Pythagoras Theorem
Trigonometry
The word "Trigonometry" is derived from the Greek word "TriGoniaMetron" 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.
Pythagoras Theorem
The relationship between the three sides of a triangle is simply known as Pythagoras Theorem. The relation was given by the popular Mathematician Pythagoras which is called Pythagoras 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 rightangled 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 = \(\sqrt{h^{2}  p^{2}}\)
Pythagoras theorem 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.
Pythagoras Theorem = Hypotenuse (h^{2}) = Perpendicular (p^{2}) + Base (b^{2})
or, h^{2} = p^{2} + b^{2}
Where, h = \(\sqrt{p^{2}+ b^{2}}\)
p = \(\sqrt{h^{2} b^{2}}\)
b = \(\sqrt{h^{2} p^{2}}\)
Solution:
Here,
\(\angle\)ABC = \(\theta\), angle of reference.
\(\angle\)ABC = 90°
Hence,
AC = h, AB = p and BC = b
Solution:
Here,
\(\angle\)QPR = α, angle of reference.
\(\angle\)PQR = 90°
Hence,
RP = h, RQ = p and PQ = b
Solution:
Here,
\(\angle\)EFG = β, angle of reference.
\(\angle\)GEF 90°
Hence,
FG = h, EG = p and EF = b.
Solution:
Here,
\(\angle\)ABC = 90°
Hence,
or, \(\theta\) + 35° = 90°
or, \(\theta\) = 90°− 35°
∴ \(\theta\) = 55°
Solution:
\(\angle\)XYZ = 90°
Here,
or, β + 56° = 90°
or, β = 90°− 56°
∴ β = 34°
Solution:
\(\angle\)QPR = 90°
Here,
or, α + 28 = 90°
or, α = 90°− 28°
∴ α = 52°
Solution,
Here,
AB = P = 3cm
BC = b = 4cm
AC = h = ?
From the right angled ΔABC we have,
or, h = \(\sqrt{p^2+b^2}\)
or, h = \(\sqrt{3^2+4^2}\)
or, h = \(\sqrt{9+16}\)
or, h = \(\sqrt{25}\)
∴ h = 5cm
Solution:
Here,
PQ = h = 13cm
RQ = p = 12cm
PR = b = ?
From right angled ΔPQR we have,
or, b = \(\sqrt{h^2−p^2}\)
or, b = \(\sqrt{13^2−12^2}\)
or, b = \(\sqrt{169−12}\)
or, b = \(\sqrt{25}\)
∴ b = 5cm
Solution:
Here,
YZ = h = 10cm
XY = b = 6cm
XZ = p = ?
From right angled ΔXYZ we have,
or, p = \(\sqrt{h^2−b^2}\)
or, p = \(\sqrt{10^2−6^2}\)
or, p = \(\sqrt{100−36}\)
or, p = \(\sqrt{64}\)
∴ p = 8cm
Solution:
The three sides of a triangle are 5cm, 12cm and 13cm
Here,
or, 13^{2} = 5^{2} + 12^{2}
or, 169 = 25 + 144
∴ 169 = 169
i.e. h^{2} = p^{2}+ b^{2}
Hence, the triangle is right angle triangle.
Use the Pythagorean theorem, with a = 12 and b = 5.
or, c^{2}= a^{2}+ b^{2}[Pythagorean theorem]
or, c^{2}= 12^{2}+ 5^{2} [Putting value of a & b]
or, c^{2}= 144 + 25
or, c^{2}= 169
or, \(\sqrt{c^2}\) = \(\sqrt{169}\) [Squaring on both sides]
\(\therefore\) c =13
Hence, The length of the hypotenuse is 13 metres.
Solution:
Use the Pythagorean theorem, with a = 3 and b = 4.
or, c^{2}= a^{2}+ b^{2}[Pythagorean theorem]
or, c^{2}= 3^{2}+ 4^{2}[Putting value of a & b]
or, c^{2}= 9 + 16
or, c^{2}= 25
or, \(\sqrt{c^2}\) = \(\sqrt{25}\) [Squaring on both side]
\(\therefore\) c = 5
Hence, The length of the hypotenuse is 5 metres.
Solution:
L.H.S = \(\sqrt{\frac{secβ + 1}{secβ− 1}}\)
= \(\sqrt{\frac{secβ + 1}{secβ− 1}× \frac{secβ + 1}{secβ + 1}}\)
= \(\sqrt{\frac{(secβ + 1)^2}{secβ^2 − 1}}\)
= \(\sqrt{\frac{(secβ + 1)^2}{tan^2β}}\)
= \(\sqrt{(\frac{sectanβ + 1}{tanβ})^2}\)
= \(\frac{secβ + 1}{tanβ}\)
= \(\frac{\frac{1}{cosβ} + 1}{\frac{sinβ}{cosβ}}\)
= \(\frac{\frac{1 + cosβ}{cosβ}}{\frac{sinβ}{cosβ}}\)
= \(\frac{1 + cosβ}{cosβ}\) × \(\frac{cosβ}{sinβ}\)
= \(\frac{1 + cosβ}{sinβ}\)
= R.H.S proved
Solution:
L.H.S = \(\frac{tanθ + tanβ}{cotθ + cotβ}\)
= \(\frac{\frac{sinθ}{cosθ} + \frac{sinβ}{cosβ}}{\frac{cosθ}{sinθ} + \frac{cosβ}{sinβ}}\)
= \(\frac{\frac{sinθ . cosβ + sinβ . cosθ}{cosθ . cosβ}}{\frac{cosθ . sinβ + cosβ . sinθ}{sinθ . sinβ}}\)
=\(\frac{sinθ . cosβ + sinβ . cosθ}{cosθ . cosβ}\) × \(\frac{sinθ . sinβ }{cosθ . sinβ + cosβ . sinθ }\)
= \(\frac{sinθ . sinβ}{cosθ . cosβ}\)
= \(\frac{sinθ}{cosθ}\) .\(\frac{sinβ}{cosβ}\)
= tanθ . tanβ
= R.H.S proved
Solution:
l.H.S = \(\frac{cosα + sinα}{cosα− sinα}\) [Dividing numerator and denominator by cosα]
= \(\frac{\frac{cosα + sinα}{cosα}}{\frac{cosα− sinα}{cosα}}\)
= \(\frac{\frac{cosα}{cosα} + \frac{sinα}{cosα}}{\frac{cosα}{cosα} − \frac{sinα}{cosα}}\)
= \(\frac{1 + tanα}{1− tanα}\) (Middle Term)
Middle Term = \(\frac{1 + tanα}{1− tanα}\)
= \(\frac{1 + \frac{1}{cotα}}{1 −\frac{1}{cotα}}\)
= \(\frac{cotα + 1}{cotα}\)× \(\frac{cotα }{cotα− 1}\)
= \(\frac{cotα + 1}{cotα− 1}\) (R.H.S)
\(\therefore\) L.H.S = M.T = R.H.S proved
Solution:
L.H.S = \(\frac{1}{1 + cosθ}\) + \(\frac{1}{1− cosθ}\)
= \(\frac{1(1− cosθ) + 1(1 + cosθ)}{(1 + cosθ)(1− cosθ)}\)
= \(\frac{2}{1− cos^2θ}\)
= \(\frac{2}{sin^2θ}\)
= \(\frac{2}{\frac{1}{cosec^2θ}}\)
= 2cosec^{2}θ
= R.H.S proved
Solution:
L.H.S = \(\frac{tanθ}{1− cotθ}\) + \(\frac{cotθ}{1− tanθ}\)
= \(\frac{\frac{sinθ}{cosθ}}{1− \frac{cosθ}{sinθ}}\) +\(\frac{\frac{cosθ}{sinθ}}{1− \frac{sinθ}{cosθ}}\)
= \(\frac{\frac{sinθ}{cosθ}}{\frac{sinθ−cosθ}{sinθ}}\)+\(\frac{\frac{cosθ}{sinθ}}{\frac{cosθ−sinθ}{cosθ}}\)
= \(\frac{sinθ}{cosθ}\)× \(\frac{sinθ}{(sinθ−cosθ)}\) + \(\frac{cosθ}{sinθ}\)×\(\frac{cosθ}{(cosθ−sinθ)}\)
= \(\frac{sin^2θ}{cosθ(sinθ−cosθ)}\) + \(\frac{cos^2θ}{sinθ(cosθ−sinθ)}\)
= \(\frac{sin^2θ}{cosθ(sinθ−cosθ)}\)−\(\frac{cos^2θ}{sinθ(sinθ−cosθ)}\)
= \(\frac{sin^2θ× sinθ−cos^2θ×cosθ}{sinθ . cosθ(sinθ−cosθ)}\)
= \(\frac{sin^3θ−cos^3θ}{sinθ . cosθ(sinθ−cosθ)}\)
= \(\frac{(sinθ−cosθ)(sin^2θ +sinθ . cosθ +cos^2θ)}{sinθ . cosθ(sinθ−cosθ)}\)
= \(\frac{(sin^2θ +sinθ . cosθ +cos^2θ)}{sinθ . cosθ}\)
= \(\frac{sin^2θ +cos^2θ +sinθ . cosθ}{sinθ . cosθ}\)
= \(\frac{1 +sinθ . cosθ}{sinθ . cosθ}\)
= \(\frac{1}{sinθ . cosθ}\) +\(\frac{sinθ . cosθ}{sinθ . cosθ}\)
= \(\frac{1}{sinθ . cosθ}\) + 1
= \(\frac{1}{sinθ}\)× \(\frac{1}{cosθ}\) + 1
= secθ . cosecθ + 1
= R.H.S proved

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