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.
In the above figure, the ratio of AB (perpendicular) to AC (base) with reference angle is called sine θ.
∴ sin θ = \(\frac{AB}{AC}\) = \(\frac{perpendicular}{hypotenuse}\) = \(\frac{p}{h}\)
In the above figure, the ratio of BC (base) to AC (base) with reference angle is called cosine θ.
∴ cos θ = \(\frac{BC}{AC}\) = \(\frac{base}{hypotenuse}\) = \(\frac{b}{h}\)
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^{2} = CA.CD |
A.A.A. axiom |
3. | ΔABC∼ΔABD | Same as above |
4. |
\(\frac{CA}{AB}\) = \(\frac{AB}{AD}\) or, AB^{2} = CA.AD |
Corresponding sides of similar triangles |
5. |
AB^{2} + BC^{2} = CA.AD + CA.CD or, AB^{2} + BC^{2} = CA (AD + CD) or, AB^{2} + BC^{2} = CA.CA or, AB^{2} + BC^{2} = CA^{2} |
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.
.
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
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 = \(\frac{p}{b}\) = \(\frac{1.2cm}{0.9 cm}\) = \(\frac{12}{9}\) = \(\frac{4}{3}\)
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 = \(\frac{p}{b}\) = \(\frac{9}{40}\) = \(\frac{9}{40}\)
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 = \(\frac{p}{b}\) = \(\frac{24}{7}\) = \(=frac{24}{7}\)
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 = \(\frac{p}{b}\) = \(\frac{16}{63}\) = \(\frac{16}{63}\)
= ( \(\frac{1}{2}\) )^{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 30^{o }
\(\therefore\) \(\beta\) = 30^{o} Ans.
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.
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:
tan^{2} 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° + cos^{2} 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:
tan^{2} 60° + 4 cos^{2 }45°
2
10
19
5
Find the value of:
sin^{2}30° - cos^{2} 60°
two
one
zero
three
Find the value of:
(tan^{2} 45° -cos^{2}60°)-(tan^{2}45°-siin^{2}30°)
two
five
nine
zero
Find the value of:
(cos^{2} 45° + tan^{2 }60°)-(sin^{2}45°-tan^{2} 0°)
3
10
6
2
If α = 60° and θ =90°, find the value of : 4cos^{2}α.sin θ.
2
4
3
1
If A = 30° and B =0°, find the value of : 4sin^{2} A. cos^{2}B.
1
10
16
5
If A =45° and B =30°, find the value of:
(sin A cos B + cosA sin B)^{2 }+ (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)^{2 }+ (sin A cos B - cos A sin B)^{2}
4
12
8
1
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