \(\beta\) = sin30o
\(\therefore\) \(\beta\) = 30o Ans.

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  • In a right-angled triangle, sin θ=(frac{1}{2}),find the numerical value of cos θ.

    (frac{sqrt 4}{2})


    (frac{sqrt 3}{2})


    (frac{sqrt 5}{4})


    (frac{sqrt 6}{3})


  • Express sec A in terms of tan A.

    (sqrt{4+sin^2 A})


    (sqrt{1+tan^2 A})


    `(sqrt{1+sin^2 A})


    (sqrt{3-sin^2 A})


  • Express cosA in terms of sin A.

    (sqrt{1-sin^2 A})


    (sqrt{1-cos^2 A})


    (sqrt{1-cos A})


    (sqrt{1-sin A})


  • Find the value of:

    tan2 45° + cos 2 45°

    (frac{1}{1})


    (frac{3}{2})


    (frac{4}{2})


    (frac{1}{3})


  • Find the value of:

    sin 45° cos 45°  tan 45°

    (frac{4}{2})


    (frac{2}{1})


    (frac{3}{2})


    (frac{1}{2})


  • Find the value of:

    Sin 30° + cos2 60°

    (frac{1}{2})


    (frac{3}{2})


    (frac{2}{2})


    (frac{2}{1})


  • Find the value of:

    Sin 30°  cos 60°  tan 45°

    (frac{2}{1})


    (frac{1}{3})


    (frac{1}{4})


    (frac{1}{2})


  • sin 90° cos 45 tan 0°

    zero


    three


    two


    one


  • Find the value of:

    tan2 60° + 4 cos45°

    2


    10


    19


    5


  • Find the value of:

    sin230° - cos2 60°

    two


    one


    zero


    three


  • Find the value of:

     (tan2 45° -cos260°)-(tan245°-siin230°)

    two


    five


    nine


    zero


  • Find the value of:

    (cos2 45° + tan60°)-(sin245°-tan2 0°)

    3


    10


    6


    2


  • If α = 60° and θ =90°, find the value of : 4cos2α.sin θ.

    2


    4


    3


    1


  • If A = 30° and B =0°, find the value of : 4sin2 A. cos2B.

    1


    10


    16


    5


  • If A =45° and B =30°, find the value of:

    (sin A cos B + cosA sin B)+ (cos A cos B - sin a sin B)2

    3


    2


    2.2


    1


  • If A =45° and B =30°, find the value of:

    (cos A cos B + sin A sin B)+ (sin A cos B - cos A sin B)2

    4


    12


    8


    1


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Pratham Koirala

SinA -3/4 find the value of cosA


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how vgj


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in triancle abc, m and n are the mid points of bc and ac respectively .if mn eual to 3 cm and angle cnm=40 degree,find the length of ab and size of angle bac

in triangle abc, m and n are the mid points of bc and ac respectively .if mn equal to 3 cm and angle cnm=40 degree,find the length of ab and size of angle ba


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sudha ghimire

If xsin^3a ycos^3a = sinacosa and xsina _ ycosa = 0, prove that: x^2 y^2 =1.


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